字符串格式中的命名占位符

Question

在Python中，当格式化字符串时，我可以按名称而不是按位置填充占位符，就像这样：

print "There's an incorrect value '%(value)s' in column # %(column)d" % \
  { 'value': x, 'column': y }

我想知道这在Java中是否可能（希望没有外部库）？

Answer 1

StringSubstitutor< Apache Commons Text 库中的 /code>是实现此目的的一种轻量级方法，前提是您的值的格式已经正确。

Map<String, String> values = new HashMap<>();
values.put("value", "1");
values.put("column","2");

StringSubstitutor sub = new StringSubstitutor(values, "%(", ")");
String result = sub.replace("There's an incorrect value '%(value)' in column # %(column)");

result 字符串将包含以下内容：

第 2 列中的值“1”不正确

使用 Maven 时，您可以将此依赖项添加到 pom.xml 中：

<dependency>
    <groupId>org.apache.commons</groupId>
    <artifactId>commons-text</artifactId>
    <version>1.10.0</version>
</dependency>

Answer 2

不完全是，但您可以使用 MessageFormat 来多次引用一个值：

MessageFormat.format("There's an incorrect value \"{0}\" in column # {1}", x, y);

上述操作也可以使用 String.format() 完成，但如果您需要构建复杂的表达式，我发现 messageFormat 语法更清晰，而且您不需要关心要放入的对象的类型字符串

Answer 3

Apache Common 的另一个示例 StringSubstitutor 用于简单的命名占位符。

String template = "Welcome to {theWorld}. My name is {myName}.";

Map<String, String> values = new HashMap<>();
values.put("theWorld", "Stackoverflow");
values.put("myName", "Thanos");

String message = StringSubstitutor.replace(template, values, "{", "}");

System.out.println(message);

// Welcome to Stackoverflow. My name is Thanos.

Answer 4

您可以使用 StringTemplate 库，它提供您想要的以及更多。

import org.antlr.stringtemplate.*;

final StringTemplate hello = new StringTemplate("Hello, $name$");
hello.setAttribute("name", "World");
System.out.println(hello.toString());

Answer 5

public static String format(String format, Map<String, Object> values) {
    StringBuilder formatter = new StringBuilder(format);
    List<Object> valueList = new ArrayList<Object>();

    Matcher matcher = Pattern.compile("\\$\\{(\\w+)}").matcher(format);

    while (matcher.find()) {
        String key = matcher.group(1);

        String formatKey = String.format("${%s}", key);
        int index = formatter.indexOf(formatKey);

        if (index != -1) {
            formatter.replace(index, index + formatKey.length(), "%s");
            valueList.add(values.get(key));
        }
    }

    return String.format(formatter.toString(), valueList.toArray());
}

例子：

String format = "My name is ${1}. ${0} ${1}.";

Map<String, Object> values = new HashMap<String, Object>();
values.put("0", "James");
values.put("1", "Bond");

System.out.println(format(format, values)); // My name is Bond. James Bond.

Answer 6

感谢您的帮助！根据您的所有线索，我编写了例程来完全执行我想要的操作——使用字典进行类似 python 的字符串格式化。由于我是 Java 新手，任何提示都会受到赞赏。

public static String dictFormat(String format, Hashtable<String, Object> values) {
    StringBuilder convFormat = new StringBuilder(format);
    Enumeration<String> keys = values.keys();
    ArrayList valueList = new ArrayList();
    int currentPos = 1;
    while (keys.hasMoreElements()) {
        String key = keys.nextElement(),
        formatKey = "%(" + key + ")",
        formatPos = "%" + Integer.toString(currentPos) + "$";
        int index = -1;
        while ((index = convFormat.indexOf(formatKey, index)) != -1) {
            convFormat.replace(index, index + formatKey.length(), formatPos);
            index += formatPos.length();
        }
        valueList.add(values.get(key));
        ++currentPos;
    }
    return String.format(convFormat.toString(), valueList.toArray());
}

