如何禁用 UITextField 中的字母字符？

Question

在我的应用程序中，我需要允许用户仅输入数字。 我如何允许 UITextField 仅接收来自用户的数字？

Answer 1

本示例中的字符是允许的，因此如果您不希望用户使用某个字符，请将其从 myCharSet 中排除。

- (BOOL)textField:(UITextField *)theTextField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string 
{
    NSCharacterSet *myCharSet = [NSCharacterSet characterSetWithCharactersInString:@"0123456789"];
    for (int i = 0; i < [string length]; i++) {
        unichar c = [string characterAtIndex:i];
        if (![myCharSet characterIsMember:c]) {
            return NO;
        }
    }

    return YES;
} 

Answer 2

我更喜欢以下解决方案，它实际上可以防止除数字和退格键之外的任何输入。出于某种原因，退格键由空字符串表示，除非空字符串返回 YES，否则无法使用。当用户输入数字以外的字符时，我还会弹出一个警报视图。

- (BOOL)textField:(UITextField *)theTextField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string 
{
    if (string.length == 0) {
        return YES;
    }
    NSCharacterSet *myCharSet = [NSCharacterSet characterSetWithCharactersInString:@"0123456789"];
    for (int i = 0; i < [string length]; i++) {
        unichar c = [string characterAtIndex:i];
        if ([myCharSet characterIsMember:c]) {
            return YES;
        }
    }
    UIAlertView *av = [[UIAlertView alloc] initWithTitle:@"Invalid Input" message:@"Only numbers are allowed for participant number." delegate:self cancelButtonTitle:@"Dismiss" otherButtonTitles:nil];
    [av show];
    return NO;
}

Answer 3

这可能是最干净、最简单的解决方案，只允许正数或负数。这也允许退格。

-(BOOL)textField:(UITextField *)textField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string{

    NSCharacterSet *allowedCharacters = [NSCharacterSet characterSetWithCharactersInString:@"-0123456789"];

    if([string rangeOfCharacterFromSet:allowedCharacters.invertedSet].location == NSNotFound){

        return YES;

    }

    return NO;

}

Answer 4

您可以做的一件事是显示数字键盘，并在文本字段旁边或其他地方添加一个动态按钮来隐藏键盘。

Answer 5

你们可能会发火..但这对我有用..只有数字（包括负数）和退格键。

NSCharacterSet *validCharSet;

if (range.location == 0) 
    validCharSet = [NSCharacterSet characterSetWithCharactersInString:@"0123456789-."];
else
    validCharSet = [NSCharacterSet characterSetWithCharactersInString:@"0123456789."];

if ([[string stringByTrimmingCharactersInSet:validCharSet] length] > 0 ) return NO;  //not allowable char


NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
[numberFormatter setNumberStyle:NSNumberFormatterDecimalStyle];

NSNumber* candidateNumber;

NSString* candidateString = [textField.text stringByReplacingCharactersInRange:range withString:string];

range = NSMakeRange(0, [candidateString length]);

[numberFormatter getObjectValue:&candidateNumber forString:candidateString range:&range error:nil];

if  (candidateNumber == nil ) {

    if  (candidateString.length <= 1) 
        return YES;
    else
        return NO;
}

return YES;

Answer 6

这是我的解决方案，使用方法 isSupersetOfSet: 应用集合代数，这也不允许粘贴带有无效字符的文本：

- (BOOL)textField:(UITextField *)theTextField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string
{
    if (string.length == 0 || [_numericCharSet isSupersetOfSet:[NSCharacterSet characterSetWithCharactersInString:string]]) {
        return YES;
    }
    else {
        UIAlertView *av = [[UIAlertView alloc] initWithTitle:@"Invalid Input"
                                                     message:@"Only numeric input allowed."
                                                    delegate:self
                                           cancelButtonTitle:@"Close"
                                           otherButtonTitles:nil];
        [av show];
        return NO;
    }
}

注意：根据 Apple 开发者库，最好缓存静态NSCharacterSet 而不是一次又一次地创建它（这里是 _numericCharSet）。

不过，我更喜欢让用户输入任何字符并验证当 textField 尝试退出第一响应者时调用的方法 textFieldShouldEndEditing: 中的值。
通过这种方式，用户可以粘贴任何文本（可能由字母和数字混合组成）并将其整理到我的文本字段中。用户不喜欢看到他们的行动受到限制。

Answer 7

在斯威夫特

func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {
    if textField.tag == 2 {  //your textField
        let invalid = NSCharacterSet(charactersInString: "aeiou")  //characters to block
        if let range = string.rangeOfCharacterFromSet(invalid) {
            return false
        }
    }

    return true
}

Answer 8

这是一个快速示例，

func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {

    var disabledCharacters:NSCharacterSet = NSCharacterSet(charactersInString: "0123456789")
        for (var i:Int = 0; i < count(string); ++i) {
            var c = (string as NSString).characterAtIndex(i)
            if (disabledCharacters.characterIsMember(c)) {
                println("Can't use that character dude :/")
                return false
            }
        }

    return true
}

不要忘记将 UITextFieldDelegate 添加到您的 UIViewController 中。

Answer 9

- (BOOL)textField:(UITextField *)textField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string
{
    char *x = (char*)[string UTF8String];
    //NSLog(@"char index is %i",x[0]);
    if([string isEqualToString:@"-"] || [string isEqualToString:@"("] || [string isEqualToString:@")"] || [string isEqualToString:@"0"] || [string isEqualToString:@"1"] ||  [string isEqualToString:@"2"] ||  [string isEqualToString:@"3"] ||  [string isEqualToString:@"4"] ||  [string isEqualToString:@"5"] ||  [string isEqualToString:@"6"] ||  [string isEqualToString:@"7"] ||  [string isEqualToString:@"8"] ||  [string isEqualToString:@"9"] || x[0]==0 || [string isEqualToString:@" "]) {

    NSUInteger newLength = [textField.text length] + [string length] - range.length;
    return (newLength > 14) ? NO : YES;
} else {
    return NO;
}
}

Answer 10

这个线程有点旧，但为了参考，我将在 swift 3 中留下一个解决方案。该解决方案将结合 decimalDigits 和实际的小数。您可以将任何您想要的组合放在一起，但对于我的情况来说，这就是要求。

// instantiate a mutable character set
let characterSet = NSMutableCharacterSet()
// assign the needed character set
characterSet.formUnion(with: NSCharacterSet.decimalDigits)
// only need the decimal character added to the character set
characterSet.addCharacters(in: ".")
// invert and return false if it's anything other than what we're looking for
if string.rangeOfCharacter(from: characterSet.inverted) != nil {
    return false
}