Javascript 只允许英文字符和数字进行文本输入
声明:本页面是StackOverFlow热门问题的中英对照翻译,遵循CC BY-SA 4.0协议,如果您需要使用它,必须同样遵循CC BY-SA许可,注明原文地址和作者信息,同时你必须将它归于原作者(不是我):StackOverFlow
原文地址: http://stackoverflow.com/questions/7144167/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
only allow English characters and numbers for text input
提问by Gereltod
Live Demo: http://jsfiddle.net/thisizmonster/DveuB/
现场演示:http: //jsfiddle.net/thisizmonster/DveuB/
How can I change this so that the input only allows the characters A-Z, a-z, 0-9 while typing, without using a regular expression?
如何更改此设置,以便在输入时只允许输入字符 AZ、az、0-9,而不使用正则表达式?
回答by Paul
Assuming you also want to accept spaces:
假设您还想接受空格:
$("#user").keypress(function(event){
var ew = event.which;
if(ew == 32)
return true;
if(48 <= ew && ew <= 57)
return true;
if(65 <= ew && ew <= 90)
return true;
if(97 <= ew && ew <= 122)
return true;
return false;
});
If you don't want to accept spaces then remove the if(ew == 32) return true;
如果您不想接受空格,请删除 if(ew == 32) return true;
回答by Siya Qalistic
<input type="text" id="firstName" onkeypress="return (event.charCode >= 65 && event.charCode <= 90) || (event.charCode >= 97 && event.charCode <= 122) || (event.charCode >= 48 && event.charCode <= 57)" />
The ASCII Character Set : https://www.w3schools.com/charsets/ref_html_ascii.asp
ASCII 字符集:https: //www.w3schools.com/charsets/ref_html_ascii.asp
回答by 321X
You can do something like this: http://jsfiddle.net/DveuB/1/
你可以这样做:http: //jsfiddle.net/DveuB/1/
Then it's only 0-9, a-z and A-Z
然后就只有0-9了,az和AZ
$(function(){
$("#user").keypress(function(event){
if ((event.charCode >= 48 && event.charCode <= 57) || // 0-9
(event.charCode >= 65 && event.charCode <= 90) || // A-Z
(event.charCode >= 97 && event.charCode <= 122)) // a-z
alert("0-9, a-z or A-Z");
});
});
Update:http://jsfiddle.net/DveuB/4/
To prevent what @mu is talking about:
更新:http : //jsfiddle.net/DveuB/4/
为了防止@mu 在说什么:
$("#user").keyup(function(event){
if (event.altKey == false && event.ctrlKey == false)
if ((event.keyCode >= 48 && event.keyCode <= 57 && event.shiftKey== false) ||
(event.keyCode >= 65 && event.keyCode <= 90) ||
(event.keyCode >= 97 && event.keyCode <= 122))
alert("0-9, a-z or A-Z");
});
回答by SajithG
<html>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
$(document).ready(function () {
$('.clsAlphaNoOnly').keypress(function (e) { // Accept only alpha numerics, no special characters
var regex = new RegExp("^[a-zA-Z0-9 ]+$");
var str = String.fromCharCode(!e.charCode ? e.which : e.charCode);
if (regex.test(str)) {
return true;
}
e.preventDefault();
return false;
});
})
</script>
<body>
<input class='clsAlphaNoOnly' type='text'>
</body>
</html>
回答by Vladimir Ishenko
I'm using this function.
我正在使用这个功能。
By modifying RegExp you can get rid of any characters you don't like personally.
通过修改 RegExp,您可以删除您个人不喜欢的任何字符。
$(function(){
$("#user").bind('keypress',function(e){
var regex = new RegExp("^[a-zA-Z0-9 ]+$");
var str = String.fromCharCode(!e.charCode ? e.which : e.charCode);
if (regex.test(str)) return true;
e.preventDefault();
return false;
});
});
回答by Bas Slagter
Think you have to write some if statements or something like that, that takes the keycode and validates it against some numbers that represent other keycodes:
认为您必须编写一些 if 语句或类似的东西,它采用键码并根据代表其他键码的一些数字对其进行验证:
for example:
if(keycode < 80)
return false;
例如:
if(keycode < 80)
return false;