JSON Object To Javascript Array?

Question

I'm having problems decoding a JSON object. I call a PHP script using an Ajax call. The PHP script returns a JSON encoded object which I can read but only the first record. How can I decode the JSON object? I am using JQUERY Mobile.

PHP script:

<?php

    $host = "localhost";
    $user = "root";
    $pass = "[password]";

    $databaseName = "zombieSurvival";
    $tableName = "TBLusers";

    //Connect to mysql database

    include 'DB.php';
    $con = mysql_connect($host,$user,$pass);
    $dbs = mysql_select_db($databaseName, $con);

    // 2) Query database

    $result = mysql_query("SELECT * FROM $tableName");
    $array = mysql_fetch_array($result);
    echo json_encode($array);

?>

and here is the Ajax call:

$.ajax({
       url: '/PHP/getUserMarkers.php',
       data: "",
       dataType: 'json',
       success: function(data){

       //How can I treat 'data' variable to make it a
       //javascript array?


       }
       });

Even if this is a `localhost` machine, posting your password is not advisable. — Giacomo1968, Jun 17 '14 at 20:54
You can see an answer in this duplicate: [Converting JSON Object into Javascript array](http://stackoverflow.com/questions/20881213/converting-json-object-into-javascript-array) — Iulian Miu, Jun 17 '14 at 20:56
Thanks Jake - oversight on my part... Gonna take a looking at the duplicate posted by lulian... — user3535074, Jun 17 '14 at 20:57
"*How can I decode the JSON object?*" The `dataType: 'json'` setting tells jQuery to decode it for you. So, as long as the response is valid JSON, `data` should already be an `Array`. Regarding "*but only the first record*," you'll have to loop with `mysql_fetch_array()` to gather a list of all records. [mysql_fetch_array add all rows?](http://stackoverflow.com/questions/4643340/mysql-fetch-array-add-all-rows) — Jonathan Lonowski, Jun 17 '14 at 21:06
Thanks a lot Jonathan! After checking all the advice in the posted suggestion from lulian I can see that the issue seems to be exactly that, that the object passed from the php script only includes the first record. Checking your link... — user3535074, Jun 17 '14 at 21:20

score 0 · Answer 1 · answered Jun 17 '14 at 21:30

The issue was the JSON object returned by the PHP script. Adjusted the PHP script as below and now I can access all returned rows and individual field names.

<?php

    $host = "localhost";
    $user = "root";
    $pass = "muertealregueton";

    $databaseName = "zombieSurvival";
    $tableName = "TBLusers";


    include 'DB.php';
    $con = mysql_connect($host,$user,$pass);
    $dbs = mysql_select_db($databaseName, $con);


    $result = mysql_query("SELECT * FROM $tableName");
    $array = array();
    while(($row = mysql_fetch_array($result))) {
        $array[] = $row;
    }
    echo json_encode($array);

?>

and the ajax code which calls an alert on one of the later records returned:

$.ajax({
       url: '/PHP/getUserMarkers.php',
       data: "",
       dataType: 'json',
       success: function(data){


       //now I can access each row and field
       //like here I access the 3rd field of the 15th record
       alert(data[15][2]);



       }
       });

JSON Object To Javascript Array?

1 Answers1