Answer 7

这是一个旧线程，但仅供记录，您也可以使用 Java 8 样式，如下所示：

public static String replaceParams(Map<String, String> hashMap, String template) {
    return hashMap.entrySet().stream().reduce(template, (s, e) -> s.replace("%(" + e.getKey() + ")", e.getValue()),
            (s, s2) -> s);
}

用法：

public static void main(String[] args) {
    final HashMap<String, String> hashMap = new HashMap<String, String>() {
        {
            put("foo", "foo1");
            put("bar", "bar1");
            put("car", "BMW");
            put("truck", "MAN");
        }
    };
    String res = replaceParams(hashMap, "This is '%(foo)' and '%(foo)', but also '%(bar)' '%(bar)' indeed.");
    System.out.println(res);
    System.out.println(replaceParams(hashMap, "This is '%(car)' and '%(foo)', but also '%(bar)' '%(bar)' indeed."));
    System.out.println(replaceParams(hashMap, "This is '%(car)' and '%(truck)', but also '%(foo)' '%(bar)' + '%(truck)' indeed."));
}

输出将是：

This is 'foo1' and 'foo1', but also 'bar1' 'bar1' indeed.
This is 'BMW' and 'foo1', but also 'bar1' 'bar1' indeed.
This is 'BMW' and 'MAN', but also 'foo1' 'bar1' + 'MAN' indeed.

Answer 8

Apache Commons Text 的 StringSubstitutor 可以使用。 （请参阅 依赖信息 了解如何将其包含在项目中.) 请注意，StrSubstitutor 已弃用。

import org.apache.commons.text.StringSubstitutor;
// ...
Map<String, String> values = new HashMap<>();
values.put("animal", "quick brown fox");
values.put("target", "lazy dog");
StringSubstitutor sub = new StringSubstitutor(values);
String result = sub.replace("The ${animal} jumped over the ${target}.");
// "The quick brown fox jumped over the lazy dog."

此类支持为变量提供默认值。

String result = sub.replace("The number is ${undefined.property:-42}.");
// "The number is 42."

要使用递归变量替换，请调用 setEnableSubstitutionInVariables(true);。

Map<String, String> values = new HashMap<>();
values.put("b", "c");
values.put("ac", "Test");
StringSubstitutor sub = new StringSubstitutor(values);
sub.setEnableSubstitutionInVariables(true);
String result = sub.replace("${a${b}}");
// "Test"

Answer 9

我是一个小型库的作者，它完全可以满足您的需求：

Student student = new Student("Andrei", 30, "Male");

String studStr = template("#{id}\tName: #{st.getName}, Age: #{st.getAge}, Gender: #{st.getGender}")
                    .arg("id", 10)
                    .arg("st", student)
                    .format();
System.out.println(studStr);

或者您可以链接参数:

String result = template("#{x} + #{y} = #{z}")
                    .args("x", 5, "y", 10, "z", 15)
                    .format();
System.out.println(result);

// Output: "5 + 10 = 15"

Answer 10

在撰写本文时，Java 中还没有内置任何内容。我建议您编写自己的实现。我更喜欢简单流畅的构建器界面，而不是创建映射并将其传递给函数 - 您最终会得到一个很好的连续代码块，例如：

String result = new TemplatedStringBuilder("My name is {{name}} and I from {{town}}")
   .replace("name", "John Doe")
   .replace("town", "Sydney")
   .finish();

这是一个简单的实现：

class TemplatedStringBuilder {

    private final static String TEMPLATE_START_TOKEN = "{{";
    private final static String TEMPLATE_CLOSE_TOKEN = "}}";

    private final String template;
    private final Map<String, String> parameters = new HashMap<>();

    public TemplatedStringBuilder(String template) {
        if (template == null) throw new NullPointerException();
        this.template = template;
    }

    public TemplatedStringBuilder replace(String key, String value){
        parameters.put(key, value);
        return this;
    }

    public String finish(){

        StringBuilder result = new StringBuilder();

        int startIndex = 0;

        while (startIndex < template.length()){

            int openIndex  = template.indexOf(TEMPLATE_START_TOKEN, startIndex);

            if (openIndex < 0){
                result.append(template.substring(startIndex));
                break;
            }

            int closeIndex = template.indexOf(TEMPLATE_CLOSE_TOKEN, openIndex);

            if(closeIndex < 0){
                result.append(template.substring(startIndex));
                break;
            }

