Question

我尝试使用ajax从数据库获取数据，我认为我在ajax网址中存在问题

我的json结果在这个网址

http://127.0.0.1:8888/manage/JournalVente.php?method=jason&popup=1

结果是

[{"IDCUSTOMER":"255","NBORDER":"10","DATE_SAVE":"0000-00-00 00:00:00","DATE_SEND":"0000-00-00 00:00:00","COMPANY":"haboussa","PAYMENT_TYPE":"CHQ","TOTAL_TTC":"0.00"}]

我的代码是

function Load() {
    $('#update_panel').html("<img src='../static/loading.gif' />");

    $.ajax({
        url: 'http://127.0.0.1:8888/manage/JournalVente.php?method=jason&popup=1',
        type: 'GET',
        dataType: 'json',
        success: function (d) {
            // alert(d.length);
            if (d.length > 0) {
                var $data = $('<table></table').addClass('table table-responsive table-striped');
                var header = "<thead><tr><th width='5px' class='no_sort'></th><th width='5px' class='no_sort'></th><th width='5px' class='no_sort'></th><th>R&eacute;f&eacute;rence</th><th>Date saisie</th><th>Date &eacute;ch&eacute;ance</th><th>Client</th><th>Type de paiement</th><th>Montant</th></tr></thead>";
                $data.append(header);       

                $.each(d, function (i, row) {
                    //alert(row.FIRSTNAME);
                    var $row = $('<tr/>');
                    $row.append($('<td/>').html("<a href='invoice.php?idcustomer=" + row.IDCUSTOMER + "'&idorder=" + row.IDORDER + "'&print=1' target='_blank' ><img src='../static/css/actions/pdf.png' border='0' /></a>"));
                    $row.append($('<td/>').html("<a href='?method=update_details&idcustomer=" + row.IDCUSTOMER + "'&idorder=" + row.IDORDER + "'><img src='../static/css/actions/edit.png' border='0' /></a>"));
                    $row.append($('<td/>').html("<a  href='?method=delete&idorder=" + row.IDORDER + "'&popup=1' class='popup-delete'><img src='../static/css/actions/delete.png' border='0' /></a>"));
                    $row.append($('<td/>').html(row.NBORDER));
                    $row.append($('<td/>').html(row.DATE_SAVE));
                    $row.append($('<td/>').html(row.DATE_SEND));
                    $row.append($('<td/>').html(row.COMPANY));
                    $row.append($('<td/>').html(row.PAYMENT_TYPE));
                    $row.append($('<td/>').html(row.TOTAL_TTC));
                    $data.append($row);
                });
                alert($data)
                $('#update_panel').html($data);
            }
            else {
                var $noData = $('<div/>').html('No Data Found!');
                $('#update_panel').html($noData);
            }
        },
        error: function () {
            alert('Error! Please try again.')
        }
    });
}

它总是给我警报错误！请再试一次。

如何将参数传递给ajax网址

0 个答案: