PHP array to jquery array via JSON

Question

Im having a little confusion why the following is not working.

get.php

<?php

$username="root";
$password="root";
$database="testing";

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$name= $_GET['name'];

$query="SELECT * FROM tableone ";
$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();

$array = array();

$i=0;
while ($i < $num) {

    $first=mysql_result($result,$i,"firstname");
    $last=mysql_result($result,$i,"lastname");
    $date=mysql_result($result,$i,"date");
    $ID=mysql_result($result,$i,"id");

    $array[] = $first;

    $i++;
}

echo json_encode($array);

?>

jQuery

var arr = new Array();

    $.get("get.php", function(data){
         arr = data;
         alert(arr);
    }, "json");

When I run the following I get a list of names that looks like this

["James","Lydia","John"]

But when I try to single out an entry such as arr[2], i am give just a 'J', why is it that the elements arnt single entries like I would expect?

Can anyone lend a hand?

Thanks!

Update

$.get("get.php", function(data){

     arr = $.parseJSON(data);
     alert(arr);
}, "json");

does not seem to return results?

It seems the data was not parsed to JavaScript array yet and arr is still a string. Otherwise, alert(arr) would output James, Lydia, John. — Felix Kling
– Felix Kling, Commented Apr 13, 2012 at 15:07
As a side note - var arr = new Array(); doesn't do anything, since you'll assign another value to that variable afterwards anyway. — Dmytro Shevchenko
– Dmytro Shevchenko, Commented Apr 13, 2012 at 15:09

BenM · Accepted Answer · 2012-04-13 15:07:22Z

4

data contains a string of the JSON, so arr[2] will be the third character in the string. You need to use $.parseJSON(data).

answered Apr 13, 2012 at 15:07

BenM

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4 Comments

Felix Kling Over a year ago

So the question is why does passing the data type argument not work?

James Dunay Over a year ago

would this be how I would use it in this instance? arr = ($.parseJSON(data));

BenM Over a year ago

Yep, that's correct. No need for the wrapper, though: arr = $.parseJSON(data);

James Dunay Over a year ago

Hmmm, it doesnt seem to be working, would you take a look at the updated code in the question?

Quentin · Accepted Answer · 2012-04-13 15:08:20Z

3

Your PHP is claiming that the JSON is HTML. You need to explicitly say it is JSON to get jQuery to handle it as such automatically.

header("Content-type: application/json");

answered Apr 13, 2012 at 15:08

Quentin

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1 Comment

Felix Kling Over a year ago

Shouldn't it work nevertheless if the data type argument is provided?

sod · Accepted Answer · 2012-04-13 15:11:00Z

0

Send the json header in php via header('Content-type: application/json'); so jQuery should automatically decode your json data.

answered Apr 13, 2012 at 15:11

sod

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Comments

ninean · Accepted Answer · 2012-04-13 15:27:28Z

0

you can't gave the data value to the array.

 $.get("get.php", function(data){
     alert(data[2]);
}, "json");

answered Apr 13, 2012 at 15:27

ninean

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