字符串之间的Preg_Replace字符

Question

Im trying to write the most efficient way to escape double quote marks (") from a a json feed that contains quote marks in incorrect places.

ie

{ "count": "1", "query": "www.mydomain.com/watchlive/type/livedvr/event/69167/"%20%20sTyLe=X:eX/**/pReSsIoN(window.location=56237)%20"", "error": "500"}

there are three keys above - count, query and error. The value in "query" is invalid as the extra double quotes are rendering a invalid json.

If I escape it using \" then the json is valid and can be parsed by the PHP engine, but since the json can have over 5000 sets of data, I cant just manually go and change the offending line(s).

I know that using a combination of preg_match and str_replace will work, but its very messy and not maintainable code. I need the reg_ex to use in something like this

$buffer = '{ "count": "1", "query": "www.mydomain.com/watchlive/type/livedvr/event/69167/"%20%20sTyLe=X:eX/**/pReSsIoN(window.location=56237)%20"", "error": "500"}'

preg_match('/(query": ")(.*)(", "error)/', $buffer , $match);

Thanks in advance

Answer 1

Match and replace using this expression:

(?:"query"\s*:\s*"|(?<!\A)\G)[^"]*\K"(?=.*?",)
\"

In PHP, this would use preg_replace():

$buffer = preg_replace('/(?:"query"\s*:\s*"|(?<!\A)\G)[^"]*\K"(?=.*?",)/', '\"', $buffer);
var_dump($buffer);

---

Explanation:

(?:                # Start non-capturing group
  "query"\s*:\s*"  # Match "query":" literally, with optional whitespace  
 |                 # OR
  (?<!\A)          # Make sure we are not at the beginning of the string
  \G               # Start at the end of last match
)                  # End non-capturing
[^"]*              # Go through non-" characters
\K                 # Remove everything to the left from the match
"                  # Match " (this will be the only thing matched and replaced)
(?=                # Start lookahead group
  .*?",            # Lazily match up until the ", (this is the end of the JSON value)
)                  # End lookahead group