Question

我有这个JavaScript函数，用于强制用户只在文本框中键入数字。现在，我想修改此功能，以便用户输入加号（+）符号。怎么做到这一点？

//To only enable digit in the user input

function isNumberKey(evt)
{
    var charCode = (evt.which) ? evt.which : event.keyCode
    if (charCode > 31 && (charCode < 48 || charCode > 57))
        return false;
    return true;
}

Answer 1

由于'+'符号的十进制ASCII代码为43，因此您可以将其添加到您的条件中。

例如：

function isNumberKey(evt)
{
    var charCode = (evt.which) ? evt.which : event.keyCode
    if (charCode != 43 && charCode > 31 && (charCode < 48 || charCode > 57))
        return false;
    return true;
}

这样，允许加号。

Answer 2

此代码可能有效。我添加了对SHIFT + (equal sign)和小键盘+的支持。

function isNumberKey(evt)
{
  var charCode = (evt.which) ? evt.which : event.keyCode;
  var shiftPressed = (window.Event) ? e.modifiers & Event.SHIFT_MASK : e.shiftKey;

  if ((shiftPressed && charCode == 187) || (charCode == 107))
  {
    return true;
  } else if ((charCode > 95) && (charCode < 106)) {
    return true;
  } else if (charCode > 31 && (charCode < 48 || charCode > 57))) {
    return false;
  } else {
    return true;
  }
}

Answer 3

这是愚蠢的...根本不是答案。我建议你这样做。

function isNumberKey(evt)
{
    console.log(evt.keyCode);
    return false;
}

找出所有键的范围，并实现它。

Answer 4

这是工作。 Javascript键码仅允许数字和加号

<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>JavaScript form validation</title>
</head>
<body>

<form name="form1" action="#">

Mobile Number:&nbsp;&nbsp;<input type='text'  id='PhoneNumber' maxlength="10" onKeyPress="return IsNumeric3(event);" ondrop="return false;" onpaste="return false;"/>
<span id="error3" style="color: Red; display: none">* Input digits (0 - 9)</span>

</form>
 <script type="text/javascript">
        var specialKeys = new Array();
        specialKeys.push(8); 
		specialKeys.push(43); 
		specialKeys.push(37); 
		specialKeys.push(39); 
		//Backspace
        function IsNumeric3(e) {
            var keyCode = e.which ? e.which : e.keyCode
            var ret = (keyCode != 37 && keyCode != 8 && keyCode != 46 && (keyCode >= 48 && keyCode <= 57) || specialKeys.indexOf(keyCode) != -1);
            document.getElementById("error3").style.display = ret ? "none" : "inline";
            return ret;
        }
    </script>

 <script>
function stringlength(inputtxt, minlength, maxlength)
{ 
var field = inputtxt.value; 
var mnlen = minlength;
var mxlen = maxlength;

if(field.length<mnlen || field.length> mxlen)
{ 
alert("Please input the 10 digit mobile number");
return false;
}
else
{ 

return true;
}
}
</script>
</body>
</html>

谢谢朋友们

Answer 5

以下是修改后的代码：

function isNumberKey(evt)
{
    var charCode = (evt.which) ? evt.which : event.keyCode
    if ( (charCode >= 48  && charCode <= 57) || charCode == 43) 
        return true;
    return false;
}

Answer 6

这是代码。工作正常与数字和加号+电话字段。你将不得不在keydown函数上实现代码。定位特定手机字段的ID /类并使用keydown函数。

    //allows only these keys
    // backspace, delete, tab, escape, and enter
    if ( event.keyCode == 107 || event.keyCode == 46 || event.keyCode == 8 || event.keyCode == 9 || event.keyCode == 27 || event.keyCode == 13 || 
         // Ctrl+A
        (event.keyCode == 65 && event.ctrlKey === true) || 
         // home, end, left, right
        (event.keyCode >= 35 && event.keyCode <= 39)) {
             return;
    }
    else {
        // Ensure that it is a number and stop the keypress
        if (event.shiftKey || (event.keyCode < 48 || event.keyCode > 57) && (event.keyCode < 96 || event.keyCode > 105 )) {
            event.preventDefault(); 
        }   
    }

Answer 7

利用上面同事的经验，编写一个适合我的功能。它会过滤除数字，箭头和退格之外的所有内容。也许对某人有用。

function isKeyCorrect(keyEvent) {
    var charCode = keyEvent.which ? keyEvent.which : keyEvent.keyCode;
    var isNotNumber = charCode < 48 || charCode > 57;
    var isNotArrow = charCode < 37 || charCode > 40;
    var isNotBackspace = charCode !== 8;

    return isNotNumber && isNotArrow && isNotBackspace;
}

Answer 8

<script type="text/javascript">
 $(document).ready(function() {
    `enter code here`  $('#form-1').submit(function(msg) {  
        $.post("action.php?act=login",$(this).serialize(),function(data){      
            if (data == 'ERR:A3001004') { alert("Güvenlik hatası, sayfayı yenileyin."); }
            else if (data == 'TIMEEND') { alert("Anahtarınızın süresi dolmuş.");    }
            else if (data == 'INVALID') { alert("Geçersiz anahtar şifresi girdiniz.");  }
            else if (data == 'OK') { alert("Başarıyla giriş yaptınız. Yetişkinlere göre içerik barındıran sitelere erişim sağlayabilirsiniz."); }
        });

        return false; 
    });
});
      function isNumberKey(evt)
      {
         var charCode = (evt.which) ? evt.which : event.keyCode
         if (charCode > 31 && (charCode < 48 || charCode > 57))
            return false;

         return true;
      }
</script>

JavaScript密钥代码仅允许数字和加号

8 个答案: