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JSON Parsing

Posted on 2011-03-10
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Last Modified: 2013-11-08
Hello, am having some difficulty parsing the following JSON:
{
    "references": [{
        "type": "user",
        "full_name": "Jim Bo Bob",
        "id": 5,
        "url": "https://yahoo.com",
    }, {
        "type": "user",
        "full_name": "Bob Joe Jim",
        "id": 11,
        "url": "https://google.com",
    }, {
        "type": "topic",
        "name": "Muscular Arms",
        "id": 28
    }, {
        "voted_user_id": 5,
        "type": "approved",
        "approved_user_id": 7,
        "topic_id": 28,
        "comment": "His arms are massive",
        "id": 297
    }, {
        "voted_user_id": 11,
        "type": "approved",
        "approved_user_id": 7,
        "topic_id": 28,
        "comment": "Very moist",
        "id": 89
    }]
}

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I need to be able to loop through each object and determine when references.id == references.voter_user_id.  For example, parsing through the JSON above, I should find a match between the first object (references.id = 5) and the fourth object (references.voter_user_id: 5).

Am currently using the following jquery js to loop:

$.each(data.references, function (i, ref) {
	if(ref.id == ref.voter_user_id) {
		// do stuff
	}
});

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That seems to check *within* each object whether ref.id == ref.voter_user_id and this is not what I want.
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Question by:aznprncritic
  • 3
5 Comments
 
LVL 82

Expert Comment

by:leakim971
ID: 35094460
voted_user_id versus
voter_user_id

no ?
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Author Comment

by:aznprncritic
ID: 35094544
Sorry, must have been a typo.  The problem is my js is checking *within* each object for a match between ref.id and ref.voted_user_id.
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LVL 82

Accepted Solution

by:
leakim971 earned 500 total points
ID: 35094621
Check this (we can do better if both are not in the same array) :
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>json find peer same array</title>
<script language="javascript" src="http://code.jquery.com/jquery-1.5.1.min.js"></script>
<script language="javascript">
	var data = {"references":[{"type":"user","full_name":"Jim Bo Bob","id":5,"url":"https://yahoo.com",},{"type":"user","full_name":"Bob Joe Jim","id":11,"url":"https://google.com",},{"type":"topic","name":"Muscular Arms","id":28},{"voted_user_id":5,"type":"approved","approved_user_id":7,"topic_id":28,"comment":"His arms are massive","id":297},{"voted_user_id":11,"type":"approved","approved_user_id":7,"topic_id":28,"comment":"Very moist","id":89}]};
	$(document).ready(function() {
		var json = data.references;
		$.each(data.references, function (i, ref) {
			for(var j in json) {
				if(ref.id == json[j]["voted_user_id"]) {
					alert( "full name is : " + ref["full_name"] + "\ntype is : "+ ref.type );
				}
			}
		});
	});
</script>
</head>
<body>
</body>
</html>

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LVL 19

Expert Comment

by:Albert Van Halen
ID: 35094675
Your references array contains different objects : users, topic and approved.
You need to filter them all out by their type -> all users, the topic and approved items.
var users = references.filter(function(ref) { return ref.type=="user"; });
var approvedItems = references.filter(function(ref) { return ref.type=="approved"; });
var topics = references.filter(function(ref) { return ref.type=="topic"; });

// now you can loop through all your approved items
var numApproved = approvedItems.length;
for(var x = 0; x < numApproved; x++) {
   var approvedItem = approvedItems[x];
   var user = users.filter(function(user) { return user.id == approvedItem.approved_user_id; });
   if(user.length > 0) {
      // user array containing all the users with id equel to approvedItem.approved_user_id
   }
}

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0
 
LVL 82

Expert Comment

by:leakim971
ID: 35097786
Check this useful online tools : http://jsonviewer.stack.hu/
Paste you json object in the window, you can format or parse it easily graphically (click on the Viewer tab)
Thanks for the points!

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