我经常对诽谤作如下表述:
/^(d{1,3})(.{0,1}d{0,2})$/
但允许“1”
How could I fix this?
下面的缩略语为1-3位数,可选择后加1位数和1-2位数。
/^(d{1,3})(.d{1,2})?$/
请注意,我也修改了。 它是一条与任何th相匹配的果园,因此必须逃脱。
/^(d{1,3})(.d{1,2})?$/
I assume you re trying to match between 0 and 999.99, with 0, 1, or 2 decimals. If there are no decimals, you want no period separator. If that s the case, you want to the above.
如果你不想在编号一开始就设一个<0>>>>,你可以使用以下表述:
(d*.)?d+
这与:
1
.1
1.1
This will not match:
1.
.
在作一些修改后,这一表述仍可用于强迫一定数量的数位数:
(d{1,3}.)?d{1,2}
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