javascript 数字和一位小数的正则表达式
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Regular expression for numbers and one decimal
提问by Matthew Meppiel
I can't seem to get a simple regular expression to work. Here's what I have at the moment:
我似乎无法让一个简单的正则表达式起作用。这是我目前所拥有的：
$(".Hours").on('input', function (e) {
var regex = /^\d+(\.\d{0,2})?$/g;
if (!regex.test(this.value)) {
if (!regex.test(this.value[0]))
this.value = this.value.substring(1, this.value.length);
else
this.value = this.value.substring(0, this.value.length - 1);
}
});
I need the user to be able to only enter numbers and one decimal (with only two numbers after the decimal). It's working properly now, with the exception that a user cannot start with a decimal.
我需要用户只能输入数字和一位小数（小数点后只有两个数字）。它现在工作正常，除了用户不能以小数开头。
Acceptable: 23.53 0.43 1111.43 54335.34 235.23 .53 <--- Not working
可接受：23.53 0.43 1111.43 54335.34 235.23 .53 <--- 不工作
Unacceptable: 0234.32 <--- The user can currently do this 23.453 1.343 .234.23 1.453.23
不可接受：0234.32 <--- 用户目前可以这样做 23.453 1.343 .234.23 1.453.23
Any help on this?
有什么帮助吗？
回答by Tushar Gupta - curioustushar
RegExp -
正则表达式 -
^(\d+)?([.]?\d{0,2})?$
Explanation
解释
Assert position at the beginning of the string ?^?
Match the regular expression below and capture its match into backreference number 1 ?(\d+)??
Between zero and one times, as many times as possible, giving back as needed (greedy) ???
Match a single digit 0..9 ?\d+?
Between one and unlimited times, as many times as possible, giving back as needed (greedy) ?+?
Match the regular expression below and capture its match into backreference number 2 ?([.]?\d{0,2})??
Between zero and one times, as many times as possible, giving back as needed (greedy) ???
Match the character “.” ?[.]??
Between zero and one times, as many times as possible, giving back as needed (greedy) ???
Match a single digit 0..9 ?\d{0,2}?
Between zero and 2 times, as many times as possible, giving back as needed (greedy) ?{0,2}?
Assert position at the end of the string (or before the line break at the end of the string, if any) ?$?
回答by leaf
Here is a suggestion : /^((\d|[1-9]\d+)(\.\d{1,2})?|\.\d{1,2})$/.
这里有一个建议：/^((\d|[1-9]\d+)(\.\d{1,2})?|\.\d{1,2})$/。
Allows : 0, 0.00, 100, 100.1, 100.10, .1, .10...
允许：0, 0.00, 100, 100.1, 100.10, .1, .10...
Rejects : 01, 01.1, 100., .100, ....
拒绝：01, 01.1, 100., .100, ....
回答by katosh
Would this meet your needs:
这是否满足您的需求：
var regex = /^\d+([.]?\d{0,2})?$/g;
回答by Brian Stephens
Your regex:
var regex = /^\d+(\.\d{0,2})?$/g;
你的正则表达式：
var regex = /^\d+(\.\d{0,2})?$/g;
What you need:
var regex = /^\d*(\.\d{1,2})?$/;
你需要什么：
var regex = /^\d*(\.\d{1,2})?$/;
You were requiring at least one digit before the decimal (\d+). I have also changed it so if you include a decimal, there must be at least one digit after it.
您要求小数点 ( \d+)前至少有一位数字。我也改变了它，所以如果你包含一个小数，它后面必须至少有一个数字。
