Javascript and PHP array values

Question

Hey i sent a JSON encoded array to my JS page with two links.

{"download":"http://JapSeyz.dk/Test/Medarbejder/fileDownload.php?filename=test___1847598251_26-04-2013.png","view":"http://JapSeyz.dk/Test/Medarbejder/uploads/test___1847598251_26-04-2013.png"}

that is the output

it is in the array called status and if i try status[download] it is undefined, and status[0,1,2] and so on, just gives me a single character.... How do i select the whole link from download at once?

Thanks :)

Heres all the code.. Sorry (Some danish text, but i just want to display a Force DL link and a view link)

var links = JSON.parse(status);
    var download = status.download;
    var view = status.view;
    var status = "Brug dette link hvis folk kun skal se dette <br /><div class='link-copy'>http://JapSeyz.dk/Test/Medarbejder/"+view+"</div><br /><br />Hvis du vil have folk til at downloade den skal du bruge dette link <br /> <div class='link-copy'>"+download;
   //set the status message to that returned by the server
   document.getElementById('status').innerHTML=status;

And the PHP:

$test = json_encode(array("download" => $link, "view" => "http://JapSeyz.dk/Test/Medarbejder/".$move));

echo returnStatus($test);

Check [json_decode](http://php.net/manual/en/function.json-decode.php) and [json_encode](http://php.net/manual/en/function.json-encode.php) PHP function to convert json to php array/object and php to json.. — Svetoslav, Apr 26 '13 at 10:58
Please post also relevant Javascript code how to read the json object. Our magic globes do not wort today. — Reporter, Apr 26 '13 at 11:00
Well i send it from at PHP page, but i am retrieving it again in a script of JS, not jQuery, just standard JS — Jesper Jacobsen, Apr 26 '13 at 11:02
You are passing JSON response as array so you can't access it using status[download]. — Kailash Yadav, Apr 26 '13 at 11:07

score 1 · Accepted Answer · answered Apr 26 '13 at 10:59

1

Parse the json and display it like this edit.

var obj = jQuery.parseJSON('{"download":"http://JapSeyz.dk/Test/Medarbejder/fileDownload.php?filename=test___1847598251_26-04-2013.png","view":"http://JapSeyz.dk/Test/Medarbejder/uploads/test___1847598251_26-04-2013.png"}');
alert( obj.download);

Javascript

var obj = JSON.parse('{"download":"http://JapSeyz.dk/Test/Medarbejder/fileDownload.php?filename=test___1847598251_26-04-2013.png","view":"http://JapSeyz.dk/Test/Medarbejder/uploads/test___1847598251_26-04-2013.png"}');
alert( obj .download);

answered Apr 26 '13 at 10:59

Yogesh Suthar

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What if the user does not use jQuery? – Felix Kling Apr 26 '13 at 11:00
oh okay, i just remember somebody told me that i did not need to parse it in JS as it was natively understood there.. – Jesper Jacobsen Apr 26 '13 at 11:00
@Jesper: That person might have confused object literals with JSON. – Felix Kling Apr 26 '13 at 11:02
No.. I added more code to my main thread.. This works, kinda... If i alert the data it works fine, but if i use it as a variable it does not work – Jesper Jacobsen Apr 26 '13 at 11:22
@JesperJacobsen Have a look on my answer if your status is an array – Prasath K Apr 26 '13 at 11:23
1

This actually worked for me, i just messed it up first... sorry for all... It works now – Jesper Jacobsen Apr 26 '13 at 11:31

score 1 · Answer 2 · answered Apr 26 '13 at 11:01

var jsonstring = $.parseJSON('{"download":"http://JapSeyz.dk/Test/Medarbejder/fileDownload.php?filename=test___1847598251_26-04-2013.png","view":"http://JapSeyz.dk/Test/Medarbejder/uploads/test___1847598251_26-04-2013.png"}');

for (var k in jsonstring) {
  alert("key is "+k);
   alert("value at key "+k+ " is "+ jsonstring[k]);
}

score 0 · Answer 3 · answered Apr 26 '13 at 11:00

0

var jsonobj = JSON.parse('{"download":"http://JapSeyz.dk/Test/Medarbejder/fileDownload.php?filename=test___1847598251_26-04-2013.png","view":"http://JapSeyz.dk/Test/Medarbejder/uploads/test___1847598251_26-04-2013.png"}');
alert( jsonobj .download);

answered Apr 26 '13 at 11:00

Praveen kalal

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Prasath K · Answer 4 · 2013-04-26T11:32:19.850

0

status1 = [
          {"download":"http://JapSeyz.dk/Test/Medarbejder/fileDownload.php?filename=test___1847598251_26-04-2013.png","view":"http://JapSeyz.dk/Test/Medarbejder/uploads/test___1847598251_26-04-2013.png"}
         ];

alert(status1[0].download);

Status is a array of objects so you should use status[index].yourPropName

SEE FIDDLE

edited Apr 26 '13 at 11:32

answered Apr 26 '13 at 11:00

Prasath K

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Why are you using a different data structure than the OP has? – Felix Kling Apr 26 '13 at 11:01
@FelixKling it's not a different .. He clearly said status is an array – Prasath K Apr 26 '13 at 11:03
No he didn't. He only said that if he access `status` with numeric indexes, he gets back single characters, which indicates that `status` is a string. – Felix Kling Apr 26 '13 at 11:22
This just prints undefined for me :/ – Jesper Jacobsen Apr 26 '13 at 11:25
@FelixKling He said he used status[0],status[1].. not status[0][download] – Prasath K Apr 26 '13 at 11:26
@JesperJacobsen Is your status an array ...?? – Prasath K Apr 26 '13 at 11:26
I think it is an array... truth is i have not worked much with arrays, but i JSON encoded it from my PHP page, and to me it looks like an array... but all the code is in the top post – Jesper Jacobsen Apr 26 '13 at 11:27
See http://jsfiddle.net/spjp5/ – Prasath K Apr 26 '13 at 11:32
Again: If `status` was an array, then `status[0]` or `status[1]` would **not** return a single character. But it did, which proofs that `status` is a string. The actual problem is that the OP did not parse the JSON, something you didn't even mention in your answer. – Felix Kling Apr 26 '13 at 12:18

Javascript and PHP array values

4 Answers4