            String key = template.substring(openIndex + TEMPLATE_START_TOKEN.length(), closeIndex);

            if (!parameters.containsKey(key)) throw new RuntimeException("missing value for key: " + key);

            result.append(template.substring(startIndex, openIndex));
            result.append(parameters.get(key));

            startIndex = closeIndex + TEMPLATE_CLOSE_TOKEN.length();
        }

        return result.toString();
    }
}

Answer 11

您应该查看官方 ICU4J 库。它提供了一个 MessageFormat 类，类似于可用的类JDK，但前者支持命名占位符。

与本页提供的其他解决方案不同。 ICU4j 是由 IBM 维护并定期更新的 ICU 项目的一部分。此外，它还支持高级用例，例如多元化等。

这是一个代码示例：

MessageFormat messageFormat =
        new MessageFormat("Publication written by {author}.");

Map<String, String> args = Map.of("author", "John Doe");

System.out.println(messageFormat.format(args));

Answer 12

Apache Commons Lang 的 replaceEach 方法可能会根据您的具体需求派上用场。您可以轻松地使用它通过以下单个方法调用按名称替换占位符：

StringUtils.replaceEach("There's an incorrect value '%(value)' in column # %(column)",
            new String[] { "%(value)", "%(column)" }, new String[] { x, y });

给定一些输入文本，这会将第一个字符串数组中出现的所有占位符替换为第二个字符串数组中的相应值。

Answer 13

截至 2022 年，最新的解决方案是 Apache Commons Text StringSubstitutor

来自文档：

// Build map
 Map<String, String> valuesMap = new HashMap<>();
 valuesMap.put("animal", "quick brown fox");
 valuesMap.put("target", "lazy dog");
 String templateString = "The ${animal} jumped over the ${target} ${undefined.number:-1234567890} times.";

 // Build StringSubstitutor
 StringSubstitutor sub = new StringSubstitutor(valuesMap);

 // Replace
 String resolvedString = sub.replace(templateString)

;

Answer 14

如果不需要使用映射，那么您可以使用 Java 的字符串模板功能。
它在 JEP 430 中进行了描述，并作为预览功能出现在 JDK 21 中。下面是一个使用示例：

String s = STR."There's an incorrect value '\{x}' in column # \{y}";

Java 的字符串模板比其他语言（例如 C# 的字符串插值和 Python 的 f 字符串）中的功能更通用，也更安全。
例如，字符串连接或插值使 SQL 注入攻击成为可能：

String query = "SELECT * FROM Person p WHERE p.last_name = '" + name + "'";
ResultSet rs = conn.createStatement().executeQuery(query);

但此变体（来自 JEP 430）可以防止 SQL 注入：

PreparedStatement ps = DB."SELECT * FROM Person p WHERE p.last_name = \{name}";
ResultSet rs = ps.executeQuery();

Answer 15

你可以在字符串助手类上有类似的东西

/**
 * An interpreter for strings with named placeholders.
 *
 * For example given the string "hello %(myName)" and the map <code>
 *      <p>Map<String, Object> map = new HashMap<String, Object>();</p>
 *      <p>map.put("myName", "world");</p>
 * </code>
 *
 * the call {@code format("hello %(myName)", map)} returns "hello world"
 *
 * It replaces every occurrence of a named placeholder with its given value
 * in the map. If there is a named place holder which is not found in the
 * map then the string will retain that placeholder. Likewise, if there is
 * an entry in the map that does not have its respective placeholder, it is
 * ignored.
 *
 * @param str
 *            string to format
 * @param values
 *            to replace
 * @return formatted string
 */
public static String format(String str, Map<String, Object> values) {

