EditText 中带有 numberDecimal inputType 的小数分隔符逗号 (',')

Question

EditText 中的 inputType numberDecimal 使用点 . 作为小数点分隔符。在欧洲，通常使用逗号 , 代替。即使我的语言环境设置为德语，小数点分隔符仍然是 .

有没有办法将逗号作为小数点分隔符？

Answer 1

此处提供的“数字”解决方案的变体：

char separator = DecimalFormatSymbols.getInstance().getDecimalSeparator();
input.setKeyListener(DigitsKeyListener.getInstance("0123456789" + separator));

考虑区域设置分隔符。

Answer 2

以下代码用于 EditText ($ 123,125.155)

Xml 布局

  <EditText
    android:inputType="numberDecimal"
    android:layout_height="wrap_content"
    android:layout_width="200dp"
    android:digits="0123456789.,$" />

代码的货币掩码

EditText testFilter=...
testFilter.addTextChangedListener( new TextWatcher() {
        boolean isEdiging;
        @Override public void onTextChanged(CharSequence s, int start, int before, int count) { }
        @Override public void beforeTextChanged(CharSequence s, int start, int count, int after) { }

        @Override public void afterTextChanged(Editable s) {
            if(isEdiging) return;
            isEdiging = true;

            String str = s.toString().replaceAll( "[^\\d]", "" );
            double s1 = Double.parseDouble(str);

            NumberFormat nf2 = NumberFormat.getInstance(Locale.ENGLISH);
            ((DecimalFormat)nf2).applyPattern("$ ###,###.###");
            s.replace(0, s.length(), nf2.format(s1));

            isEdiging = false;
        }
    });

Answer 3

这是 Android SDK 中的已知错误。
唯一的解决方法是创建您自己的软键盘。您可以找到实现示例

Answer 4

您可以将以下内容用于不同的区域设置

private void localeDecimalInput(final EditText editText){

    DecimalFormat decFormat = (DecimalFormat) DecimalFormat.getInstance(Locale.getDefault());
    DecimalFormatSymbols symbols=decFormat.getDecimalFormatSymbols();
    final String defaultSeperator=Character.toString(symbols.getDecimalSeparator());

    editText.addTextChangedListener(new TextWatcher() {

        @Override
        public void beforeTextChanged(CharSequence s, int start, int count, int after) {

        }

        @Override
        public void onTextChanged(CharSequence s, int start, int before, int count) {

        }

        @Override
        public void afterTextChanged(Editable editable) {
            if(editable.toString().contains(defaultSeperator))
                editText.setKeyListener(DigitsKeyListener.getInstance("0123456789"));
            else
                editText.setKeyListener(DigitsKeyListener.getInstance("0123456789" + defaultSeperator));
        }
    });
}

Answer 5

您可以使用以下解决方法将逗号作为有效输入：-

通过 XML：

<EditText
    android:inputType="number"
    android:digits="0123456789.," />

以编程方式：

EditText input = new EditText(THE_CONTEXT);
input.setKeyListener(DigitsKeyListener.getInstance("0123456789.,"));

这样 Android 系统将显示数字键盘并允许输入的逗号。希望这能回答这个问题:)

Answer 6

如果您以编程方式实例化 EditText，马丁的答案将不起作用。我继续修改了 API 14 中包含的 DigitsKeyListener 类，以允许逗号和句点作为小数分隔符。

要使用此功能，请在 EditText 上调用 setKeyListener()，例如

// Don't allow for signed input (minus), but allow for decimal points
editText.setKeyListener( new MyDigitsKeyListener( false, true ) );

，但是，您仍然需要在 TextChangedListener 中使用 Martin 的技巧，其中替换逗号有句点

import android.text.InputType;
import android.text.SpannableStringBuilder;
import android.text.Spanned;
import android.text.method.NumberKeyListener;
import android.view.KeyEvent;

class MyDigitsKeyListener extends NumberKeyListener {

    /**
     * The characters that are used.
     *
     * @see KeyEvent#getMatch
     * @see #getAcceptedChars
     */
    private static final char[][] CHARACTERS = new char[][] {
        new char[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' },
        new char[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '-' },
        new char[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '.', ',' },
        new char[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '-', '.', ',' },
    };

    private char[] mAccepted;
    private boolean mSign;
    private boolean mDecimal;

    private static final int SIGN = 1;
    private static final int DECIMAL = 2;

    private static MyDigitsKeyListener[] sInstance = new MyDigitsKeyListener[4];

    @Override
    protected char[] getAcceptedChars() {
        return mAccepted;
    }

    /**
     * Allocates a DigitsKeyListener that accepts the digits 0 through 9.
     */
    public MyDigitsKeyListener() {
        this(false, false);
    }

    /**
     * Allocates a DigitsKeyListener that accepts the digits 0 through 9,
     * plus the minus sign (only at the beginning) and/or decimal point
     * (only one per field) if specified.
     */
    public MyDigitsKeyListener(boolean sign, boolean decimal) {
        mSign = sign;
        mDecimal = decimal;

        int kind = (sign ? SIGN : 0) | (decimal ? DECIMAL : 0);
        mAccepted = CHARACTERS[kind];
    }

    /**
     * Returns a DigitsKeyListener that accepts the digits 0 through 9.
     */
    public static MyDigitsKeyListener getInstance() {
        return getInstance(false, false);
    }

    /**
     * Returns a DigitsKeyListener that accepts the digits 0 through 9,
     * plus the minus sign (only at the beginning) and/or decimal point
     * (only one per field) if specified.
     */
    public static MyDigitsKeyListener getInstance(boolean sign, boolean decimal) {
        int kind = (sign ? SIGN : 0) | (decimal ? DECIMAL : 0);

        if (sInstance[kind] != null)
            return sInstance[kind];

        sInstance[kind] = new MyDigitsKeyListener(sign, decimal);
        return sInstance[kind];
    }

    /**
     * Returns a DigitsKeyListener that accepts only the characters
     * that appear in the specified String.  Note that not all characters
     * may be available on every keyboard.
     */
    public static MyDigitsKeyListener getInstance(String accepted) {
        // TODO: do we need a cache of these to avoid allocating?

        MyDigitsKeyListener dim = new MyDigitsKeyListener();

        dim.mAccepted = new char[accepted.length()];
        accepted.getChars(0, accepted.length(), dim.mAccepted, 0);

        return dim;
    }

    public int getInputType() {
        int contentType = InputType.TYPE_CLASS_NUMBER;
        if (mSign) {
            contentType |= InputType.TYPE_NUMBER_FLAG_SIGNED;
        }
        if (mDecimal) {
            contentType |= InputType.TYPE_NUMBER_FLAG_DECIMAL;
        }
        return contentType;
    }

    @Override
    public CharSequence filter(CharSequence source, int start, int end,
                               Spanned dest, int dstart, int dend) {
        CharSequence out = super.filter(source, start, end, dest, dstart, dend);

        if (mSign == false && mDecimal == false) {
            return out;
        }

        if (out != null) {
            source = out;
            start = 0;
            end = out.length();
        }

        int sign = -1;
        int decimal = -1;
        int dlen = dest.length();

        /*
         * Find out if the existing text has '-' or '.' characters.
         */

        for (int i = 0; i < dstart; i++) {
            char c = dest.charAt(i);

            if (c == '-') {
                sign = i;
            } else if (c == '.' || c == ',') {
                decimal = i;
            }
        }
        for (int i = dend; i < dlen; i++) {
            char c = dest.charAt(i);

            if (c == '-') {
                return "";    // Nothing can be inserted in front of a '-'.
            } else if (c == '.' ||  c == ',') {
                decimal = i;
            }
        }

        /*
         * If it does, we must strip them out from the source.
         * In addition, '-' must be the very first character,
         * and nothing can be inserted before an existing '-'.
         * Go in reverse order so the offsets are stable.
         */

        SpannableStringBuilder stripped = null;

        for (int i = end - 1; i >= start; i--) {
            char c = source.charAt(i);
            boolean strip = false;

            if (c == '-') {
                if (i != start || dstart != 0) {
                    strip = true;
                } else if (sign >= 0) {
                    strip = true;
                } else {
                    sign = i;
                }
            } else if (c == '.' || c == ',') {
                if (decimal >= 0) {
                    strip = true;
                } else {
                    decimal = i;
                }
            }

            if (strip) {
                if (end == start + 1) {
                    return "";  // Only one character, and it was stripped.
                }

                if (stripped == null) {
                    stripped = new SpannableStringBuilder(source, start, end);
                }

                stripped.delete(i - start, i + 1 - start);
            }
        }

        if (stripped != null) {
            return stripped;
        } else if (out != null) {
            return out;
        } else {
            return null;
        }
    }
}

Answer 7

恕我直言，解决这个问题的最佳方法是仅使用输入过滤器。这里有一个很好的要点 DecimalDigitsInputFilter。然后你就可以：

editText.setInputType(TYPE_NUMBER_FLAG_DECIMAL | TYPE_NUMBER_FLAG_SIGNED | TYPE_CLASS_NUMBER)
editText.setKeyListener(DigitsKeyListener.getInstance("0123456789,.-"))
editText.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(5,2)});

Answer 8

我有一个解决方案，允许用户输入点和逗号（如果键盘上可用），但只显示区域设置默认分隔符。此外，它不允许用户输入超过 1 个分隔符。对 EditText 的引用或无限循环没有问题。它是该线程中几个答案的组合，适合我的需求。

与接受的答案一样，相应地配置 EditText：

android:inputType="numberDecimal"
android:digits="0123456789.,"

然后在 EditText 上设置自定义 TextWatcher：

myEditText.addTextChangedListener(FlexibleDecimalSeparatorTextWatcher())

并包含自定义 TextWatcher：

import android.text.Editable
import android.text.SpannableStringBuilder
import android.text.TextWatcher
import android.widget.EditText
import java.text.DecimalFormatSymbols

/**
 * The [FlexibleDecimalSeparatorTextWatcher] allows the user to input both the comma (,) and dot (.) as a decimal separator,
 * and will then automatically convert each entered separator into the locale default separator.
 * If the user were to enter multiple separators - every separator but the first will be removed.
 *
 * To provide comma and dot support, set the [EditText] inputType to 'numberDecimal' and its digits to '0123456789.,'.
 */
class FlexibleDecimalSeparatorTextWatcher : TextWatcher {
companion object {
    private val DECIMAL_SEPARATORS = listOf('.', ',')
    private val LOCALE_DEFAULT_DECIMAL_SEPARATOR = DecimalFormatSymbols.getInstance().decimalSeparator
}

override fun beforeTextChanged(s: CharSequence?, start: Int, count: Int, after: Int) {}
override fun onTextChanged(s: CharSequence?, start: Int, before: Int, count: Int) {}

override fun afterTextChanged(s: Editable?) {
    if (s != null) {
        val textWithConvertedSeparators = convertSeparatorsToLocaleDefault(s.toString())
        val textWithoutMultipleSeparators = removeAdditionalSeparators(textWithConvertedSeparators)

        // Make the change if required. This only triggers one additional afterTextChanged call if there were changes.
        if(s.toString() != textWithoutMultipleSeparators) {
            s.replace(0, s.length, SpannableStringBuilder(textWithoutMultipleSeparators))
        }
    }
}

