使用 php mysql 错误显示和更新数据库中的选定值

Question

**请帮我解决这个问题。这是我的代码。 **

<?php
    $query = "SELECT * FROM `computer_equipment` WHERE ce_category = 'Mouse' AND ce_location = 'Left Room'";
    $result1 = mysqli_query($connect, $query);
    $options = "";
    while($row2 = mysqli_fetch_array($result1)){
        $options = $options."<option>$row2[1] : $row2[2]</option>";
    }
?>
<div class="row form-group">
    <div class="col-sm-3">
        <label class="control-label modal-label">Mouse</label>
    </div>
    <div class="col-sm-9">
        <select class="form-control" name = "mouse">
            <option value="" hidden>Select</option>
            <?php echo $options;?>
            <option value="">N/A</option>
        </select>
    </div>
</div>

它显示从数据库中选择的选项并更新