Jquery: Ajax call to a php script returning JSON data

Question

i'm new to JSON. I have a php with a multidimensional array which gets finally encoded in JSON format:

<?php
header('Content-Type: application/json');
$lista = array (
    'Conoscenti'=>array (
        0=>array(
            "Nome"=>"John",
            "Cognome"=>"Doe",
            "Nato"=>array(
                "Giorno"=>16,
                "Mese"=>"Febbraio",
                "Anno"=>1972
            )
        ),
        1=>array(
            "Nome"=>"Elvis",
            "Cognome"=>"Presley",
            "Nato"=>array(
                "Giorno"=>12,
                "Mese"=>"Luglio",
                "Anno"=>1984
            )
        ),
        2=>array(
            "Nome"=>"Mick",
            "Cognome"=>"Jagger",
            "Nato"=>array(
                "Giorno"=>13,
                "Mese"=>"Novembre",
                "Anno"=>1984
            )
        )
     ),
    "Amici"=>array(
        0=>array(
            "Nome"=>"Michael",
            "Cognome"=>"Myers",
            "Nato"=>array(
                "Giorno"=>8,
                "Mese"=>"Dicembre",
                "Anno"=>1986
            )
        ),
        1=>array(
            "Nome"=>"Jim",
            "Cognome"=>"Fear",
            "Nato"=>array(
                "Giorno"=>4,
                "Mese"=>"Febbraio",
                "Anno"=>1985
            )
        )
    )
);
echo json_encode($lista);
?>

I want to load this through Ajax: to do this, i wrote some JQuery code:

var output ="<ul>";
$.ajax({
    url:"lista.php",
    dataType:"json"
}).done(function(data) {
    $.each(data.Conoscenti, function() {
        output +="<li>"+this.Nome+" "+this.Cognome+" è un mio conoscente. &Egrave; nato il "+this.Nato.Giorno+" "+this.Nato.Mese+" "+this.Nato.Anno+"</li>";
    });
});
output += "</ul>";
$("body").html(output);

But the html page shows blank, without even errors. Analyzing the source code it shows only the ul tag, since it's out of the ajax call. Anyone knows how can i fix it? Thank you.

Rory McCrossan · Accepted Answer · 2015-04-08 15:10:34Z

1

The issue is because the promise is executed after the request completes. This means that $('body').html(output) has already been run before you loop over the returned JSON.

To solve this you need to execute all code which depends on the response within the done() handler. Try this:

$.ajax({
    url:"lista.php",
    dataType:"json"
}).done(function(data) {
    var output = "<ul>";
    $.each(data.Conoscenti, function() {
        output += "<li>" + this.Nome + " " + this.Cognome + " è un mio conoscente. &Egrave; nato il " + this.Nato.Giorno + " " + this.Nato.Mese + " " + this.Nato.Anno + "</li>";
    });
    output += "</ul>";
    $("body").html(output);
});

answered Apr 8, 2015 at 15:10

Rory McCrossan

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1 Comment

Daniele Squalo Over a year ago

Another nice explaination: i was reading about that but interrupted. Now iknow why is it important and i'll read very carefully now. Thanks =)

user4764833 · Accepted Answer · 2015-04-08 15:31:16Z

0

    $.ajax({
        url:"lista.php",
        dataType:"json"
    })
    .done(function(data) {
        var output ="<ul>";
        $.each( data.Conoscenti, function() {
           output +="<li>"+ this.Nome+" "+this.Cognome+" è un mio conoscente. &Egrave; nato il "+ this.Nato.Giorno+" "+ this.Nato.Mese+" "+ this.Nato.Anno+"</li>";
        });
        output += "</ul>";
        $("body").html(output); 
    });

answered Apr 8, 2015 at 15:31

user4764833

1

Comments

sircamp · Accepted Answer · 2015-04-08 15:33:47Z

0

The problem occurs because the printing of the variable "out" occurs before the execution of the ajax request. You have to put the printing of the variable within the callback "done".

A good idea, then, would be to have always a branch "done" and a branch "fail" (possibly also the branch "always") so as to manage the success or failure of the request.

Try this:

$.ajax({
 url:"lista.php",
 dataType:"json"
})
.done(function(data) {
   var output = "<ul>";
   $.each(data.Conoscenti, function() {
      output += "<li>" + this.Nome + " " + this.Cognome + " è     un mio conoscente. &Egrave; nato il " + this.Nato.Giorno + " " + this.Nato.Mese + " " + this.Nato.Anno + "</li>";
});
output += "</ul>";
$("body").html(output);
})
.fail(function( jqXHR, textStatus, errorThrown ){
   /*manage your error*/
})
.always(function()){
  console.log("completed");
}

answered Apr 8, 2015 at 15:33

sircamp

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1 Comment

Daniele Squalo Over a year ago

Nice...i suppose i did this because of something i still don't know even after working with JQuery for long time. Well, thank you :D

Eisson · Accepted Answer · 2015-04-08 17:42:38Z

0

Daniele, the problem is that you insert the html out of the done function, please try with the code below:

var output ="<ul>";
$.ajax({
    url:"lista.php",
    dataType:"json"
}).done(function(data) {
    $.each(data.Conoscenti, function() {
        output +="<li>"+this.Nome+" "+this.Cognome+" è un mio conoscente. &Egrave; nato il "+this.Nato.Giorno+" "+this.Nato.Mese+" "+this.Nato.Anno+"</li>";
    });
    output += "</ul>";
    $("body").html(output);
});

answered Apr 8, 2015 at 17:42

Eisson

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