How to use loop for multiple arrays imported from Database

Question

In server side I give table data from MySql and I send it with json_encode to JQuery:

<?php
include 'DB.php';

$result20 = mysql_query("SELECT * FROM Gallery WHERE Section = 'Chosen' AND ID = 19");          
$array20 = mysql_fetch_row($result20);                          

$result19 = mysql_query("SELECT * FROM Gallery WHERE Section = 'Chosen' AND ID = 19");          
$array19 = mysql_fetch_row($result19);  

$data = array();
$data['Div20'] = $array20;
$data['Div19'] = $array19;
echo json_encode($data);
?>

json_encode export this arrays: {"Div20":["Image20","20.jpg"],"Div19":["Image19","19.jpg"]}

but, in client side I need use a loop for use all arrays in events. When I use for, it's not work with multiple arrays, how to do it?

$(function() {
  $.get('data.php' ,function(response)
  {
     var data = jQuery.parseJSON(response);
     var array;
     for(array in data)
     {
       var ImageID = data.array[0];
       var ImageSrc = data.array[1];

       $('#'+ImageID ).click(function(){
            //some codes
       })
     }
   })
})

check with console...it showing any error ?

rynhe
– rynhe

2013-12-19 07:30:48 +00:00
Commented Dec 19, 2013 at 7:30 — rynhe
– rynhe, Commented Dec 19, 2013 at 7:30
no it not have any error, but it's not work

MojtabaSh
– MojtabaSh

2013-12-19 07:35:45 +00:00
Commented Dec 19, 2013 at 7:35 — MojtabaSh
– MojtabaSh, Commented Dec 19, 2013 at 7:35

rynhe · Accepted Answer · 2013-12-19 07:37:07Z

1

Try this

$(function() {
  $.get('data.php' ,function(response) {
     var data = jQuery.parseJSON(response);
     $.each( data, function( key, value) {        
       var ImageID  = value[0];
       var ImageSrc = value[1];

       $("#"+ImageID ).click(function(){
            //some codes
       })     
   })
})

answered Dec 19, 2013 at 7:37

rynhe

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1 Comment

MojtabaSh Over a year ago

Thanks for everyone, but this code works very well. many thanks rynhe

Rohan Kumar · Accepted Answer · 2013-12-19 07:36:06Z

1

It will work if you add # to your imageid like,

$('#'+ImageID ).click(function(){
    //some codes
});

I've tried some code for you,

var json={"Div20":["Image20","20.jpg"],"Div19":["Image19","19.jpg"]};
for(div in json){
    ImageID=json[div][0];
    ImageSRC=json[div][1];
    $('#'+ImageID)
           .attr('src',ImageSRC)
           .click(function(){
               alert(this.src);            
           });
}

Demo

answered Dec 19, 2013 at 7:36

Rohan Kumar

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Comments

Magnus Engdal · Accepted Answer · 2013-12-19 07:39:05Z

0

Replace .array with [array] in your for loop:

 for(array in data)
 {
   var ImageID = data[array][0];
   var ImageSrc = data[array][1];
 }

answered Dec 19, 2013 at 7:39

Magnus Engdal

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1 Comment

MojtabaSh Over a year ago

Thanks Mr.Magnus Engdal

Akaryatrh · Accepted Answer · 2013-12-19 07:39:46Z

0

"array" is not an attribute of your json object, it's just an argument of your for ... in loop. So you must use it as a dynamic value. Moreover you missed using a # to target properly the element id.

var data = {"Div20":["Image20","20.jpg"],"Div19":["Image19","19.jpg"]};
for(array in data)
{
    var ImageID = "#"+data[array][0];
    var ImageSrc = data[array][1];
    $(ImageID).on("click",function(){
        //some codes
    });
};

Avoid using $.each() loop as someone recommended above, jQuery loops are usually slower than native loops.

answered Dec 19, 2013 at 7:39

Akaryatrh

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Comments

Yahoo - Jungly · Accepted Answer · 2013-12-19 09:32:32Z

0

Try this..

$(function() {

$.get('data.php' ,function(response) {

var data = jQuery.parseJSON(response);
 $.each(data,function(k, v){
       var ImageID = v[0]; 
       var ImageSrc = v[1];
  $('#'+ImageID ).click(function(){
     //some codes`enter code here`
     })

}) })

answered Dec 19, 2013 at 9:32

Yahoo - Jungly

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