Modify input fields with JSON data

Question

I've got the following code :

$.post("docsel.php", {id : id}, function(data) {
   $('input#titre').val(data.titre);
   $('input#content').val(data.content);
});

docsel.php returns a JSON encoded array with titre and content and their values. I checked that returned data is correct.

Now I'd like to modify input fields with results, but my code doesn't work. What am I doing wrong?

EDIT:

Here is the data (i put auteur and not content to simplify) :

"[{"titre":"Dipt\u00e9rosodomanie et t\u00e9trapilotomie radiaire, pourquoi ?","auteur":"Alex"}]"

I modified my code like this :

$.post("docsel.php", {id : id}, showResult(data));
                });
            });

function showResult(data)

{
  var obj = JSON.parse(data);

  var titre = obj.titre[0];

$('#titre').val(titre);

but it doesn't work, I mean input field "titre" is not modified. Please take into account that I am an absolute beginner in JQuery / Php .

Not an answer, but you don't need to select the input, because id is always unique: $('#titre'). Or are you saying it isn't unique? — PeeHaa
– PeeHaa, Commented Feb 20, 2012 at 17:15
What do you mean by "doesn't work"? That doesn't help us at all. — gen_Eric
– gen_Eric, Commented Feb 20, 2012 at 17:18
Did you edit Mick Hansen's answer with your code? Don't do that. Add it to the question. — gen_Eric
– gen_Eric, Commented Feb 20, 2012 at 21:12

gen_Eric · Accepted Answer · 2012-02-20 21:12:56Z

2

$.post("docsel.php", {id : id}, function(data) {
   $('input#titre').val(data.titre);
   $('input#content').val(data.content);
}, "json");

I believe you do need to specify when you want the returned data to be parsed as JSON. You could test the data returned via console.log(data) if it outputs a string and not an object, it hasn't been parsed.

edited Feb 20, 2012 at 21:12

gen_Eric

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answered Feb 20, 2012 at 17:19

Mick Hansen

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1 Comment

gen_Eric Over a year ago

+1, this is (probably) the issue, I've done this myself way too many times.

gen_Eric · Accepted Answer · 2012-02-20 21:19:16Z

0

$.post("docsel.php", {id : id}, showResult(data));

This runs the showResult function, and then uses its return value (undefined) as the callback.

It should be:

$.post("docsel.php", {id : id}, function(data){
  showResult(data)
});

Or, better yet:

$.post("docsel.php", {id : id}, showResult);

You don't need to use JSON.parse yourself. Just pass 'json' to $.post and it'll do that for you.

$.post("docsel.php", {id : id}, showResult, 'json');

function showResult(data){
    var titre = data[0].titre; // Data is an array of objects
    $('#titre').val(titre);
}

answered Feb 20, 2012 at 21:19

gen_Eric

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1 Comment

Alex Leguevaques Over a year ago

It works ! For some unknown reason, the syntax : ` $('#titre').val(titre); ` doesn't work for me. I tried $('[name=titre]').val(titre); and my problem was solved. Thank you very much for your help.

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