转义除标签属性之外的匹配引号

Question

我想转义除标签属性中的匹配引号之外的匹配引号，例如：

输入：

xyz <test foo='123 abc' bar="def 456"> f00 'escape me' b4r "me too" but not this </tEsT> blah 'escape " me'

预期输出：

xyz <test foo='123 abc' bar="def 456"> f00 \'escape me\' b4r \"me too\" but not this </tEsT> blah \'escape " me\'

我有以下正则表达式：

$result = preg_replace('/(([\'"])((\\\2|.)*?)\2)/', "\\\\$2$3\\\\$2", $input);

返回：

xyz <test foo=\'123 abc\' bar=\"def 456\"> f00 \'escape me\' b4r \"me too\" but not this </tEsT> blah \'escape " me\'

现在我想使用正则表达式零宽度负向后查找来跳过具有等号的匹配引号前面：

$result = preg_replace('/((?<=[^=])([\'"])((\\\2|.)*?)\2)/', "\\\\$2$3\\\\$2", $input);

但结果仍然不符合预期：

xyz <test foo='123 abc\' bar="def 456"> f00 \'escape me\' b4r "me too" but not this </tEsT> blah \'escape " me'

您能给我建议如何跳过整个不需要的块（“blah blah blah”）而不是只跳过第一个引用吗？

Answer 1

不要向后看来建立背景，而要向前看。通常会容易得多。

$result = preg_replace('/([\'"])(?![^<>]*>)((?:(?!\1).)*)\1/',
                       '\\\\$1$2\\\\$1',
                        $subject);

(['"])            # capture the open quote
(?![^<>]*>)       # make sure it's not inside a tag
(                 # capture everything up to the next quote
  (?:             # ...after testing each character to
    (?!\1|[<>]).  # ...to be sure it's not the opening quote
  )*              # ...or an angle bracket
)
\1                # match another quote of the same type as the first one

我假设属性值中不会有任何尖括号。

Answer 2

这是另一张。

$str = "xyz <test foo='123 abc' bar=\"def 456\"> f00 'escape me' b4r \"me too\" but not this <br/> <br/></tEsT> blah 'escape \" me'";

$str_escaped = preg_replace_callback('/(?<!\<)[^<>]+(?![^<]*\>)/','escape_quotes',$str);
// check all the strings outside every possible tag
// and replace each by the return value of the function below

function escape_quotes($str) {
    if (is_array($str)) $str = $str[0];
    return preg_replace('/(?<!\\\)(\'|")/','\\\$1',$str);
    // escape all the non-escaped single and double quotes
    // and return the escaped block
}