Question

这是我调用ajax函数的js。

function gettaluka(id)
{
    if(id){
            $.ajax({
                type:'POST',
                url:'ajaxData.php',
                data:{'id':id},
                success:function(data){  //console.log(html); return;
                    $('#talukaplace').html(data);
                    storetaluka();
                }
            }); 
        }else{
            $('#talukaplace').html('<option value="">Select Taluka</option>');

        }
}

在此，我进行了2个下拉菜单。基于对第一个下拉列表的选择，即赋值，我正在第二个下拉列表中从db中获取数据。

如果我选择选项值1，则必须将其各自的数据存储在数据库表中，例如assign，其col字段为“ cost_assignment”；当我选择选项值2时，必须将其各自的数据插入到同一表中但在不同的列字段为“ proj_assignment”。

<td align="left">
    Assignment
    <span class="redtext">
        *
    </span>
</td>
<td>
    <div class="">
        <div class="form-group">
            <!-- <input type="text" class="form-control" name="distplace1" id="distplace1" placeholder="District Preference 1*" > -->
            <select class="form-control" id="distplace1" name="distplace1" onchange="gettaluka(this.value);" required="">
                <option value="0">
                    Select District
                </option>
                <option value="1">
                    Cost Chargeable
                </option>
                <option value="2">
                    Project Name
                </option>
                <option value="3">
                    Other Recovery
                </option>
            </select>
        </div>
    </div>
</td>
<td>
    <div class="">
        <div class="form-group">
            <select class="form-control" id="talukaplace" maxlength="10" name="talukaplace" required="">
                <option maxlength="10" value="0">
                    Select Taluka
                </option>
            </select>
        </div>
    </div>
</td>

ajaxData.php

$dist_id = $_POST["id"];
if ($dist_id == '1') {
    ?>
        <html>
              <?php
$query   = "select cost_name from hris_cost_centers";
    $results = mysqli_query($conn, $query);
// $rowCount = $query->num_rows;
    if (mysqli_num_rows($results) > 0) {
        echo '<option value="" maxlength="100">Select Taluka</option>';
        while ($row = mysqli_fetch_assoc($results)) {
            echo '<option value="' . $row['cost_name'] . '">' . $row['cost_name'] . '</option>';
        }
    } else {
        echo '<option value="">Select Taluka</option>';
    }
}
;
elseif ($dist_id == '2') {
    $query   = "select bta_proj_name from bta_project";
    $results = mysqli_query($conn, $query);
    if (mysqli_num_rows($results) > 0) {
        echo '<option value="">Select Project</option>';
        while ($row = mysqli_fetch_assoc($results)) {
            echo '<option value="' . $row['bta_proj_name'] . '">' . $row['bta_proj_name'] . '</option>';
        }
    } else {

    }
}

根据要插入到数据库中的下拉列表值的选择，我有2个下拉列表。我可以插入它，但值会插入1列中

如何根据下拉值的选择插入表格列？

0 个答案: