Json with PHP mysql and Jquery

Question

I have a strange problem..

When i try to access my json code with the corresponding number [0], [1] etc.. I just get the first character of the object.

First of my code:

test2.php

if(isset($_POST['getCustomersArray'])){
$runQuery = mysql_query($Query) or
 die("SQL: $Query)<br />".mysql_error()); 
$numrows = mysql_num_rows($runQuery); 
$array = array(array());
for($i = 0;$i <= 2; $i++){
 $row = mysql_fetch_array($runQuery);
 $array[$i]['namn'] = $row['fornamn'];
}
 print json_encode($array);
}

scriptfile.js

$.ajax({
 type:"POST",
 url: "test2.php",
 data: "getCustomersArray=true",
 datatype: "JSON",
 cache: false,
 success: function(json) {
 console.log(json[0]);
 }
});

The result (from console.log(json[0])):

[

The result from just console.log(json):

[{"namn":"the first name"},{"namn":"The secound name"},{"namn":"the third name"}]

Im not sure why the squarebrackets are there but maybe they should be?

Ive been fuzzing with this problem for a while now and im sure its something stupid. Please help.

Answer 1

You have an incorrect option in the AJAX settings,

datatype: "json",

It should be:

dataType: "json",

• Martin: try json in lowercase. dataType: "json"
• MacMac: @Martin Aha, I thought jQuery lowercases it by default.
• Martin: I also thought that to be honest, but it seems not, haha!

Answer 2

datatype: "JSON",

supposed to be

dataType: "json", // json in lowercase and T has to be captalized

Answer 3

Make sure you have the following code:

if(isset($_POST['getCustomersArray'])){
$runQuery = mysql_query($Query) or
 die("SQL: $Query)<br />".mysql_error()); 
$numrows = mysql_num_rows($runQuery); 
$array = array(array());
for($i = 0;$i <= 2; $i++){
 $row = mysql_fetch_array($runQuery);
 $array[$i]['namn'] = $row['fornamn'];
}
header("Content-Type: application/json; charset=UTF-8");
print json_encode($array);
}

Here you need to set the content-type to application/json and set the correct charset to avoid any cross-browser issues. Take a look at the following tutorial from my website which should cover all of this and perhaps some improvements to your code: PHP jQuery Search Tutorial - using JSON object the proper way

Hope this helps :)