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Parsing json

Posted on 2013-11-05
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Last Modified: 2013-11-05
Hi,

I have a service call in my application which returns jsaon data. The call is like so:

            $.ajax({
                type: "GET",
                dataType: "json",
                url: 'http://rest.smartdoc.se/api/users/' + username + '/' + password,
                success: function (Data) {
                    var obj = $.parseJSON(Data);
                    alert(obj.FullName);
                }

            });

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The service returns this:

{"Data":[{"FullName":"Doesn't exist","username":"sdf","password":"sdf"}],"Errors":null}

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There obviously is something wrong because it is not working. Please help me.

Best regards
RTSol
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Question by:RTSol
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6 Comments
 
LVL 82

Expert Comment

by:leakim971
ID: 39623947
why :
url: 'http://rest.smartdoc.se/api/users/' + username + '/' + password,

and not :
url: '/api/users/' + username + '/' + password,
0
 

Author Comment

by:RTSol
ID: 39624012
Hi,

The application is running on a smartphone. I need to provide the complete url to the service which is running on a web server. The call to the service works fine and it returns the desired json. My problem is just to parse it.

-RTSol
0
 
LVL 43

Expert Comment

by:Chris Stanyon
ID: 39624016
Look at your JSON, you probably need this:

var obj = $.parseJSON(Data);
alert(obj.Data.FullName);
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LVL 30

Expert Comment

by:Alexandre Simões
ID: 39624025
Well... I looked at your code and really bad is this line:

'http://rest.smartdoc.se/api/users/' + username + '/' + password

You're passing username and password in clear text and not even https?
And I hope you're not doing this for each request...
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LVL 43

Accepted Solution

by:
Chris Stanyon earned 2000 total points
ID: 39624027
Actually, no - you need this:

success: function (Data) {
    var person = Data.Data[0]; //grab the first person from the Data array
    alert(person.FullName);
}

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The Data returned contains a Data key, holding an array or people. If you are likely to be passing back more than one person then you'll need to loop through Data.Data
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Author Closing Comment

by:RTSol
ID: 39624042
Thanks a lot - that is what I needed.
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