Regex在一个可选字符串上匹配non-greedy而在另一个字符串上匹配greedy

Question

我研究了一段时间，还没有找到匹配以下模式的线索（不过，我对regex也很陌生），它看起来像

/abc/foo/bar(/*) 

或

/abc/foo/bar/stop

所以我想将上面的字符串匹配并捕获为/abc/foo/bar。现在，“/stop”是一个可选字符串，可以附加在模式的末尾。目标是获得所需的捕获，同时忽略“stop”（如果“stop”存在多次，则在第一个“stop”处停止），同时允许在中间设置尽可能多的斜线（行尾的斜线除外）。

^(/.*[^/])/*$

/abc/foo/bar
/abc/foo/bar/
/abc/foo/bar///
/abc/foo/bar/stop
/abc/foo/bar/stop/foo/bar/stop/stop
/abc/foo/bar//stop

但它与以下任何项都不匹配：

/abc/foo/bar/sto (will match the whole "/abc/foo/bar/sto" instead)
/abc/foo/bar/abc/foo/bar (it will catch "/abc/foo/bar/abc/foo/bar" instead)

让我知道这是否足够清楚。谢谢！

Answer 1

试试看：

/^(?:\/+(?!$|(?:stop\/?))[^\/]+)*/

Regex101演示

说明：

null

/^.+?(?=\/*$|\/+stop\b)/

null

var re = /^.+?(?=\/*$|\/+stop\b)/,
    test_strings = ["/abc/foo/bar",
                    "/abc/foo/bar/",
                    "/abc/foo/bar///",
                    "/abc/foo/bar/stop",
                    "/abc/foo/bar/stop/foo/bar/stop/stop",
                    "/abc/foo/bar//stop",
                    "/abc/foo/bar/sto",
                    "/abc/foo/bar/abc/foo/bar"];
for(var s = 0; s < test_strings.length; s++) {
    console.log(test_strings[s].match(re)[0]);
}

/*
Results:

/abc/foo/bar
/abc/foo/bar
/abc/foo/bar
/abc/foo/bar
/abc/foo/bar
/abc/foo/bar
/abc/foo/bar/sto
/abc/foo/bar/abc/foo/bar 

*/