Php 用preg_replace删除外部引号

我的文件中有以下行：

normal line    
"line in quotes"
line with "quoted" word
line with quotes word at "end"
"quoted line with quoted word at "end""

我只需要删除外部引号

结果:

normal line    
line in quotes
line with "quoted" word
line with quotes word at "end"
quoted line with quoted word at "end"

这可以用一个

preg\u replace（）

来完成吗？

类似这样的东西：

$result = preg_replace('~^"((?:[^"\r\n]+|"[^\r\n"]*")*+)"$~m', '$1', $text);

图案详情：

~                    # pattern delimiter
^                    # anchor for the begining of the line
"                    #
(                    # open the capture group 1
    (?:              # non-capturing group, content of a line between quotes:
        [^"\r\n]+    #    - all that is not a quote or a newline
      |              # OR
        "[^\r\n"]*"  #    - a substring between quotes
    )*+              # repeat the group zero or more times
)                    # close the capture group
"                    #
$                    # anchor for the end of the line
~m                   # multiline modifier to change ^ and $ 

---

Preg_replace没有“如果”之类的词来表示“如果开头和结尾都有引用”。因此，您必须表达两种可能性：

$r = preg_replace('/^"(.*)"$|^(.*)$/', '$1$2', $b);

|的左边匹配一行，开头和结尾都是“1”。如果不匹配，$1为空。|的右边匹配任何其他内容，其中$2为内容。如果左边匹配，则不会转到右边