    StringBuilder builder = new StringBuilder(str);

    for (Entry<String, Object> entry : values.entrySet()) {

        int start;
        String pattern = "%(" + entry.getKey() + ")";
        String value = entry.getValue().toString();

        // Replace every occurence of %(key) with value
        while ((start = builder.indexOf(pattern)) != -1) {
            builder.replace(start, start + pattern.length(), value);
        }
    }

    return builder.toString();
}

Answer 16

根据答案我创建了MapBuilder类：

public class MapBuilder {

    public static Map<String, Object> build(Object... data) {
        Map<String, Object> result = new LinkedHashMap<>();

        if (data.length % 2 != 0) {
            throw new IllegalArgumentException("Odd number of arguments");
        }

        String key = null;
        Integer step = -1;

        for (Object value : data) {
            step++;
            switch (step % 2) {
                case 0:
                    if (value == null) {
                        throw new IllegalArgumentException("Null key value");
                    }
                    key = (String) value;
                    continue;
                case 1:
                    result.put(key, value);
                    break;
            }
        }

        return result;
    }

}

然后我创建了类StringFormat 用于字符串格式化：

public final class StringFormat {

    public static String format(String format, Object... args) {
        Map<String, Object> values = MapBuilder.build(args);

        for (Map.Entry<String, Object> entry : values.entrySet()) {
            String key = entry.getKey();
            Object value = entry.getValue();
            format = format.replace("$" + key, value.toString());
        }

        return format;
    }

}

您可以这样使用：

String bookingDate = StringFormat.format("From $startDate to $endDate"), 
        "$startDate", formattedStartDate, 
        "$endDate", formattedEndDate
);

Answer 17

我还创建了一个 util/helper 类（使用 jdk 8），它可以格式化字符串并替换变量的出现。

为此，我使用了 Matchers“appendReplacement”方法，该方法仅对格式字符串的受影响部分进行所有替换和循环。

目前，帮助程序类没有得到很好的 javadoc 文档记录。我将来会改变这一点；）
无论如何，我评论了最重要的几行（我希望如此）。

    public class FormatHelper {

    //Prefix and suffix for the enclosing variable name in the format string.
    //Replace the default values with any you need.
    public static final String DEFAULT_PREFIX = "${";
    public static final String DEFAULT_SUFFIX = "}";

    //Define dynamic function what happens if a key is not found.
    //Replace the defualt exception with any "unchecked" exception type you need or any other behavior.
    public static final BiFunction<String, String, String> DEFAULT_NO_KEY_FUNCTION =
            (fullMatch, variableName) -> {
                throw new RuntimeException(String.format("Key: %s for variable %s not found.",
                                                         variableName,
                                                         fullMatch));
            };
    private final Pattern variablePattern;
    private final Map<String, String> values;
    private final BiFunction<String, String, String> noKeyFunction;
    private final String prefix;
    private final String suffix;

    public FormatHelper(Map<String, String> values) {
        this(DEFAULT_NO_KEY_FUNCTION, values);
    }

    public FormatHelper(
            BiFunction<String, String, String> noKeyFunction, Map<String, String> values) {
        this(DEFAULT_PREFIX, DEFAULT_SUFFIX, noKeyFunction, values);
    }

    public FormatHelper(String prefix, String suffix, Map<String, String> values) {
        this(prefix, suffix, DEFAULT_NO_KEY_FUNCTION, values);
    }

    public FormatHelper(
            String prefix,
            String suffix,
            BiFunction<String, String, String> noKeyFunction,
            Map<String, String> values) {
        this.prefix = prefix;
        this.suffix = suffix;
        this.values = values;
        this.noKeyFunction = noKeyFunction;

