jquery and ajax, how to return an array and display it

Question

I met some trouble in my ajax request.

In fact I do not know how to ou many records.

I've tried this :
    $rqt = "SELECT a,b,c from table";
    $res = mysql_query($rqt);
    while ($data = mysql_fetch_assoc($res)):
    $objet = $data;
    endwhile;
//requettage
    return $objet;

but it out to php only one rows even if I have many records in database.

Second trouble is for display that records.

I've tried this in jquery

$.ajax({
                    type: "POST",
                    url: "requetes_ajax/fetch_debours_detail.json.php",
                    data: "groupe="+groupe_debours,
                    success: function(data){
                    $('.remove').remove()
                    console.log(data);
                    $.each(data, function (key, value) {
                        $('#fetchdebours tr:last').after('<tr class="remove"><td>'+data.libelle_debours+'</td><td>'+data.date+'</td><td>'+data.debours_montant_ht_tva+'</td><td>'+data.debours_montant_ht_no_tva+'</td><td>'+data.debours_montant_ttc+'</td></tr>')
                        //alert(key + ': ' + value);
                    });
                    }
                });

for that things the trouble I have is that instead of displaying all records in one row, it fill all items horizontally and also vertically so I have as many rows as I have column.

Anykind of help will be much appeciated

Mate Solymosi · Accepted Answer · 2013-11-04 11:27:40Z

1

In the while loop you are forever replacing the $objet variable with the latest row, so in the end the latest row is all what you get.

To build an array, use this code:

$objet = array();
while ($data = mysql_fetch_assoc($res)):
    $objet[] = $data;
endwhile;

To pass the data as JSON you should first declare dataType: "json" in your $.ajax call. The second step is to actually output the PHP array as JSON. Try the following code:

$json = json_encode($objet);
print($json);

Since now you are passing an entire array instead of just one object, your display code needs a small adjustment:

$.each(data, function (k, obj) {
    $('#fetchdebours tr:last').after('<tr class="remove"><td>' + obj.libelle_debours + '</td> ... </tr>')
});

I haven't tested the code, use with caution.

edited Nov 4, 2013 at 11:27

answered Nov 4, 2013 at 11:20

Mate Solymosi

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1 Comment

Stanislas Piotrowski Over a year ago

I have encoded the json to output I just had some trouble with the php loop to populate the array and with the each of jquery now it works fine because of you Thank you so much for your help

AbstractChaos · Accepted Answer · 2013-11-04 11:26:00Z

0

From the syntax of you ajax request, you want json to comeback as a response, try this:

$rqt = "SELECT a,b,c from table";
$res = mysql_query($rqt);
$data = mysql_fetch_assoc($res);
return json_encode($data);

see json_encode

requires PHP >= 5.2.0

answered Nov 4, 2013 at 11:26

AbstractChaos

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