php jquery iterate php array in success function

Question

I have jquery pop form . It takes one input from the user ,mapping_key , Once the user enters the mapping key ,i make an ajax call to check if there is a user in the database with such a key. This is my call .

Javascript:

$.ajax({       
    url : base_url+'ns/config/functions.php',
    type: 'POST',
    data : {"mapping_key":mapping_key} ,
    success: function(response) {
            alert(response)                        
    }
});

PHP:

$sql = "select first_name,last_name,user_email,company_name from registered_users where mapping_key = '$mapping_key'";    
$res = mysql_query($sql);
$num_rows = mysql_num_rows($res);
if($num_rows == 0)
{
    echo $num_rows;
}
else{
    while($result = mysql_fetch_assoc($res))
    {
       print_r($result);
    }
}

Now i want to loop through the returned array and add those returned values for displaying in another popup form. Would appreciate any advice or help.

Damien Legros · Accepted Answer · 2013-04-10 07:43:32Z

2

In your php, echo a json_encoded array:

$result = array();
while($row = mysql_fetch_assoc($res)) {
   $result[] = $row;
}
echo json_encode($result);

In your javascript, set the $.ajax dataType property to 'json', then you will be able to loop the returned array:

$.ajax({       
    url : base_url+'ns/config/functions.php',
    type: 'POST',
    data : {"mapping_key":mapping_key} ,
    dataType : 'json',
    success: function(response) {
        var i;
        for (i in response) {
          alert(response[i].yourcolumn);
        }                      
    }
});

answered Apr 10, 2013 at 7:43

Damien Legros

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Comments

thumber nirmal · Accepted Answer · 2013-04-10 07:38:37Z

0

change

data : {"mapping_key":mapping_key} ,

to

data: "mapping_key=" + mapping_key,

answered Apr 10, 2013 at 7:38

thumber nirmal

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Comments

Jai · Accepted Answer · 2013-04-10 07:49:23Z

0

You have to take the posted mapping_key:

$mapping_key = $_POST['mapping_key'];

$sql = "select first_name,last_name,user_email,company_name from registered_users
        where mapping_key = '$mapping_key'";

or this:

$sql = "select first_name,last_name,user_email,company_name from registered_users
        where mapping_key = $_POST['mapping_key']";

answered Apr 10, 2013 at 7:49

Jai

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