正则表达式问题：将此模式与硬引号或软引号匹配-RegexQuestion:Matchingthispatternwithhardorsoftquotes

Here's another approach:

这是另一种方法:

$rgx='~name()|\w+)=(?|"([^"]*)"|\'([^\']*)\'))+?\1~i';

I know this question is old, but when it resurfaced just now I thought up another use for the "empty capturing groups as checkboxes" idiom from the Cookbook. The first, non-capturing group handles the matching of all "name=value" pairs under the control of a reluctant plus (+?). If the attribute name is literally name, the empty group (()) matches nothing, then the backreference (\1) matches nothing again, breaking out of the loop. (The backreference succeeds because the group participated in the match, even though it didn't consume any characters.)

我知道这个问题已经过时了,但是当它刚刚重新浮出水面时,我想到了另一个用于“空白捕获组作为复选框”的习惯用法。第一个非捕获组在不情愿的加号(+?)的控制下处理所有“名称=值”对的匹配。如果属性名称实际上是名称,则空组(())不匹配任何内容,然后反向引用(\ 1)再次匹配任何内容,从而突破循环。 (后向引用成功,因为该组参与了匹配,即使它没有消耗任何字符。)

The attribute value is captured each time in group #2, overwriting whatever was captured on the previous iteration. (The branch-reset construct ((?|(...)|(...)) enables us to "re-use" group #2 to capture the value inside the quotes, whichever kind of quotes they were.) Since the loop quits after the name name comes up, the final captured value corresponds to that attribute.

每次在组#2中捕获属性值,覆盖在先前迭代中捕获的任何内容。 (branch-reset构造((?|(...)|(...))使我们能够“重用”组#2来捕获引号内的值,无论它们是什么类型的引号。)名称出现后循环退出,最终捕获的值对应于该属性。

See a demo on Ideone

查看Ideone上的演示