        //Create the Pattern and quote the prefix and suffix so that the regex don't interpret special chars.
        //The variable name is a "\w+" in an extra capture group.
        variablePattern = Pattern.compile(Pattern.quote(prefix) + "(\\w+)" + Pattern.quote(suffix));
    }

    public static String format(CharSequence format, Map<String, String> values) {
        return new FormatHelper(values).format(format);
    }

    public static String format(
            CharSequence format,
            BiFunction<String, String, String> noKeyFunction,
            Map<String, String> values) {
        return new FormatHelper(noKeyFunction, values).format(format);
    }

    public static String format(
            String prefix, String suffix, CharSequence format, Map<String, String> values) {
        return new FormatHelper(prefix, suffix, values).format(format);
    }

    public static String format(
            String prefix,
            String suffix,
            BiFunction<String, String, String> noKeyFunction,
            CharSequence format,
            Map<String, String> values) {
        return new FormatHelper(prefix, suffix, noKeyFunction, values).format(format);
    }

    public String format(CharSequence format) {

        //Create matcher based on the init pattern for variable names.
        Matcher matcher = variablePattern.matcher(format);

        //This buffer will hold all parts of the formatted finished string.
        StringBuffer formatBuffer = new StringBuffer();

        //loop while the matcher finds another variable (prefix -> name <- suffix) match
        while (matcher.find()) {

            //The root capture group with the full match e.g ${variableName}
            String fullMatch = matcher.group();

            //The capture group for the variable name resulting from "(\w+)" e.g. variableName
            String variableName = matcher.group(1);

            //Get the value in our Map so the Key is the used variable name in our "format" string. The associated value will replace the variable.
            //If key is missing (absent) call the noKeyFunction with parameters "fullMatch" and "variableName" else return the value.
            String value = values.computeIfAbsent(variableName, key -> noKeyFunction.apply(fullMatch, key));

            //Escape the Map value because the "appendReplacement" method interprets the $ and \ as special chars.
            String escapedValue = Matcher.quoteReplacement(value);

            //The "appendReplacement" method replaces the current "full" match (e.g. ${variableName}) with the value from the "values" Map.
            //The replaced part of the "format" string is appended to the StringBuffer "formatBuffer".
            matcher.appendReplacement(formatBuffer, escapedValue);
        }

        //The "appendTail" method appends the last part of the "format" String which has no regex match.
        //That means if e.g. our "format" string has no matches the whole untouched "format" string is appended to the StringBuffer "formatBuffer".
        //Further more the method return the buffer.
        return matcher.appendTail(formatBuffer)
                      .toString();
    }

    public String getPrefix() {
        return prefix;
    }

    public String getSuffix() {
        return suffix;
    }

    public Map<String, String> getValues() {
        return values;
    }
}

您可以为具有值（或后缀前缀或 noKeyFunction）的特定 Map 创建类实例
喜欢：

    Map<String, String> values = new HashMap<>();
    values.put("firstName", "Peter");
    values.put("lastName", "Parker");


    FormatHelper formatHelper = new FormatHelper(values);
    formatHelper.format("${firstName} ${lastName} is Spiderman!");
    // Result: "Peter Parker is Spiderman!"
    // Next format:
    formatHelper.format("Does ${firstName} ${lastName} works as photographer?");
    //Result: "Does Peter Parker works as photographer?"

此外，您还可以定义如果值映射中的键丢失会发生什么（两种方式都有效，例如格式字符串中的变量名称错误或映射中缺少键）。
默认行为是抛出未经检查的异常（未经检查，因为我使用默认的 jdk8 函数，它无法处理检查的异常），例如：

    Map<String, String> map = new HashMap<>();
    map.put("firstName", "Peter");
    map.put("lastName", "Parker");


    FormatHelper formatHelper = new FormatHelper(map);
    formatHelper.format("${missingName} ${lastName} is Spiderman!");
    //Result: RuntimeException: Key: missingName for variable ${missingName} not found.

您可以在构造函数调用中定义自定义行为，例如：

Map<String, String> values = new HashMap<>();
values.put("firstName", "Peter");
values.put("lastName", "Parker");


FormatHelper formatHelper = new FormatHelper(fullMatch, variableName) -> variableName.equals("missingName") ? "John": "SOMETHING_WRONG", values);
formatHelper.format("${missingName} ${lastName} is Spiderman!");
// Result: "John Parker is Spiderman!"