/**
 * This function converts all entered separators (in [DECIMAL_SEPARATORS]) to the [LOCALE_DEFAULT_DECIMAL_SEPARATOR].
 */
private fun convertSeparatorsToLocaleDefault(original: String): String {
    var result = original
    DECIMAL_SEPARATORS.forEach { separator ->
        if (separator != LOCALE_DEFAULT_DECIMAL_SEPARATOR && result.contains(separator)) {
            result = result.replace(separator, LOCALE_DEFAULT_DECIMAL_SEPARATOR)
        }
    }
    return result
}

/**
 * Strip out all separators but the first.
 * In this function we assume all separators are already converted to the locale default.
 */
private fun removeAdditionalSeparators(original: String): String {
    var result = original
    var separatorCount = result.count { c -> c == LOCALE_DEFAULT_DECIMAL_SEPARATOR }
    if(separatorCount > 1) {
        // We will reverse the text so we can keep stripping the last (first in reverse) separator off.
        var textReversed = result.reversed()
        val separatorRegex = Regex.fromLiteral(LOCALE_DEFAULT_DECIMAL_SEPARATOR.toString())
        while (separatorCount > 1) {
            textReversed = textReversed.replaceFirst(separatorRegex, "")
            separatorCount--
        }
        // And finally we reverse it back to the original order.
        result = textReversed.reversed()
    }
    return result
}

}

Answer 9

对于 Mono(Droid) 解决方案：

decimal decimalValue = decimal.Parse(input.Text.Replace(",", ".") , CultureInfo.InvariantCulture);

Answer 10

这里的所有其他帖子都有重大漏洞，所以这里有一个解决方案：

• 根据区域强制使用逗号或句号，不会让您键入相反的逗号或句号。
• 如果 EditText 以某个值开头，它将根据需要替换正确的分隔符。

在XML中：

<EditText
    ...
    android:inputType="numberDecimal" 
    ... />

类变量：

private boolean isDecimalSeparatorComma = false;

在onCreate中，找到当前语言环境中使用的分隔符：

if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.N) {
    NumberFormat nf = NumberFormat.getInstance();
    if (nf instanceof DecimalFormat) {
        DecimalFormatSymbols sym = ((DecimalFormat) nf).getDecimalFormatSymbols();
        char decSeparator = sym.getDecimalSeparator();
        isDecimalSeparatorComma = Character.toString(decSeparator).equals(",");
    }
}

还有onCreate，用这个来更新如果您加载当前值，则为：

// Replace editText with commas or periods as needed for viewing
String editTextValue = getEditTextValue(); // load your current value
if (editTextValue.contains(".") && isDecimalSeparatorComma) {
    editTextValue = editTextValue.replaceAll("\\.",",");
} else if (editTextValue.contains(",") && !isDecimalSeparatorComma) {
    editTextValue = editTextValue.replaceAll(",",".");
}
setEditTextValue(editTextValue); // override your current value

此外，在创建时添加侦听器

editText.addTextChangedListener(editTextWatcher);

if (isDecimalSeparatorComma) {
    editText.setKeyListener(DigitsKeyListener.getInstance("0123456789,"));
} else {
    editText.setKeyListener(DigitsKeyListener.getInstance("0123456789."));
}

editTextWatcher

TextWatcher editTextWatcher = new TextWatcher() {
    @Override
    public void beforeTextChanged(CharSequence s, int start, int count, int after) { }

    @Override
    public void onTextChanged(CharSequence s, int start, int before, int count) { }

    @Override
    public void afterTextChanged(Editable s) {
        String editTextValue = s.toString();

        // Count up the number of commas and periods
        Pattern pattern = Pattern.compile("[,.]");
        Matcher matcher = pattern.matcher(editTextValue);
        int count = 0;
        while (matcher.find()) {
            count++;
        }

