json string is not converting into object

Question

Here is the server side code I am trying to send json

$x = array();
$timestamp = strtotime('22-09-2008'); 
$x["x"] = $timestamp;
$x["y"] = 22;

$val = '[{ "name": "weight", "dataPoints": ['.json_encode($x).'] }]';
echo json_encode($val);

So output for above code looks like

   "[{ \"name\": \"weight\", \"dataPoints\": [{\"x\":1222041600,\"y\":22}] }]"

Below is the client side code I get the data via Jquery getJSON

var jqxhr = $.getJSON( "https://domain/gettracker.php?id="+id, function(data) {

 console.log(data);

})

I suppose getJson converts json to object automatically , but it logs the raw json like below

"[ {  name: "weight", dataPoints: [{"x":1222041600,"y":22}] } ]"

I tried to do json parse , but i get error.

I guess I am not sending the data properly via php.Can some one guide me ?

Uhm. $val is a string. json_encode($val) encodes that string as JSON, so you get "....." which is valid JSON. jQuery is calling JSON.parse, which returns the .... part. It works perfectly fine, just not like you intended because you are not preparing your data correctly. I recommend to read the documentation and examples about json_decode: php.net/manual/en/function.json-encode.php — Felix Kling
– Felix Kling, Commented Feb 20, 2015 at 16:30
json_encode converts the value to JSON (and returns a string). If you are already having a string containing JSON, you don't want to call json_encode on it. Makes sense? — Felix Kling
– Felix Kling, Commented Feb 20, 2015 at 16:37

Rory McCrossan · Accepted Answer · 2015-02-20 16:29:58Z

3

Your JSON string is not valid - the property names should be enclosed in double quotes - ", and you don't need to encode the string again.

$val = '[{ "name": "weight", "dataPoints": ['.json_encode($x).'] }]';
echo $val;

Or better yet, use json_encode to create the string for you:

$data = array(
    'name' => 'weight',
    'dataPoints' => $x
);

echo json_encode($data);

answered Feb 20, 2015 at 16:29

Rory McCrossan

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4 Comments

Felix Kling Over a year ago

Just to be clear (for future readers): It does not not work because the string value is incorrect (at least that's not the main issue). The issue is that the OP tries to JSON encode a string.

Vishnu Over a year ago

edited question , your first solution not working, 2nd solution is working fine ,,thank you

Felix Kling Over a year ago

@Vishnu: I think you missed the part where echo json_encode($val); becomes echo $val;. But yeah, the second solution is the way to go.

Vishnu Over a year ago

oh yea , both solution working...but 2nd solution looks perfect :)

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