javascript 只允许输入文本上的数字、退格和删除键
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Only allow numbers, backspace and delete keys on input text
提问by VansFannel
I have this input text in a HTML5 page:
我在 HTML5 页面中有这个输入文本：
<input type="text" class="quantity"
onkeypress='return (event.charCode >= 48 && event.charCode <= 57) || event.charCode == 8 || event.charCode == 46' required />
Because I need an input text that does allow only numbers, but I could need also delete a number. When I press backspace or delete nothing happens.
因为我需要一个只允许数字的输入文本，但我可能还需要删除一个数字。当我按退格键或删除时，什么也没有发生。
The code above only allows numbers. How can I allow also backspace and delete?
上面的代码只允许数字。我怎样才能允许退格和删除？
回答by steo
keypresseventgives only output of letters code. Use keydowninstead.
keypressevent只给出字母代码的输出。使用keydown来代替。
The keypress event is fired when a key is pressed down and that key normally produces a character value (use input instead).
当一个键被按下并且该键通常会产生一个字符值时会触发 keypress 事件（使用 input 代替）。
<input type="text" class="quantity"
onkeydown='return (event.which >= 48 && event.which <= 57)
|| event.which == 8 || event.which == 46' required />
I'm using e.whichbecause keydownproduce it but, as the doc says, whichis deprecated and keyshould be used instead ( even if not fully implemented )
我正在使用e.which因为keydown生产它，但正如文档所说，which已被弃用，key应改为使用（即使没有完全实现）
Check out the Fiddleand keypress docs
回答by Johnny Blaze
this jquery function does it
这个 jquery 函数做到了
$('#id').keypress(function(e) {
var a = [46];
var k = e.which;
console.log( k );
a.push(8);
a.push(0);
for (i = 48; i < 58; i++)
a.push(i);
if (!($.inArray(k,a)>=0))
e.preventDefault();
});
回答by R4nc1d
You shouldn't use the keypress event, but the keyup or keydown event because the keypress event is intented for real (printable) characters. "keydown" is handled at a lower level so it will capture all non-printing keys like DEL and ENTER
您不应使用 keypress 事件，而应使用 keyup 或 keydown 事件，因为 keypress 事件旨在用于真实（可打印）字符。“keydown”在较低级别处理，因此它将捕获所有非打印键，如 DEL 和 ENTER