        // Don't let it put more than one comma or period
        if (count > 1) {
            s.delete(s.length()-1, s.length());
        } else {
            // If there is a comma or period at the end the value hasn't changed so don't update
            if (!editTextValue.endsWith(",") && !editTextValue.endsWith(".")) {
                doSomething()
            }
        }
    }
};

doSomething() 示例，转换为数据的标准周期操纵

private void doSomething() {
    try {
        String editTextStr = editText.getText().toString();
        if (isDecimalSeparatorComma) {
            editTextStr = editTextStr.replaceAll(",",".");
        }
        float editTextFloatValue = editTextStr.isEmpty() ?
                0.0f :
                Float.valueOf(editTextStr);

        ... use editTextFloatValue
    } catch (NumberFormatException e) {
        Log.e(TAG, "Error converting String to Double");
    }
}

Answer 11

您可以执行以下操作：

DecimalFormatSymbols d = DecimalFormatSymbols.getInstance(Locale.getDefault());
input.setFilters(new InputFilter[] { new DecimalDigitsInputFilter(5, 2) });
input.setKeyListener(DigitsKeyListener.getInstance("0123456789" + d.getDecimalSeparator()));

然后您可以使用输入过滤器：

    public class DecimalDigitsInputFilter implements InputFilter {

Pattern mPattern;

public DecimalDigitsInputFilter(int digitsBeforeZero, int digitsAfterZero) {
    DecimalFormatSymbols d = new DecimalFormatSymbols(Locale.getDefault());
    String s = "\\" + d.getDecimalSeparator();
    mPattern = Pattern.compile("[0-9]{0," + (digitsBeforeZero - 1) + "}+((" + s + "[0-9]{0," + (digitsAfterZero - 1) + "})?)||(" + s + ")?");
}

@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {

    Matcher matcher = mPattern.matcher(dest);
    if (!matcher.matches())
        return "";
    return null;
}

}

Answer 12

要本地化您的输入，请使用：

char sep = DecimalFormatSymbols.getInstance().getDecimalSeparator();

然后添加：

textEdit.setKeyListener(DigitsKeyListener.getInstance("0123456789" + sep));

不要忘记将“，”替换为“。”所以 Float 或 Double 可以毫无错误地解析它。

Answer 13

我可以确认建议的修复程序不适用于三星 IME（至少在 S6 和 S9 上），也许还有 LG。无论区域设置如何，它们仍然显示一个点作为小数分隔符。切换到 Google 的 IME 可以解决这个问题，但对于大多数开发人员来说这并不是一个选择。

Oreo 中也没有修复这些键盘的问题，因为这是三星和/或 LG 必须做的修复，然后甚至推送到他们古老的手机上。

相反，我分叉了 number-keyboard 项目 并添加了一种其行为类似于 IME 的模式： 分叉。有关详细信息，请参阅项目示例。这对我来说效果很好，类似于您在银行应用程序中看到的许多“PIN 输入”假 IME。

Answer 14

您可以使用 inputType="phone"，但是在这种情况下，您将必须处理多个 , 或 . 存在，因此需要进行额外的验证将是必要的。

Answer 15

我对 KOTLIN 的修复

遇到了同样的错误，我用以下方法修复了它：

val separator = DecimalFormatSymbols.getInstance().decimalSeparator
mEditText.keyListener = DigitsKeyListener.getInstance("0123456789$separator")

并且效果很好。 ！但！在三星键盘上，不显示分隔符，因此您无法输入十进制数字。

所以我必须通过检查是否使用三星键盘来解决这个问题：

    val x = Settings.Secure.getString(getContentResolver(), Settings.Secure.DEFAULT_INPUT_METHOD);
    if (x.toLowerCase().contains("samsung")) {}
  

但是你仍然有“。”作为小数点分隔符。因此，如果分隔符是逗号，则必须用逗号替换点：

val separator: Char = DecimalFormatSymbols.getInstance().decimalSeparator
 
  if (separator == ',') {
     mEditText.addTextChangedListener(object : TextWatcher {
                        override fun beforeTextChanged(s: CharSequence?, start: Int, count: Int, after: Int) = Unit
    
                        override fun onTextChanged(s: CharSequence?, start: Int, before: Int, count: Int) = Unit
    
                        override fun afterTextChanged(s: Editable?) {
    
                            if (!s.isNullOrEmpty()) {
                                
                                    if (s.toString().contains(".")) {
                                        val replaced = s.toString().replace('.', separator)
                                        mEditText.setText(replaced)
                                        mEditText.setSelection(replaced.length)
                                    }
    
                                
                            }
                        }
                    })
                }

但是，您必须检查是否没有人在 EditText 字段中键入更多“，”。这可以通过正则表达式来完成。

我的整个解决方案：

val x = Settings.Secure.getString(getContentResolver(), Settings.Secure.DEFAULT_INPUT_METHOD);
  if (x.toLowerCase().contains("samsung")) {
         val Number_REGEX: Pattern =  Pattern.compile("^([1-9])*([.,]{1}[0-9]{0,10})?$")
         val separator: Char = DecimalFormatSymbols.getInstance().decimalSeparator
         if (separator == ',') {
            mEditText.addTextChangedListener(object : TextWatcher {
                override fun beforeTextChanged(s: CharSequence?, start: Int, count: Int, after: Int) = Unit

                override fun onTextChanged(s: CharSequence?, start: Int, before: Int, count: Int) = Unit

                override fun afterTextChanged(s: Editable?) {

                    if (!s.isNullOrEmpty()) {
                        val matcherMail = Number_REGEX.matcher(s.toString())
                        if (!matcherMail.matches()) {
                            val length: Int = s.length

                            s.delete(length - 1, length);
                        } else {
                            if (s.toString().contains(".")) {
                                val replaced = s.toString().replace('.', separator)
                                mEditText.setText(replaced)
                                mEditText.setSelection(replaced.length)
                            }

                        }
                    }
                }
            })
        }
    } else {
        val separator = DecimalFormatSymbols.getInstance().decimalSeparator
        mEditText.keyListener = DigitsKeyListener.getInstance("0123456789$separator")
       
    }

xml文件：

 <com.google.android.material.textfield.TextInputEditText
        android:id="@+id/tEditText"
        android:layout_width="match_parent"
        android:layout_height="wrap_content"
        android:hint="Input"
        android:inputType="numberDecimal"
        android:imeOptions="actionDone"/>

如果你想使用号码，请确保获得正确的格式：

val x = NumberFormat.getInstance().parse(mEditText.text.toString()).toDouble()

Answer 16

Android 有一个内置的数字格式化程序。

您可以将其添加到 EditText 以允许小数点和逗号：
android:inputType="numberDecimal" 和 android:digits="0123456789.,"

然后在代码中的某个位置，当用户单击保存时或输入文本后（使用听众）。

// Format the number to the appropriate double
try { 
    Number formatted = NumberFormat.getInstance().parse(editText.getText().toString());
    cost = formatted.doubleValue();
} catch (ParseException e) {
    System.out.println("Error parsing cost string " + editText.getText().toString());
    cost = 0.0;
}

Answer 17

我决定仅在编辑时将逗号更改为点。这是我棘手且相对简单的解决方法：

    editText.setOnFocusChangeListener(new View.OnFocusChangeListener() {
        @Override
        public void onFocusChange(View v, boolean hasFocus) {
            EditText editText = (EditText) v; 
            String text = editText.getText().toString();
            if (hasFocus) {
                editText.setText(text.replace(",", "."));
            } else {
                if (!text.isEmpty()) {
                    Double doubleValue = Double.valueOf(text.replace(",", "."));
                    editText.setText(someDecimalFormatter.format(doubleValue));
                }
            }
        }
    });

someDecimalFormatter 将使用逗号或点取决于区域设置

Answer 18

我的解决方案是：

• 在主要活动中：

  字符分隔符 =DecimalFormatSymbols.getInstance().getDecimalSeparator();
textViewPitchDeadZone.setKeyListener(DigitsKeyListener.getInstance("0123456789" +分隔符));

• 在xml文件中：
  <代码>
  android:imeOptions="flagNoFullscreen"
  android:inputType="numberDecimal"

我将 editText 中的双精度数作为字符串。

Answer 19

8 年多过去了，我很惊讶这个问题还没有解决...
我在这个简单的问题上遇到了困难，因为@Martin 得到最多支持的答案允许输入多个分隔符，即用户可以输入“12,,,,,,12,1,,21,2,”< br>
另外，第二个问题是，在某些设备上，数字键盘上不显示逗号（或需要多次按点按钮）

这是我的解决方案，它解决了上述问题和让用户输入“.”和 ','，但在 EditText 中他将看到与当前语言环境相对应的唯一小数分隔符：

editText.apply { addTextChangedListener(DoubleTextChangedListener(this)) }

以及文本观察器：

  open class DoubleTextChangedListener(private val et: EditText) : TextWatcher {

    init {
        et.inputType = InputType.TYPE_CLASS_NUMBER or InputType.TYPE_NUMBER_FLAG_DECIMAL
        et.keyListener = DigitsKeyListener.getInstance("0123456789.,")
    }

    private val separator = DecimalFormatSymbols.getInstance().decimalSeparator

    override fun beforeTextChanged(s: CharSequence?, start: Int, count: Int, after: Int) {
        //empty
    }

    @CallSuper
    override fun onTextChanged(s: CharSequence, start: Int, before: Int, count: Int) {
        et.run {
            removeTextChangedListener(this@DoubleTextChangedListener)
            val formatted = toLocalizedDecimal(s.toString(), separator)
            setText(formatted)
            setSelection(formatted.length)
            addTextChangedListener(this@DoubleTextChangedListener)
        }
    }

    override fun afterTextChanged(s: Editable?) {
        // empty
    }

    /**
     * Formats input to a decimal. Leaves the only separator (or none), which matches [separator].
     * Examples:
     * 1. [s]="12.12", [separator]=',' -> result= "12,12"
     * 2. [s]="12.12", [separator]='.' -> result= "12.12"
     * 4. [s]="12,12", [separator]='.' -> result= "12.12"
     * 5. [s]="12,12,,..,,,,,34..,", [separator]=',' -> result= "12,1234"
     * 6. [s]="12.12,,..,,,,,34..,", [separator]='.' -> result= "12.1234"
     * 7. [s]="5" -> result= "5"
     */
    private fun toLocalizedDecimal(s: String, separator: Char): String {
        val cleared = s.replace(",", ".")
        val splitted = cleared.split('.').filter { it.isNotBlank() }
        return when (splitted.size) {
            0 -> s
            1 -> cleared.replace('.', separator).replaceAfter(separator, "")
            2 -> splitted.joinToString(separator.toString())
            else -> splitted[0]
                    .plus(separator)
                    .plus(splitted.subList(1, splitted.size - 1).joinToString(""))
        }
    }
}

Answer 20

简单的解决方案，制作自定义控件。 （这是在 Xamarin android 中制作的，但应该可以轻松移植到 java）

public class EditTextDecimalNumber:EditText
{
    readonly string _numberFormatDecimalSeparator;

    public EditTextDecimalNumber(Context context, IAttributeSet attrs) : base(context, attrs)
    {
        InputType = InputTypes.NumberFlagDecimal;
        TextChanged += EditTextDecimalNumber_TextChanged;
        _numberFormatDecimalSeparator = System.Threading.Thread.CurrentThread.CurrentUICulture.NumberFormat.NumberDecimalSeparator;

