PHP / Javascript RegExp - 非捕获组

I have three variations of a string:

我有三个字符串变体:

1. view=(edit:29,30)
2. view=(edit:29,30;)
3. view=(edit:29,30;x:100;y:200)

I need a RegExp that:

我需要一个RegExp:

1. capture up to and including ",30"

捕获到包括“,30”

2. capture "x:100;y:200" - whenever there's a semicolon after the first match;

捕获“x:100; y:200” - 只要在第一场比赛后有分号;

3. WILL NOT include leftmost semicolon in any of the groups;

不包括任何组中最左边的分号;

4. entire string on the right of the first semicolon and up to ')' can/should be in the same group.

第一个分号右边的整个字符串和最多')'可以/应该在同一个组中。

I came up with:

我提出了:

$pat = '/view=\((\w+)(:)([\d,]+)((;[^)]+){0,}|;)\)/';

Applied to 'view=(edit:29,30;x:100;y:200)' it yields:

应用于'view =(编辑:29,30; x:100; y:200)'它产生:

Array
(
    [0] => view=(edit:29,30;x:100;y:200)
    [1] => edit
    [2] => :
    [3] => 29,30
    [4] => ;x:100;y:200
    [5] => ;x:100;y:200
)

THE QUESTION. How do I remove ';' from matches [4] and [5]?

问题。如何删除';'来自比赛[4]和[5]?

IMPORTANT. The same RegExp should work with a string when no semicolons are present, as: 'view=(edit:29,30)'.

重要。当没有分号存在时,相同的RegExp应该使用字符串,如:'view =(edit:29,30)'。

$pat = '/view=\((\w+)(:)([\d,]+)((;[^)]+){0,}|;)\)/';
$str = 'view=(edit:29,30;x:100;y:200)';
preg_match($pat, $str, $m);
print_r($m);

Thanks!

2 个解决方案

#1

You don’t need to group everything. Try this regular expression:

您不需要对所有内容进行分组。试试这个正则表达式:

/view=\((\w+):([\d,]+)(?:;([^)]+)?)?\)/

#2

I guess you want something like this:

我想你想要这样的东西:

$pattern = '/view=\\((\\w+):(\\d+,\\d+)(?:;((?:\\w+:\\d+;?)*))?\\)/';

Should return

[0] view=(edit:29,30;x:100;y:200)
[1] edit
[2] 29,30
[3] x:100;y:200