如何丢弃正则表达式组后的所有字符
我只想允许 2 位正双精度数。
我得到了这个，它正在工作，但是在我成功匹配模式后，它会重新开始，而不是丢弃传入的字符。
([1-9]+[\d]*(\.|,)?[\d]{0,2})
Input
12.34 //Pass Filter
12.340 //Rejected - Zero cannot be the first character
12.345 //Pass Filter - It pass, but it is supposed to be rejected since it has 3 decimal digits
I want to only allow a positive double with 2 digits.
I got this, which is working, but after I successfully match the pattern it starts over again instead of discarding incoming characters.
([1-9]+[\d]*(\.|,)?[\d]{0,2})
Input
12.34 //Pass Filter
12.340 //Rejected - Zero cannot be the first character
12.345 //Pass Filter - It pass, but it is supposed to be rejected since it has 3 decimal digits
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如果您不想部分匹配，可以锚定字符串，并将零和逗号放入字符类中。
如果您想匹配
.或,零次或 1 次，您可以使用?而不是*请参阅正则表达式演示。
If you don't want partial matches, you can anchor the string, and put the zero and comma in a character class.
If you want to match the
.or,zero or 1 times you can use?instead of*See a regex demo.