        KeyListener = DigitsKeyListener.GetInstance($"0123456789{_numberFormatDecimalSeparator}");
    }

    private void EditTextDecimalNumber_TextChanged(object sender, TextChangedEventArgs e)
    {
        int noOfOccurence = this.Text.Count(x => x.ToString() == _numberFormatDecimalSeparator);
        if (noOfOccurence >=2)
        {
            int lastIndexOf = this.Text.LastIndexOf(_numberFormatDecimalSeparator,StringComparison.CurrentCulture);
            if (lastIndexOf!=-1)
            {
                this.Text = this.Text.Substring(0, lastIndexOf);
                this.SetSelection(this.Text.Length);
            }

        }
    }
}

Answer 21

我必须想出一个解决办法，它由不同的答案组成。
这将允许使用“,”或“.”。最多一位小数

这是我的 editText:

val separator: Char = DecimalFormatSymbols.getInstance().decimalSeparator
editTextBox.filters = arrayOf<InputFilter>(DecimalDigitsInputFilter(5, 1, separator))
editTextBox.keyListener = DigitsKeyListener.getInstance("0123456789$separator")

和我处理特定正则表达式的类：

class DecimalDigitsInputFilter(
    digitsBeforeZero: Int, 
    digitsAfterZero: Int, 
    separator: Char
) : InputFilter {
    private val mPattern: Pattern =
        Pattern.compile("[0-9]{0," + (digitsBeforeZero - 1) + "}+((\\$separator[0-9]{0," + (digitsAfterZero - 1) + "})?)||(\\$separator)?")

    override fun filter(source: CharSequence, start: Int, end: Int, dest: Spanned, dstart: Int, dend: Int): CharSequence? {
        val matcher = mPattern.matcher(dest)
        return if (!matcher.matches()) "" else null
    }
}

Answer 22

EditText.inputType =
    TYPE_NUMBER_FLAG_DECIMAL or TYPE_NUMBER_FLAG_SIGNED or TYPE_CLASS_NUMBER

EditText.keyListener = DigitsKeyListener.getInstance("0123456789,-")

EditText.addTextChangedListener(object : TextWatcher {
    var beforeText = ""

    override fun beforeTextChanged(s: CharSequence?, start: Int, count: Int, after: Int) {
        beforeText = s.toString()
    }

    override fun onTextChanged(s: CharSequence?, start: Int, before: Int, count: Int) {}

    override fun afterTextChanged(s: Editable?) {
        if (s != null) {
            if (s.length > 1 && s.last() == '-') {
                s.replace(0, s.length, beforeText)
            }

            if (s.length == 1 && s.last() == ',') {
                s.replace(0, s.length, beforeText)
            }

            if (s.length > 1 && beforeText.length < s.length && beforeText.contains(',') && s.last() == ',') {
                s.replace(0, s.length, beforeText)
            }
        }
    }
})

Answer 23

我不知道为什么你的答案这么复杂。如果 SDK 中存在错误，您必须覆盖它或绕过它。

我选择了第二种方法来解决这个问题。如果将字符串格式化为 Locale.ENGLISH，然后将其放入 EditText（即使是空字符串）。示例：

String.format(Locale.ENGLISH,"%.6f", yourFloatNumber);

追求该解决方案，您的结果与显示的键盘兼容。然后浮点数和双精度数以典型的编程语言方式工作，使用点而不是逗号。

Answer 24

我认为这个解决方案比这里写的其他解决方案简单：

<EditText
    android:inputType="numberDecimal"
    android:digits="0123456789," />

这样当你按“.”时在软键盘中没有任何反应；只允许使用数字和逗号。

Answer 25

解决方法（直到 Google 修复此错误）是使用 EditText 和 android:inputType="numberDecimal" 和 android:digits="0123456789.,"< /代码>。

然后使用以下 afterTextChanged 将 TextChangedListener 添加到 EditText：

public void afterTextChanged(Editable s) {
    double doubleValue = 0;
    if (s != null) {
        try {
            doubleValue = Double.parseDouble(s.toString().replace(',', '.'));
        } catch (NumberFormatException e) {
            //Error
        }
    }
    //Do something with doubleValue
}