Writing JSON object to .json file on server

Question

I'm trying to write my JSON object to a .json file on the server. The way I'm doing this now is:

JavaScript:

function createJsonFile() {
 var jsonObject = {
 "metros" : [],
 "routes" : []
 };
 // write cities to JSON Object
 for ( var index = 0; index < graph.getVerticies().length; index++) {
 jsonObject.metros[index] = JSON.stringify(graph.getVertex(index).getData());
 }
 // write routes to JSON Object
 for ( var index = 0; index < graph.getEdges().length; index++) {
 jsonObject.routes[index] = JSON.stringify(graph.getEdge(index));
 }
 // some jQuery to write to file
 $.ajax({
 type : "POST",
 url : "json.php",
 dataType : 'json',
 data : {
 json : jsonObject
 }
 });
};

PHP:

<?php
 $json = $_POST['json'];
 $info = json_encode($json);
 $file = fopen('new_map_data.json','w+');
 fwrite($file, $info);
 fclose($file);
?>

It is writing fine and the information seems to be correct, but it is not rendering properly. It is coming out as:

{"metros":["{\\\"code\\\":\\\"SCL\\\",\\\"name\\\":\\\"Santiago\\\",\\\"country\\\":\\\"CL\\\",\\\"continent\\\":\\\"South America\\\",\\\"timezone\\\":-4,\\\"coordinates\\\":{\\\"S\\\":33,\\\"W\\\":71},\\\"population\\\":6000000,\\\"region\\\":1}",

... but I'm expecting this:

"metros" : [
 {
 "code" : "SCL" ,
 "name" : "Santiago" ,
 "country" : "CL" ,
 "continent" : "South America" ,
 "timezone" : -4 ,
 "coordinates" : {"S" : 33, "W" : 71} ,
 "population" : 6000000 ,
 "region" : 1
 } ,

Why am I getting all of these slashes and why it is all on one line?

• RobertPitt: Are you using a framework or performing any sanitization in POST before hand, try turning magic_qoutes of in php ini and restart.

Answer 1

You are double-encoding. There is no need to encode in JS and PHP, just do it on one side, and just do it once.

// step 1: build data structure
var data = {
 metros: graph.getVerticies(),
 routes: graph.getEdges()
}
// step 2: convert data structure to JSON
$.ajax({
 type : "POST",
 url : "json.php",
 data : {
 json : JSON.stringify(data)
 }
});

Note that the dataType parameter denotes the expected response type, not the the type you send the data as. Post requests will be sent as application/x-www-form-urlencoded by default.

I don't think you need that parameter at all. You could trim that down to:

$.post("json.php", {json : JSON.stringify(data)});

Then (in PHP) do:

<?php
 $json = $_POST['json'];
 /* sanity check */
 if (json_decode($json) != null)
 {
 $file = fopen('new_map_data.json','w+');
 fwrite($file, $json);
 fclose($file);
 }
 else
 {
 // user has posted invalid JSON, handle the error 
 }
?>

• Francesco: should'nt be fwrite($file, $json); ??

Answer 2

Don't JSON.stringify. You get a double JSON encoding by doing that.

You first convert your array elements to a JSON string, then you add them to your full object, and then you encode your big object, but when encoding the elements already encoded are treated as simple strings so all the special chars are escaped. You need to have one big object and encode it just once. The encoder will take care of the children.

For the on row problem try sending a JSON data type header: Content-type: text/json I think (didn't google for it). But rendering will depend only on your browser. Also it may be possible to encode with indentation.

• Hristo: I guess that worked... I thought I need to do JSON.stringify so that it formats my object properly? I defined a toJSON function for my object so that is why i was using stringify. I took it out and it got rid of the \\\ but it is still all on one line...
• Tomalak: @Hristo: What's wrong with all of it being on one line? I don't think you should care.
• Alin P.: He may need to check if while in development and this might help him. If it's just for storing there's nothing wrong.
• Tomalak: jsonlint.com does that better than visually scanning a formatted string.

Answer 3

Probably too late to answer the question. But I encountered the same issue. I resolved it by using "JSON_PRETTY_PRINT"

Following is my code:

<?php
if(isset($_POST['object'])) {
 $json = json_encode($_POST['object'],JSON_PRETTY_PRINT);
 $fp = fopen('results.json', 'w');
 fwrite($fp, $json);
 fclose($fp);
} else {
 echo "Object Not Received";
}
?>

Answer 4

<html>
<head>
 <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" integrity="sha384-BVYiiSIFeK1dGmJRAkycuHAHRg32OmUcww7on3RYdg4Va+PmSTsz/K68vbdEjh4u" crossorigin="anonymous">
 <script type="text/javascript" src="http://code.jquery.com/jquery-1.4.3.min.js" ></script>
</head>
<body>
 <?php
 $str = file_get_contents('data.json');//get contents of your json file and store it in a string
 $arr = json_decode($str, true);//decode it
 $arrne['name'] = "sadaadad";
 $arrne['password'] = "sadaadad";
 $arrne['nickname'] = "sadaadad";
 array_push( $arr['employees'], $arrne);//push contents to ur decoded array i.e $arr
 $str = json_encode($arr);
 //now send evrything to ur data.json file using folowing code
 if (json_decode($str) != null)
 {
 $file = fopen('data.json','w');
 fwrite($file, $str);
 fclose($file);
 }
 else
 {
 // invalid JSON, handle the error 
 }
 ?>
 <form method=>
</body>

data.json

{ 
 "employees":[ 
 { 
 "email":"11BD1A05G9",
 "password":"INTRODUCTION TO ANALYTICS",
 "nickname":4
 },
 { 
 "email":"Betty",
 "password":"Layers",
 "nickname":4
 },
 { 
 "email":"Carl",
 "password":"Louis",
 "nickname":4
 },
 { 
 "name":"sadaadad",
 "password":"sadaadad",
 "nickname":"sadaadad"
 },
 { 
 "name":"sadaadad",
 "password":"sadaadad",
 "nickname":"sadaadad"
 },
 { 
 "name":"sadaadad",
 "password":"sadaadad",
 "nickname":"sadaadad"
 }
 ]
 }

• sudheer: evrything you need in one file,this is the most inefficient code i've ever writeen in my life ,but it solves ur problem!