Question

有几天我忙着解决以下问题。我想通过点击加载引导表到<div>。我的情况：

的index.php

<div id="page-content">
<div id="data" class="animated"></div> on page load, loaddata
<div id="table" class="animated">
<table id="report-table" data-toggle="table"></table>                
</div> hidden, onclick hide #data and show #table
</div>

loaddata.php

$tab_id = $_POST['tab_id'];
$tab_name = $_POST['tab_name'];

$selectTabbladen = $gebruiker_data->runQuery("SELECT * FROM documenten LEFT JOIN relaties ON documenten.relatie_id = relaties.relatie_id LEFT JOIN clienten ON documenten.clienten_id = clienten.clienten_id WHERE documenten.tab_id = $tab_id ORDER BY document_datum"); 

if (!$selectTabbladen->execute()) return false;

if ($selectTabbladen->rowCount() > 0) {
    $tabblad_data = array();
    while ($tabdata = $selectTabbladen->fetch()) {
            $tabblad_data[] = array(
              "id"                  => $tabdata['id'],
              "document_soort"      => $tabdata['document_soort'],
              "voornaam"            => $tabdata['voornaam'],
              "relatie_naam"        => $tabdata['relatie_naam'],
              "tabblad"             => $tabdata['document_status'],
              "status"              => $tabdata['document_status'],
              "aanmaak_datum"       => $tabdata['document_datum'] = date('d M Y H:i:s'),
              "laatst_gewijzigd"    => $tabdata['document_datumgewijzigd'] = date('d M Y H:i:s'),
              );
    }

                print '</tbody>
                    </table>

                ';


    $json_data = json_encode($tabblad_data);

    print_r ($json_data);
}

AJAX

    $('body').on('click', '.tab_data', function () {
    content.hide();
    $('#dataa').show();

    var tab_id = $(this).attr("id");

    $.ajax({
        type: "POST",
        url: "loaddata.php",
        data: {
            tab_id: tab_id
        },
        dataType:"json",
        success : function(data) {   
            $('#report-table').bootstrapTable({
                data: data

            });
        }
    });
});

结果我得到了：找不到匹配的记录

你能帮帮我吗？我做错了什么？

Answer 1

如果您已正确传递了tab_id，那么问题就在于您的查询

而不是

$selectTabbladen = $gebruiker_data->runQuery("SELECT * FROM documenten LEFT JOIN relaties ON documenten.relatie_id = relaties.relatie_id LEFT JOIN clienten ON documenten.clienten_id = clienten.clienten_id WHERE documenten.tab_id = $tab_id ORDER BY document_datum");

试试这个

$selectTabbladen = $gebruiker_data->runQuery("SELECT * FROM documenten LEFT JOIN relaties ON documenten.relatie_id = relaties.relatie_id LEFT JOIN clienten ON documenten.clienten_id = clienten.clienten_id WHERE documenten.tab_id = ".$tab_id." ORDER BY document_datum");

<强>更新

试试这个

$('#report-table').bootstrapTable({
   'load': data
});

BootstrapTable，找不到匹配的记录

1 个答案: