这个jQuery代码可以过滤掉按住Shift, Ctrl或Alt时键入的字符。

$('#AmountText').keydown(function (e) {
    if (e.shiftKey || e.ctrlKey || e.altKey) { // if shift, ctrl or alt keys held down
        e.preventDefault();         // Prevent character input
    } else {
        var n = e.keyCode;
        if (!((n == 8)              // backspace
        || (n == 46)                // delete
        || (n >= 35 && n <= 40)     // arrow keys/home/end
        || (n >= 48 && n <= 57)     // numbers on keyboard
        || (n >= 96 && n <= 105))   // number on keypad
        ) {
            e.preventDefault();     // Prevent character input
        }
    }
});

这是我用来验证整数或浮点值的数字输入(不显眼的jQuery风格):

$('input[name="number"]').keyup(function(e) { var float = parseFloat($(this).attr('data-float')); /* 2 regexp for validating integer and float inputs ***** > integer_regexp : allow numbers, but do not allow leading zeros > float_regexp : allow numbers + only one dot sign (and only in the middle of the string), but do not allow leading zeros in the integer part *************************************************************************/ var integer_regexp = (/[^0-9]|^0+(?!$)/g); var float_regexp = (/[^0-9\.]|^\.+(?!$)|^0+(?=[0-9]+)|\.(?=\.|.+\.)/g); var regexp = (float % 1 === 0) ? integer_regexp : float_regexp; if (regexp.test(this.value)) { this.value = this.value.replace(regexp, ''); } }); <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <input type="text" data-float="1" id="number" name="number" placeholder="integer"> <input type="text" data-float="0.1" id="number" name="number" placeholder="float">

我也想回答:)

    $('.justNum').keydown(function(event){
        var kc, num, rt = false;
        kc = event.keyCode;
        if(kc == 8 || ((kc > 47 && kc < 58) || (kc > 95 && kc < 106))) rt = true;
        return rt;
    })
    .bind('blur', function(){
        num = parseInt($(this).val());
        num = isNaN(num) ? '' : num;
        if(num && num < 0) num = num*-1;
        $(this).val(num);
    });

就是这样……只是数字。:)几乎只用“模糊”就可以了，但是……

我想帮点忙，我做了我的版本，onlyNumbers函数…

function onlyNumbers(e){
    var keynum;
    var keychar;

    if(window.event){  //IE
        keynum = e.keyCode;
    }
    if(e.which){ //Netscape/Firefox/Opera
        keynum = e.which;
    }
    if((keynum == 8 || keynum == 9 || keynum == 46 || (keynum >= 35 && keynum <= 40) ||
       (event.keyCode >= 96 && event.keyCode <= 105)))return true;

    if(keynum == 110 || keynum == 190){
        var checkdot=document.getElementById('price').value;
        var i=0;
        for(i=0;i<checkdot.length;i++){
            if(checkdot[i]=='.')return false;
        }
        if(checkdot.length==0)document.getElementById('price').value='0';
        return true;
    }
    keychar = String.fromCharCode(keynum);

    return !isNaN(keychar);
}

只需添加输入标签“…输入……id="price" onkeydown="return onlyNumbers(event)"…"然后你就完成了;)

需要确保你有数字键盘和tab键工作

 // Allow only backspace and delete
            if (event.keyCode == 46 || event.keyCode == 8  || event.keyCode == 9) {
                // let it happen, don't do anything
            }
            else {
                // Ensure that it is a number and stop the keypress
                if ((event.keyCode >= 48 && event.keyCode <= 57) || (event.keyCode >= 96 && event.keyCode <= 105)) {

                }
                else {
                    event.preventDefault();
                }
            }

为了更详细地说明#3的答案，我将执行以下操作(注意:仍然不支持通过键盘或鼠标粘贴操作):

$('#txtNumeric').keypress(
            function(event) {
                //Allow only backspace and delete
                if (event.keyCode != 46 && event.keyCode != 8) {
                    if (!parseInt(String.fromCharCode(event.which))) {
                        event.preventDefault();
                    }
                }
            }
        );