[英]regex ignore capturing group inside capturing group
I'm trying to capture a group but at the same time ignore a sub group but need the subgroup to be true to capture it. 我正在尝试捕获一个组,但同时忽略一个子组但需要该子组才能捕获它。 For example: 例如:
/^(?:\s*)([abc])?(?:\s*)((?:d){1}[ef]*)(.*)/.exec(' a deee ghi')
// is equal to [" a deee ghi", "a", "deee", " ghi"]
// but I really want equal to [" a deee ghi", "a", "eee", " ghi"]
// and I want this to fail:
/^(?:\s*)([abc])?(?:\s*)((?:d){1}[ef]*)(.*)/.exec(' a eee ghi')
// [" a eee ghi", "a", "", "eee ghi"]
I'm trying to ignore the 'd'
in the capturing group but only want it captured if there's a 'd'
before any 'e'
s of 'f'
s in the third group 我试图忽略捕获组中的'd'
,但只想在第三组'f'
的任何'e'
之前有'd'
时捕获它
How would I do this? 我该怎么做?
EDIT: 编辑:
To clairify, I want the third group to be filled only if there's a 'd'
in the string there. 为了要求,我希望第三组只有在字符串中有'd'
才能填充。 else -> don't capture the 'e'
s or 'f'
s 否则 - >不要捕捉'e'
或'f'
If I understand you correctly: if the d
is there, then the capture-group should be eee
, but if the d
is not there, then the capture-group should be blank or null. 如果我理解正确:如果d
在那里,那么捕获组应该是eee
,但如果d
不存在,那么捕获组应该为空或空。 Is that correct? 那是对的吗? If so, you can write: 如果是这样,你可以写:
/^(?:\s*)([abc])?(?:\s*)(?:d([ef]*)|[ef]*)(.*)/.exec(' a deee ghi')
to match ([ef]*)
when d
is present, and [ef]*
when it is not. 匹配([ef]*)
当d
是存在,并且[ef]*
时,它不是。
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