मेरे पास एक jsonFixer() फ़ंक्शन है - यह दो चरणों में काम करता है:कचरा हटाना (असंगत स्वरूपण की समानता के लिए) और सुधार करना।
<?php
function jsonFixer($json){
$patterns = [];
/** garbage removal */
$patterns[0] = "/([\s:,\{}\[\]])\s*'([^:,\{}\[\]]*)'\s*([\s:,\{}\[\]])/"; //Find any character except colons, commas, curly and square brackets surrounded or not by spaces preceded and followed by spaces, colons, commas, curly or square brackets...
$patterns[1] = '/([^\s:,\{}\[\]]*)\{([^\s:,\{}\[\]]*)/'; //Find any left curly brackets surrounded or not by one or more of any character except spaces, colons, commas, curly and square brackets...
$patterns[2] = "/([^\s:,\{}\[\]]+)}/"; //Find any right curly brackets preceded by one or more of any character except spaces, colons, commas, curly and square brackets...
$patterns[3] = "/(}),\s*/"; //JSON.parse() doesn't allow trailing commas
/** reformatting */
$patterns[4] = '/([^\s:,\{}\[\]]+\s*)*[^\s:,\{}\[\]]+/'; //Find or not one or more of any character except spaces, colons, commas, curly and square brackets followed by one or more of any character except spaces, colons, commas, curly and square brackets...
$patterns[5] = '/["\']+([^"\':,\{}\[\]]*)["\']+/'; //Find one or more of quotation marks or/and apostrophes surrounding any character except colons, commas, curly and square brackets...
$patterns[6] = '/(")([^\s:,\{}\[\]]+)(")(\s+([^\s:,\{}\[\]]+))/'; //Find or not one or more of any character except spaces, colons, commas, curly and square brackets surrounded by quotation marks followed by one or more spaces and one or more of any character except spaces, colons, commas, curly and square brackets...
$patterns[7] = "/(')([^\s:,\{}\[\]]+)(')(\s+([^\s:,\{}\[\]]+))/"; //Find or not one or more of any character except spaces, colons, commas, curly and square brackets surrounded by apostrophes followed by one or more spaces and one or more of any character except spaces, colons, commas, curly and square brackets...
$patterns[8] = '/(})(")/'; //Find any right curly brackets followed by quotation marks...
$patterns[9] = '/,\s+(})/'; //Find any comma followed by one or more spaces and a right curly bracket...
$patterns[10] = '/\s+/'; //Find one or more spaces...
$patterns[11] = '/^\s+/'; //Find one or more spaces at start of string...
$replacements = [];
/** garbage removal */
$replacements[0] = '$1 "$2" $3'; //...and put quotation marks surrounded by spaces between them;
$replacements[1] = '$1 { $2'; //...and put spaces between them;
$replacements[2] = '$1 }'; //...and put a space between them;
$replacements[3] = '$1'; //...so, remove trailing commas of any right curly brackets;
/** reformatting */
$replacements[4] = '"$0"'; //...and put quotation marks surrounding them;
$replacements[5] = '"$1"'; //...and replace by single quotation marks;
$replacements[6] = '\\$1$2\\$3$4'; //...and add back slashes to its quotation marks;
$replacements[7] = '\\$1$2\\$3$4'; //...and add back slashes to its apostrophes;
$replacements[8] = '$1, $2'; //...and put a comma followed by a space character between them;
$replacements[9] = ' $1'; //...and replace by a space followed by a right curly bracket;
$replacements[10] = ' '; //...and replace by one space;
$replacements[11] = ''; //...and remove it.
$result = preg_replace($patterns, $replacements, $json);
return $result;
}
?>
उपयोग का उदाहरण:
<?php
// Received badly formatted json:
// {"contact1": "David "Dave" Letterman", price : 30.00, 'details' : "Greatest 'Hits' Album"}
$json_string = '{"contact1": "David "Dave" Letterman", price : 30.00, \'details\' : "Greatest \'Hits\' Album"}';
jsonFixer($json_string);
?>
परिणाम होगा:
{"contact1": "David \"Dave\" Letterman", "price" : "30.00", "details" : "Greatest \'Hits\' Album"}
नोट:यह सभी संभावित रूप से खराब स्वरूपित JSON स्ट्रिंग्स के साथ परीक्षण नहीं किया गया था, लेकिन मैं एक जटिल बहु स्तरीय JSON स्ट्रिंग पर उपयोग करता हूं और तब तक अच्छी तरह से काम कर रहा है।
