Question

我有以下代码，希望根据所选的下拉值将复选框添加到界面中

    <!--drop down list for the floors--> 
<div class="form-group">
    <label for="ddlFloorNo"><?php echo $_data['add_new_form_field_text_10']; ?> :</label>
    <select onchange="getUnitReport(this.value)" name="ddlFloorNo" id="ddlFloorNo" class="form-control">
        <option value="">--<?php echo $_data['add_new_form_field_text_11']; ?>--</option>
        <?php
        $result_floor = mysqli_query($link, "SELECT * FROM tbl_add_floor order by fid ASC");
        while ($row_floor = mysqli_fetch_array($result_floor)) {
            ?>
            <option <?php
            if ($floor_id == $row_floor['fid']) {
                echo 'selected';
            }
            ?> value="<?php echo $row_floor['fid']; ?>">
                <?php echo $row_floor['floor_no']; ?></option>
        <?php } ?>
    </select>
</div>
 <!--Check boxes to be displayed based on the selected floor inn the drop down list above--> 
    <div class="form-group">
        <label for="ChkOwnerUnit"><?php echo $_data['add_new_form_field_text_8']; ?> : </label>
        <?php
        $result_unit = mysqli_query($link, "SELECT * FROM tbl_add_unit where floor_no ='" . (int) $row_floor['fid'] . "' order by uid ASC");
        while ($row_unit = mysqli_fetch_array($result_unit)) {
            ?>
        <input type="checkbox" class="form-control" name="ChkOwnerUnit[]" value="<?php echo $row_unit['uid']; ?>"/>
            <?php } ?>
    </div>

我的问题是未显示复选框。我怎么可能做错了

Answer 1

这是因为您没有在行中获取数据

$result_unit = mysqli_query($link, "SELECT * FROM tbl_add_unit where floor_no ='" . (int) $row_floor['fid'] . "' order by uid ASC");

在此无法访问

$ row_floor 。您需要在html或创建数组中进行更改以存储ID。

从php中的数据库添加多个复选框不会显示

1 个答案: