return tables rows in json and print

Question

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?

in php file debug line by line, echo exit $variables at each line and hit getRecStudy.php?proid=study in url. Dont dubug ajax response. — Kishor Parida, Jul 11 '16 at 10:32
http://stackoverflow.com/questions/8649621/returning-a-json-object-from-php-in-ajax-call Check with this. Also use dataType: 'json', in your ajax call — Jagadeesh, Jul 11 '16 at 10:33

score 1 · Answer 1 · answered Jul 11 '16 at 10:39

1

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.

answered Jul 11 '16 at 10:39

Kishor Parida

295
1
10

can you give me any example link of ajax json example? – Vishal B Jul 11 '16 at 10:42
http://stackoverflow.com/questions/8649621/returning-a-json-object-from-php-in-ajax-call – Jagadeesh Jul 11 '16 at 10:45
it is still giving null @kishore – Vishal B Jul 11 '16 at 10:52
@VishalBorade null values wont print by the procedure i explained above, it would echo string, error or blank provided you hit `http://yourdomain.com/yourfile.php?proid=correctvalue` correct url in browser not through ajax. Before calling ajax make sure your php file works properly through (non-ajax) – Kishor Parida Jul 11 '16 at 10:55
It was problem with connection string, when i wrot full connection string it gave me result. i dont knw why... bdw Thanks. – Vishal B Jul 11 '16 at 10:59
Glad to know the issue is solved. If somehow my answer helped you then give it a vote. Happy coding. – Kishor Parida Jul 11 '16 at 11:23

score 0 · Answer 2 · answered Jul 11 '16 at 10:28

0

use parseJSON method in success function like this

var obj = jQuery.parseJSON( data );

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

answered Jul 11 '16 at 10:28

Dharmendra

127
12

even if we dont parse, it would still log some json encoded data not "null" value. Its loggin "null" value so parsing null json won't produce correct parsed value. – Kishor Parida Jul 11 '16 at 10:31

return tables rows in json and print

2 Answers2