CODEWITHSUNDEEP
javascriptphpjqueryjsonajax
Bilal Zafar
Bilal Zafar

Reputation: 457

Get JSON response in PHP using jQuery

Here is my code to call AJAX and get the response from another PHP file:

$.post('<?php echo get_site_url(); ?>/ajax-script/',{pickup:pickup,dropoff:dropoff,km:km},
        function(data){
            $('#fare').html(data);
            $('#loading_spinner').hide();

        });

ajaxscript.php file

$jsonData = '{"fare":30580,"actual_distance":1519,"city":"Islamabad","status":true}';

$json = json_decode($jsonData,true);
echo $json['fare'];

This code gives me the fare at the time of $('#fare').html(data);

But I need to extract the city from JSON, too, and for this I added an extra line in ajaxscript.php:

echo $json['city'];

After doing this, it gives me 30580Islamabad

How can I store these two values separately in JavaScript? I need them for future work.

Upvotes: 0

Views: 679

Answers (2)

RiggsFolly
RiggsFolly

Reputation: 94642

You are doing everything backwards

Your PHP should be

$jsonData = '{"fare":30580,"actual_distance":1519,"city":"Islamabad","status":true}';

//$json = json_decode($jsonData,true);
echo $jsonData;

As you already have a JSONString to send to your javascript.

Then your javascript will recieve a javascript object in the data parameter of 

$.post( '<?php echo get_site_url(); ?>/ajax-script/', 
          {pickup:pickup,dropoff:dropoff,km:km}, 
   function( data ) {
        $('#fare').html(data.fare);
        $('#city').html(data.city);
        $('#loading_spinner').hide();
}, "json");

Note the "JSON" at the end of the javascript to tell it to expect a JSON Object, it will then convert the JSONString to a javascript Object automatically for you so the data parameter will be an onbect

Upvotes: 2

krishna
krishna

Reputation: 7

Add Special characters at the end of each value and in jquery, using jquery split, cut the variable and display

like below;

$jsonData = '{"fare":30580^^,"actual_distance":1519^^,"city":"Islamabad^^","status":true}';

$json = json_decode($jsonData,true);
echo $json['fare'];


in jquery

function(data){
var tdata = data.split("^^");
            $('#fare').html(tdata[0]);
            $('#loading_spinner').hide();

        });

Upvotes: -1

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