How to use getJson to send variable to php file

Question

Here is the javascript code that has a "no" in the datastring and i want to pass that datastring to data.php using getJson So that in the end i can get json as a response. My data.php code is working but javascript code is not working so please help me.

Javascript code

dataString = 'no='+ no; 

$.getJSON("data.php", dataString,function(response){

        console.log(response);
  });

php code

$no=$_POST['no'];

$headers = array(
'Content-Type: application/json',
);

$url = 'https://www.outpan.com/api/get-product.php?barcode='.$no.'&apikey=318c91a99b132a29dda536d63e7fb8b4';


$ch = curl_init();


curl_setopt($ch, CURLOPT_URL, $url);

curl_setopt($ch, CURLOPT_POST, false);

curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);


curl_setopt($ch, CURLOPT_RETURNTRANSFER, true );


curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);




 $result = curl_exec($ch);



curl_close($ch);

//$result_arr = json_decode($result, true);

echo result;

Kamehameha · Accepted Answer · 2015-06-09 14:37:10Z

1

One possibility is response is not valid JSON (I am not sure, since you seemed to have tried printing the json_decoded version).
Try printing out the error by using error which is called when no valid JSON is received -

$.getJSON("data.php", dataString,function(response){
    console.log(response)
}).error(function(jqXHR, textStatus, errorThrown){ 
    console.log(textStatus+" - "+errorThrown); 
});

If an error is logged, then checkout the curl response in php and make sure it is valid json.

EDIT
This edit is assuming that he php response is invalid JSON as observed from the comments.
Looking at your php code, you have used this -

$headers = array(
'Content-Type: application/json',
);

Try this instead -

header('Content-Type: application/json');

EDIT 2
Looking at your php code again, there seems to be an error while printing -

echo result;

It should be -

echo $result;

edited Jun 9, 2015 at 14:37

answered Jun 9, 2015 at 14:06

Kamehameha

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6 Comments

Abdur Rehman Farooqi Over a year ago

This is the error "parsererror - SyntaxError: Unexpected token a"

Kamehameha Over a year ago

@AbdurRehmanFarooqi Yep, thats it. The PHP response is invalid json.

Abdur Rehman Farooqi Over a year ago

now i have commented the json part and just echo a statement and it is not returned as response.

Kamehameha Over a year ago

@AbdurRehmanFarooqi before you print anything, just copy paste the contents here and check if it is valid JSON.

Abdur Rehman Farooqi Over a year ago

Its not working if is send you source code can you make the changes?

|

gabo · Accepted Answer · 2015-06-09 13:49:27Z

0

use $_GET In your php code, you are using $_POST now

answered Jun 9, 2015 at 13:49

gabo

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1 Comment

Abdur Rehman Farooqi Over a year ago

i have checked it. Its not working. i think there is a problem with getjson

chandresh_cool · Accepted Answer · 2015-06-09 13:52:26Z

0

You should $_GET in your php code or change code like this :

$.getJSON("data.php", { no: no },function(response){

    console.log(response);
  });

In php use,

$no=$_REQUEST['no'];

answered Jun 9, 2015 at 13:52

chandresh_cool

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