CODEWITHSUNDEEP
javascriptphpjqueryajaxjson
Mistergrave
Mistergrave

Reputation: 179

Get data from database using PHP, JQuery and AJAX in JSON format

I'm having trouble getting data from my database. My goal is to get all groups from my database and return them in JSON (in an alert box or whatever).

Now it won't convert to JSON and I am getting weird response text from the ajax call. If you need anything else to solve this problem, please do not hesitate to ask.

Here is what I did.

PHP

$servername = "redacted";
$username = "redacted";
$password = "redacted";
$dbname = "redacted";

if(isset($_POST['action']) && !empty($_POST['action'])) {
    $action = $_POST['action'];
    switch($action) {
        case 'getGroups' : getAllGroups();break;
    }
}

function getAllGroups() {

    $mysqli = new mysqli($servername, $username, $password, $dbname);

    $query = $mysqli->query("SELECT * FROM groups");

    while($row = $query->fetch_object()) {
        $result[] = $row;
    }

    echo "{\"results\":";
    echo json_encode($result);
    echo "}";

    $mysqli->close();
 }

JS

function getPosts() {
    $.ajax({
        url: 'functions.php',
        data: {action: 'getGroups'},
        type: 'post',
        success: function(output) {
            var result = JSON.parse(output);
            result = result.resultaten;
            alert(result);
        }
    });
}

getPosts();

Thanks in advance, Mistergrave.

Upvotes: 2

Views: 3499

Answers (3)

Mistergrave
Mistergrave

Reputation: 179

Okay guys, I managed to solve everything. Apparently the php function couldn't find my credentials to log in to the database server because I defined them on top of the php file (and since javascript only executed the function, these credentials were undefined).

Solution:

I just copy-pasted the credentials at the start of each function so these were defined. And tadaah! It worked :).

Now I realize why the responseText was full of tables, because it started to return error tables about the connection.

I hope my explanation will help other people who have this issue as well.

Cheers, and thanks for all the helpfull answers,
Mistergrave.

Upvotes: 1

Pratik Joshi
Pratik Joshi

Reputation: 11693

use

echo json_encode(array('result'=>$result));

As it takes array as parameter. Check here

Just a note : If are sure you will return json data, use dataType:json , so you wont need JSON.parse(output).

Upvotes: 0

Sougata Bose
Sougata Bose

Reputation: 31749

No need for that extra echos. Try with - 

echo json_encode(array('results' => $result));

Instead of - 

echo "{\"results\":";
echo json_encode($result);
echo "}";

No need for - if(isset($_POST['action']) && !empty($_POST['action'])) {

if(!empty($_POST['action'])) { - do the all.

Define $result first.

$result = array();
while($row = $query->fetch_object()) {
    $result[] = $row;
}

Upvotes: 1

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