Question

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"

可以使用一个preg_replace()来完成吗？

Answer 1

这样的事情：

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

模式细节：

~                    # 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 $

Answer 2

Preg_replace没有＆＃34; if＆＃34;有点可以说＆＃34;如果在开头和结尾都有引号。＆＃34;所以，你必须表达两种可能性：

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

|的左边匹配一行＆＃34;在开始和结束。 $ 1之间的文字。如果它不匹配，则$ 1为空。 |的右侧匹配其他任何东西，其中2美元就在那里。如果左侧匹配，它就不会向右侧移动。

使用preg_replace删除外部引号

2 个答案: