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up vote 0 down vote favorite

I have to make a JSON ajax request and have an array in return.

i found this code on multiple other questions (edited to my problem):

var hej[];
function ajax_search(){ 
    $.getJSON('test.php',function(response){
    var data = response.get_response;
    for (i in data){
      hej.push([i,data[i]]);
    }
alert(hej[0]);
}

In the php part, i have a database from which i need the data i have have 5 columns pr. row. I only have one row, and i need all five columns in the array.

$sql = "SELECT * FROM 'table'
   LIMIT 1 OFFSET $i"; //random offset
$result = mysql_query($sql);

$storeArray = Array();
while($row = mysql_fetch_array($result))
{
    $storeArray[0]=$row['column1'];
    $storeArray[1]=$row['column2'];
    $storeArray[2]=$row['column3'];
    $storeArray[3]=$row['column4'];
    $storeArray[4]=$row['column5'];
}
$json = json_encode($storeArray);
echo $json;

I know im doing something wrong, and i am new to programming. I need to be able to access all column values in my javascript after the call and use them in other functions, but i cant get it to return.

I would really appriciate help, with both the javascript and the php if there are errors in either.

share|improve this question
 
What is the test.php script output? (Just open it in a browser and see what it prints) –  Jeroen Moons Aug 27 '12 at 8:33
 
why don't you just json_encode($row)? –  Nemoden Aug 27 '12 at 8:34
 
@Nemaden How can i access every column after that then? –  user1627114 Aug 27 '12 at 8:35
 
I don't see the output if it's not an empty string :) –  Nemoden Aug 27 '12 at 8:35
 
By it's name, e.g. response.column1 –  Nemoden Aug 27 '12 at 8:35
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3 Answers

up vote 0 down vote accepted

It should be:

function ajax_search(){ 
    $.getJSON('test2.php',function(response){
        // response.column1
        // response.column2
        // response.column3

        // for array push:
        var hej = [];
        $.each(response, function(key, val) {
             hej.push(val);
        });
        alert(hej[0]); // it will alert column1
    });

}

and MySQL:

$sql = "SELECT `column1`, `column2`, `column3`, `column4`, `column5` FROM `table` LIMIT 1 OFFSET $i"; //random offset
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
$json = json_encode($row);
echo $json;
share|improve this answer
 
It seems to work, thanks! –  user1627114 Aug 27 '12 at 9:22
add comment
up vote 0 down vote

Try this:

Javascript Code:

var hej[];
function ajax_search(){
    $.getJSON('test.php',function(response){
        $.each(response.results, function(key, val) {
            hej.push(val);
        });
        alert(hej[0]);
    });
}

PHP Code:

<?php
$sql = "SELECT * FROM 'table' LIMIT 1 OFFSET $i"; //random offset
$result = mysql_query($sql);

$storeArray = array();
while($row = mysql_fetch_array($result)) {
    $storeArray[0]= $row['column1'];
    $storeArray[1]=$row['column2'];
    $storeArray[2]=$row['column3'];
    $storeArray[3]=$row['column4'];
    $storeArray[4]=$row['column5'];
}

$json = json_encode( array('results' => $storeArray ) );
echo $json;
?>

Hope this helps.

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

Why use response.get_response?

var hej[];
function ajax_search(){ 
    $.getJSON('test.php',function(data){
    $.each(data, function (i,item) {
      hej.push([i,item]);
    }
alert(hej[0]);
}
share|improve this answer
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