CODEWITHSUNDEEP
javascriptphpjqueryajax
Stanislas Piotrowski
Stanislas Piotrowski

Reputation: 2694

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

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

Upvotes: 0

Views: 295

Answers (2)

Mate Solymosi
Mate Solymosi

Reputation: 5967

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.

Upvotes: 1

AbstractChaos
AbstractChaos

Reputation: 4211

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

Upvotes: 0

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