Explore our sites

Stack Exchange
tour help
careers 2.0
Take the 2-minute tour ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free, no registration required.
up vote 0 down vote favorite

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?

share|improve this question
 
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 Apr 13 '12 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. –  Shedal Apr 13 '12 at 15:09
add comment

4 Answers

up vote 4 down vote

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

share|improve this answer
 
So the question is why does passing the data type argument not work? –  Felix Kling Apr 13 '12 at 15:08
 
would this be how I would use it in this instance? arr = ($.parseJSON(data)); –  James Dunay Apr 13 '12 at 15:17
 
Yep, that's correct. No need for the wrapper, though: arr = $.parseJSON(data); –  BenM Apr 13 '12 at 15:17
 
Hmmm, it doesnt seem to be working, would you take a look at the updated code in the question? –  James Dunay Apr 13 '12 at 15:20
add comment
up vote 3 down vote

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");
share|improve this answer
 
Shouldn't it work nevertheless if the data type argument is provided? –  Felix Kling Apr 13 '12 at 15:09
add comment
up vote 0 down vote

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

share|improve this answer
add comment
up vote 0 down vote

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

 $.get("get.php", function(data){
     alert(data[2]);
}, "json");
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.