或将其委托回默认的无键行为：

...
    FormatHelper formatHelper = new FormatHelper((fullMatch, variableName) ->   variableName.equals("missingName") ? "John" :
            FormatHelper.DEFAULT_NO_KEY_FUNCTION.apply(fullMatch,
                                                       variableName), map);
...

为了更好地处理，还有静态方法表示，例如：

Map<String, String> values = new HashMap<>();
values.put("firstName", "Peter");
values.put("lastName", "Parker");

FormatHelper.format("${firstName} ${lastName} is Spiderman!", map);
// Result: "Peter Parker is Spiderman!"

Answer 18

有 Java 插件可以在 Java 中使用字符串插值（如 Kotlin、JavaScript）。支持 Java 8、9、10、11… https://github.com/antkorwin/better-strings< /a>

在字符串文字中使用变量：

int a = 3;
int b = 4;
System.out.println("${a} + ${b} = ${a+b}");

使用表达式：

int a = 3;
int b = 4;
System.out.println("pow = ${a * a}");
System.out.println("flag = ${a > b ? true : false}");

使用函数：

@Test
void functionCall() {
    System.out.println("fact(5) = ${factorial(5)}");
}

long factorial(int n) {
    long fact = 1;
    for (int i = 2; i <= n; i++) {
        fact = fact * i;
    }
    return fact;
}

有关详细信息，请阅读项目自述文件。

Answer 19

不幸的是，快速回答是否定的。但是，您可以非常接近合理的语法：

"""
   You are $compliment!
"""
.replace('$compliment', 'awesome');

至少它比 String.format 更具可读性和可预测性！

Answer 20

我的答案是：

a）尽可能使用 StringBuilder

b）保留“占位符”的位置（以任何形式：整数是最好的，特殊字符如美元宏等），然后使用 StringBuilder.insert() （参数的几个版本）。

使用外部库似乎有点矫枉过正，我相信当 StringBuilder 在内部转换为 String 时，性能会显着降低。

Answer 21

我尝试了一种快速的方法

public static void main(String[] args) 
{
    String rowString = "replace the value ${var1} with ${var2}";
    
    Map<String,String> mappedValues = new HashMap<>();
    
    mappedValues.put("var1", "Value 1");
    mappedValues.put("var2", "Value 2");
    
    System.out.println(replaceOccurence(rowString, mappedValues));
}

private static  String replaceOccurence(String baseStr ,Map<String,String> mappedValues)
{
    for(String key :mappedValues.keySet())
    {
        baseStr = baseStr.replace("${"+key+"}", mappedValues.get(key));
    }
    
    return baseStr;
}

Answer 22

我最终得到了下一个解决方案：
使用方法 substitute() 创建类 TemplateSubstitutor 并使用它来格式化输出
然后创建一个字符串模板并用值填充它

import java.util.*;
public class MyClass {

    public static void main(String args[]) {
    String template = "WRR = {WRR}, SRR = {SRR}\n" +
                      "char_F1 = {char_F1}, word_F1 = {word_F1}\n";
    
    Map<String, Object> values = new HashMap<>();
    values.put("WRR", 99.9);
    values.put("SRR", 99.8);
    values.put("char_F1", 80);
    values.put("word_F1", 70);
    
    String message = TemplateSubstitutor.substitute(values, template);
    
    System.out.println(message);
    }
}

class TemplateSubstitutor {
    public static String substitute(Map<String, Object> map, String input_str) {
        String output_str = input_str;
        for (Map.Entry<String, Object> entry : map.entrySet()) {
            String key = entry.getKey();
            Object value = entry.getValue();
            output_str = output_str.replace("{" + key + "}", String.valueOf(value));
        }
        return output_str;
    }
    
}

Answer 23

尝试 Freemarker，模板库。

Answer 24

https://dzone.com/articles/java-string-format-examples String.format(inputString, [listOfParams]) 将是最简单的方法。字符串中的占位符可以按顺序定义。有关更多详细信息，请查看提供的链接。