Question

以下代码有问题。我想要做的是用您按下编辑按钮的行中的数据填充模式表格。直到我尝试对表进行重新排序为止，脚本都可以正常工作：对表进行重新排序时，脚本会获取预先重新排序的表的信息。

这是单击编辑按钮时调用的脚本：

$(document).ready( function () {
        //create datatable
        //declaring a variable without var makes it global
        table_movimenti = $('#movimenti').DataTable();
        $('#movimenti table tbody tr td').on('click', function () {
                $("#transaction_ID").val($(this).find("td:eq(6)").attr('id'));
                $("#data_cont").val($(this).find("td:eq(0)").text());
                $("#data_valuta").val($(this).find("td:eq(1)").text());
                $("#importo").val($(this).find("td:eq(2)").text());
                $("#divisa").val($(this).find("td:eq(3)").text());
                $("#causale").val($(this).find("td:eq(4)").text());
                var categoria=$.trim($(this).find("td:eq(5)").text());
                $("#categoria option").each(function() {
                        if($(this).text() == categoria){
                                //$(this).prop("selected", true);
                                $(this).attr("selected", "selected");
                                return;
                        }
                });
        });

这是表格：

<?php 
echo "<table id='movimenti' class='table table-bordered table-striped table no-margin'>";
echo "<thead>";
echo "<tr>";
echo "<th>Data contabile</th>";
echo "<th>Data valuta</th>";
echo "<th>Importo</th>";
echo "<th>Divisa</th>";
echo "<th>Causale</th>";
echo "<th>Categoria</th>";
echo "<th>Edit</th>";
echo "</tr>";
echo "</thead>";
echo "<tbody>";
  while($row = mysqli_fetch_array($result)){
		echo "<tr class='table_row'>";
		echo "  <td>" . $row['Data_cont'] . "</td>";
		echo "  <td>" . $row['Data_valuta'] . "</td>";
		echo "  <td>" . $row['Importo'] . "</td>";
		echo "  <td>" . $row['Divisa'] . "</td>";
		echo "  <td>" . $row['Causale'] . "</td>";
		echo "  <td>";
		foreach ($categories as $value){
			if ($value['ID'] == $row['Categoria']){
			echo "  <span class='label label-success'>";
			echo "". $value['Descrizione'] ."";
			echo "<input type='text' class='label label-success' id='categoria' name='categoria' value='". $value['Descrizione'] ."'/>";
          echo "</span></td>";
		}
}

Answer 1

最后，我完全改变了方法，现在我在datatable中使用ajax选项构建表。这样，我可以使用相同的ajax选项检索该行中的所有数据。

数据表重新排序问题

1 个答案: