CODEWITHSUNDEEP
phpjqueryjsonajax
Vishal B
Vishal B

Reputation: 653

return tables rows in json and print

I am trying to return ajax response in json, but when I print it in log it gives null even tables has rows, my php code is: 

if(isset($_GET['proid'])){
    $projid = $_GET['proid'];
    include(db.php);
    $res = mysqli_query($con, "SELECT * FROM data WHERE project_id LIKE '%$projid%'");      
    while($row = mysqli_fetch_assoc($res)) 
    {
      $dataarray[] = $row;
    }
    echo json_encode($dataarray);
}

ajax : 

$.ajax({
        url : 'getRecStudy.php',
        type : 'GET',           
        data : {proid:study},
        success : function(data) {
            $('#tbody').empty();
            $("#tbody").append(data);
            console.log(data);
        }
    });

whats wrong?

Upvotes: 2

Views: 66

Answers (2)

Kishor Parida
Kishor Parida

Reputation: 293

I find no issue in your code except varibales. You need to debug the code in php file

if(isset($_GET['proid'])){
    echo $_GET['proid'] . " is proid";
    $projid = $_GET['proid'];
    include(db.php);
    echo "db connected";
    $res = mysqli_query($con, "SELECT * FROM data WHERE project_id LIKE '%$projid%'");      
    echo "result fetched";
    while($row = mysqli_fetch_assoc($res)) 
    {
      $dataarray[] = $row;
      echo "inside while";
    }
    echo json_encode($dataarray);
    print_r($dataarray);
    exit;
}

after all this hit http://yourdomain.com/yourfile.php?proid=correctvalue

You will get the bug.

Upvotes: 1

Dharmendra
Dharmendra

Reputation: 127

use parseJSON method in success function like this

var obj = jQuery.parseJSON( data );

alert( obj.name ); // For example name is a key

Upvotes: 0

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