How can i define a variable in Json? - ajax

Question

i want define a variable in Json callback.
Code;
$("select").change(function () {
var $variable = "";
$("select option:selected").each(function () {
$variable += $(this).text() + " ";
});
$("div.yaz").text($variable);
$('#result').html('loading...');
$.getJSON('program-bilgileri.php', function(JSON){
$('#result').empty();
$.each(JSON.$variable, function(i, program){
$('#result')
.append(program.isim +'<br />')
.append(program.bilgi+'<br />')
.append(program.adres+'<hr />');
});
});
})
.trigger('change');
program-bilgileri.php returns;
{
"programlar":[
{
"isim":"Zone Alarm",
"bilgi":"bilgisayarın güvenliğini sağlar",
"adres":"www.zonealarm.com"
},
{
"isim":"Opera",
"bilgi":"güvenli ve hızlı bir web tarayıcısıdır",
"adres":"www.opera.com"
},
{
"isim":"Photoshop",
"bilgi":"güçlü bir imaj işleme yazılımıdır",
"adres":"www.adobe.com"
}
]
}
The problem is here "$.each(JSON.$variable, function(i, program)" if I define $variable in JSON it isn't working.
Any idea?

Answer 1

The problems i see are
Inside the change event you are using $("select option:selected") which finds all select elements in the page, and not the changed one only.
use $(this).children('option:selected') instead.
I am assuming that you are allowing multiple selection on the select element and that is why you are doing += with the $variable.. (you are also adding a space at the end). That means, though, that the variable will be something like "programlar " or "programlar somethingelse".
Your returned JSON though has a key of programlar. A single word, no spaces.. so when you do JSON[$variable] which is the correct way to access an element based on the name in a variable, it does not match.
If the <select> element does not allow multiple selection then the solution is
$("select").change(function() {
var $variable = $(this).children("option:selected").text();
$("div.yaz").text( $variable );
$('#result').html('loading...');
$.getJSON('program-bilgileri.php', function(JSON) {
$('#result').empty();
$.each(JSON[$variable], function(i, program) {
$('#result')
.append(program.isim + '<br />')
.append(program.bilgi + '<br />')
.append(program.adres + '<hr />');
});
});
}).trigger('change');
If indeed it is a multiselect and each option can appear in the JSON then you must check for each option found in the variable.
$("select").change(function() {
var $variable = $(this).children("option:selected").map(function(){
return $(this).text();
}).get();
$("div.yaz").text( $variable.join(' ') );
$('#result').html('loading...');
$.getJSON('program-bilgileri.php', function(JSON) {
$('#result').empty();
for (index=0, length = $variable.length; index < length; index ++) {
$.each(JSON[$variable[index]], function(i, program) {
$('#result')
.append(program.isim + '<br />')
.append(program.bilgi + '<br />')
.append(program.adres + '<hr />');
});
}
});
}).trigger('change');

Answer 2

Try
$.each(JSON['programlar'], function(i, program) ... );
This will iterate over this part of your returned JSON object from PHP:
{
"isim":"Zone Alarm",
"bilgi":"bilgisayarın güvenliğini sağlar",
"adres":"www.zonealarm.com"
},
{
"isim":"Opera",
"bilgi":"güvenli ve hızlı bir web tarayıcısıdır",
"adres":"www.opera.com"
},
{
"isim":"Photoshop",
"bilgi":"güçlü bir imaj işleme yazılımıdır",
"adres":"www.adobe.com"
}