Jquery - Pass a json object in a global variable

Question

I'm trying to pass a json object from an ajax call in a variable that will be used in another function.

The thing is that if i try to console.log() that variable ($aR), it returns "undefined"

Here is the code:

$aR = '';

// Submit Data to ncbi.
// Sends form's data to classController.php
function NCBI_submit_data()
{
    $formData = $('#blastx_form').serialize();
    $php_method = 'ncbi_request';    
    $finalData = $formData + "&php_method=" + $php_method;
    $aR = ajaxReq('POST','../../classes/classController.php',$finalData,'json');
    console.log($aR);
}


// General Ajax function
function ajaxReq($method,$url,$data,$dataType)
{
    $.ajax({
        type: $method,
        url:  $url,
        async: 'false',
        data: $data,
        dataType: $dataType,
        success: function(json, textStatus, jqXHR)
        {
            $aR = json;
        },
        error: function(jqXHR, textStatus, errorThrown)
        {
            console.log('Ajax call error: '+jqXHR.status+' '+errorThrown)                
        }
    });
}

Did you try declaring $aR = {}; as an object instead of a string? — wrxsti
– wrxsti, Commented Jul 9, 2014 at 13:37
What are you doing here, you're setting $aR to the XHR object, then you're doing synchronous ajax and setting the same variable to the returned data, all the while thinking you're in PHP with no var declarations but dollarsigns in front of all the variables instead ? — adeneo
– adeneo, Commented Jul 9, 2014 at 13:37
While you do still need the var, you can use a $ in your variable name still. Not sure why for a json object, but I suppose you could. — wrxsti
– wrxsti, Commented Jul 9, 2014 at 13:39

Paul Roub · Accepted Answer · 2014-07-09 13:41:02Z

2

Here's a rough timeline of $aR in your code:

// initialized
$aR = '';

// set to `undefined`, since `ajaxReq` returns nothing
$aR = ajaxReq('POST','../../classes/classController.php',$finalData,'json');

// then, later, when the `ajax` call completes:
$aR = json; 
console.log($aR);  // would do something now

If you need to do something with that value, do so from within the success handler

success: function(json, textStatus, jqXHR)
{
  $aR = json;
  doSomethingWith($aR);   
},

Or use the AJAX object returned by $.ajax():

function ajaxReq($method,$url,$data,$dataType)
{
  return $.ajax({ ... });
}

// called as
ajaxReq('POST','../../classes/classController.php',$finalData,'json').done(
  function(json) {
    $aR = json;
    // whatever else you want to do
  }
);

answered Jul 9, 2014 at 13:41

Paul Roub

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Comments

Community · Accepted Answer · 2017-05-23 11:57:51Z

1

By default JQuery Ajax do asynchronous call. You try to use data before it is retrieved from backend. After Ajax call JS interpreter will not wait for response and will execute:

$aR = ajaxReq('POST', ....).

The function "ajaxReq" does not return anything. So, the "$aR" value is "undefined".

The data will be available later in "success" callback, after retrieving response:

success: function(json, textStatus, jqXHR)
{
  $aR = json;
},

The callback "success" will be executed later "asynchronously".

Read about the difference between a synchronous and an asynchronous request:

What is the difference between a synchronous and an asynchronous request? (async=true/false)

edited May 23, 2017 at 11:57

CommunityBot

11 silver badge

answered Jul 9, 2014 at 13:59

user3820695

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