CODEWITHSUNDEEP
phpjqueryajax
jibin dcruz
jibin dcruz

Reputation: 273

Using AJAX to pass variable to PHP and retrieve those using AJAX again

I want to pass values to a PHP script so i am using AJAX to pass those, and in the same function I am using another AJAX to retrieve those values.

The problem is that the second AJAX is not retrieving any value from the PHP file. Why is this? How can I store the variable passed on to the PHP script so that the second AJAX can retrieve it?

My code is as follows:

AJAX CODE:

$(document).ready(function() {    
    $("#raaagh").click(function(){    
        $.ajax({
            url: 'ajax.php', //This is the current doc
            type: "POST",
            data: ({name: 145}),
            success: function(data){
                console.log(data);
            }
        });  
        $.ajax({
            url:'ajax.php',
            data:"",
            dataType:'json',
            success:function(data1){
                var y1=data1;
                console.log(data1);
            }
        });
    });
});

PHP CODE:

<?php
$userAnswer = $_POST['name'];    
echo json_encode($userAnswer);    
?>

Upvotes: 18

Views: 138992

Answers (5)

Rohan Kumar
Rohan Kumar

Reputation: 40639

Use dataType:"json" for json data

$.ajax({
     url: 'ajax.php', //This is the current doc
     type: "POST",
     dataType:'json', // add json datatype to get json
     data: ({name: 145}),
     success: function(data){
         console.log(data);
     }
});

Read Docs http://api.jquery.com/jQuery.ajax/

Also in PHP 

<?php
  $userAnswer = $_POST['name']; 
  $sql="SELECT * FROM <tablename> where color='".$userAnswer."'" ;
  $result=mysql_query($sql);
  $row=mysql_fetch_array($result);
  // for first row only and suppose table having data
  echo json_encode($row);  // pass array in json_encode  
?>

Upvotes: 27

Juan Mart&#237;nez
Juan Mart&#237;nez

Reputation: 51

In your PhP file there's going to be a variable called $_REQUEST and it contains an array with all the data send from Javascript to PhP using AJAX.

Try this: var_dump($_REQUEST); and check if you're receiving the values.

Upvotes: 1

Sathish Kumar D
Sathish Kumar D

Reputation: 274

you have to pass values with the single quotes

$(document).ready(function() {    
    $("#raaagh").click(function(){    
        $.ajax({
            url: 'ajax.php', //This is the current doc
            type: "POST",
            data: ({name: '145'}), //variables should be pass like this
            success: function(data){
                console.log(data);
                           }
        });  
        $.ajax({
    url:'ajax.php',
    data:"",
    dataType:'json',
    success:function(data1){
            var y1=data1;
            console.log(data1);
            }
        });

    });
});

try it it may work.......

Upvotes: 0

Aks1357
Aks1357

Reputation: 1072

$(document).ready(function() {
    $("#raaagh").click(function() {
        $.ajax({
            url: 'ajax.php', //This is the current doc
            type: "POST",
            data: ({name: 145}),
            success: function(data) {
                console.log(data);
                $.ajax({
                    url:'ajax.php',
                    data: data,
                    dataType:'json',
                    success:function(data1) {
                        var y1=data1;
                        console.log(data1);
                    }
                });
            }
        });
    });
});

Use like this, first make a ajax call to get data, then your php function will return u the result which u wil get in data and pass that data to the new ajax call

Upvotes: 2

user1646111
user1646111

Reputation:

No need to use second ajax function, you can get it back on success inside a function, another issue here is you don't know when the first ajax call finished, then, even if you use SESSION you may not get it within second AJAX call.

SO, I recommend using one AJAX call and get the value with success.

example: in first ajax call

    $.ajax({
        url: 'ajax.php', //This is the current doc
        type: "POST",
        data: ({name: 145}),
        success: function(data){
            console.log(data);
            alert(data);
            //or if the data is JSON
            var jdata = jQuery.parseJSON(data);
        }
    });

Upvotes: 1

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