I try to find a regex that matches the string only if the string does not end with at least three '0' or more. Intuitively, I tried:
我试图找到一个匹配字符串的正则表达式，只有当该字符串至少以3 '0'结尾时才会匹配该字符串。凭直觉,我试着:
.*[^0]{3,}$
But this does not match when there one or two zeroes at the end of the string.
但是当字符串末尾有一两个0时，这个就不匹配了。
4 个解决方案
#1
13
If you have to do it without lookbehind assertions (i. e. in JavaScript):
如果您必须不使用lookbehind断言(即在JavaScript中):
^(?:.{0,2}|.*(?!000).{3})$
Otherwise, use hsz's answer.
否则,使用hsz的答案。
Explanation:
解释:
^ # Start of string
(?: # Either match...
.{0,2} # a string of up to two characters
| # or
.* # any string
(?!000) # (unless followed by three zeroes)
.{3} # followed by three characters
) # End of alternation
$ # End of string
#2
7
You can try using a negative look-behind, i.e.:
你可以试着用消极的目光看后面，例如:
(?<!000)$
Tests:
测试:
Test Target String Matches
1 654153640 Yes
2 5646549800 Yes
3 848461158000 No
4 84681840000 No
5 35450008748 Yes
Please keep in mind that negative look-behinds aren't supported in every language, however.
不过，请记住，并不是所有的语言都支持负面观点。
#3
2
What wrong with the no-look-behind, more general-purpose ^(.(?!.*0{3,}$))*$?
no-look-behind错什么,更多的通用^(。(? !。* 0 { 3 } $))* $ ?
The general pattern is ^(.(?!.* + not-ending-with-pattern + $))*$. You don't have to reverse engineer the state machine like Tim's answer does; you just insert the pattern you don't want to match at the end.
一般的模式是^(。(? !。* +无终止模式+ $)*$。你不需要像蒂姆的回答那样对状态机进行逆向设计;您只需插入不想在末尾匹配的模式。
#4
1
This is one of those things that RegExes aren't that great at, because the string isn't very regular (whatever that means). The only way I could come up with was to give it every possibility.
这是RegExes不太擅长的事情之一，因为字符串不是非常规则的(不管那是什么意思)。我能想到的唯一办法就是给它一切可能。
.*[^0]..$|.*.[^0].$|.*..[^0]$
which simplifies to
它简化了
.*([^0]|[^0].|[^0]..)$
That's fine if you only want strings not ending in three 0s, but strings not ending in ten 0s would be long. But thankfully, this string is a bit more regular than some of these sorts of combinations, and you can simplify it further.
这很好，如果你只需要字符串不以0结尾，但是字符串不会在10个0中结束会很长。但值得庆幸的是，这个字符串比这些组合中的一些更有规则性，你可以进一步简化它。
.*[^0].{0,2}$
#1
13
If you have to do it without lookbehind assertions (i. e. in JavaScript):
如果您必须不使用lookbehind断言(即在JavaScript中):
^(?:.{0,2}|.*(?!000).{3})$
Otherwise, use hsz's answer.
否则,使用hsz的答案。
Explanation:
解释:
^ # Start of string
(?: # Either match...
.{0,2} # a string of up to two characters
| # or
.* # any string
(?!000) # (unless followed by three zeroes)
.{3} # followed by three characters
) # End of alternation
$ # End of string
#2
7
You can try using a negative look-behind, i.e.:
你可以试着用消极的目光看后面，例如:
(?<!000)$
Tests:
测试:
Test Target String Matches
1 654153640 Yes
2 5646549800 Yes
3 848461158000 No
4 84681840000 No
5 35450008748 Yes
Please keep in mind that negative look-behinds aren't supported in every language, however.
不过，请记住，并不是所有的语言都支持负面观点。
#3
2
What wrong with the no-look-behind, more general-purpose ^(.(?!.*0{3,}$))*$?
no-look-behind错什么,更多的通用^(。(? !。* 0 { 3 } $))* $ ?
The general pattern is ^(.(?!.* + not-ending-with-pattern + $))*$. You don't have to reverse engineer the state machine like Tim's answer does; you just insert the pattern you don't want to match at the end.
一般的模式是^(。(? !。* +无终止模式+ $)*$。你不需要像蒂姆的回答那样对状态机进行逆向设计;您只需插入不想在末尾匹配的模式。
#4
1
This is one of those things that RegExes aren't that great at, because the string isn't very regular (whatever that means). The only way I could come up with was to give it every possibility.
这是RegExes不太擅长的事情之一，因为字符串不是非常规则的(不管那是什么意思)。我能想到的唯一办法就是给它一切可能。
.*[^0]..$|.*.[^0].$|.*..[^0]$
which simplifies to
它简化了
.*([^0]|[^0].|[^0]..)$
That's fine if you only want strings not ending in three 0s, but strings not ending in ten 0s would be long. But thankfully, this string is a bit more regular than some of these sorts of combinations, and you can simplify it further.
这很好，如果你只需要字符串不以0结尾，但是字符串不会在10个0中结束会很长。但值得庆幸的是，这个字符串比这些组合中的一些更有规则性，你可以进一步简化它。
.*[^0].{0,2}$