我想匹配一个小于或等于 100 的数字,它可以是 0-100 之间的任何数字,但正则表达式不应匹配大于 100 的数字,例如 120、130、150、999 等。
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7 回答
55
尝试这个
\b(0*(?:[1-9][0-9]?|100))\b
解释
"
\b # Assert position at a word boundary
( # Match the regular expression below and capture its match into backreference number 1
0 # Match the character “0” literally
* # Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
(?: # Match the regular expression below
# Match either the regular expression below (attempting the next alternative only if this one fails)
[1-9] # Match a single character in the range between “1” and “9”
[0-9] # Match a single character in the range between “0” and “9”
? # Between zero and one times, as many times as possible, giving back as needed (greedy)
| # Or match regular expression number 2 below (the entire group fails if this one fails to match)
100 # Match the characters “100” literally
)
)
\b # Assert position at a word boundary
"
于 2012-06-13T09:26:57.457 回答
6
我的实用建议。
就个人而言,我会完全避免编写如此复杂的正则表达式。如果您的号码在不久的将来从 100 变为 200 怎么办。您的正则表达式将不得不进行重大更改,并且可能更难编写。上述所有解决方案都不是自我解释的,您必须在代码中添加注释来补充它。那是一种气味。
可读性很重要。代码是给人类的,而不是给机器的。
为什么不围绕它编写一些代码并保持正则表达式简单易懂。
于 2016-09-19T19:29:11.727 回答
4
如果您需要正则表达式(最终),请使用代码断言:
/^(.+)$(??{$^N>=0 && $^N<=100 ? '':'^'})/
测试:
my @nums = (-1, 0, 10, 22, 1e10, 1e-10, 99, 101, 1.001e2);
print join ',', grep
/^(.+)$(??{$^N>=0 && $^N<=100 ? '':'^'})/,
@nums
结果:
0,10,22,1e-010,99
(==>这是了解代码断言的东西)。
于 2012-06-13T13:27:00.283 回答
1
此正则表达式匹配数字 0-100 音调并禁止数字 001:
\b(0|[1-9][0-9]?|100)\b
于 2014-11-18T05:59:47.667 回答
1
正则表达式
perl -le 'for (qw/0 1 19 32.4 100 77 138 342.1/) { print "$_ is ", /^(?:100|\d\d?)$/ ? "valid input" : "invalid input"}'
于 2012-06-13T09:19:23.550 回答
1
这匹配 0 到 100
^0*([0-9]|[1-8][0-9]|9[0-9]|100)$
于 2017-02-06T12:45:09.993 回答