php - 正在更新错误的数据库记录

Question

也许我累了，也许我只是做错了，但不管怎样，有些事情是不对的。

勾选该复选框时，数据库会更新，但 $id 会被忽略，并且只会更新 pu_time 列为空的第一条记录。应该发生的是当复选框被勾选时，当前时间被插入到具有提供的 id 的相应记录中。我可以回显 $id，但它仍然被忽略。

我已经用文本框成功完成了这项工作....

$sql = "SELECT * FROM  dispatch WHERE driver = '' OR pu_time = '' OR do_time = ''";
    $result = mysql_query($sql);
            $i = 0;
        $color1 = "#2b2823";
        $color2 = "#333333";
        while($row = mysql_fetch_array($result)){
            $id = $row['id'];
            if($i%2 == 1) {$color = $color1;}
                else {$color = $color2;}

                echo "<tr style='background-color: $color'>";
                echo '<td><input name="id" type="hidden" value="'.$row['id'].'" />'.$row['call_time'].'</td>';
                echo "<td>".$row['start_address']."</td>";
                echo "<td>".$row['end_address']."</td>";
                echo "<td>".$row['phone']."</td>";
                echo "<td>".$row['puat']."</td>";
if (isset($row['pu_time']) === true && empty($row['pu_time']) === true){

                        echo '<td>
                             <form id = "pu_time">
                              <input name="pick_up_time" type="checkbox" value="" onchange="this.form.submit()" />
                            </form></td>';
                                if (isset($_GET['pick_up_time'])){
                                    $pick_up = date('g:ia');
                                    $update_p = "UPDATE dispatch SET pu_time = '{$pick_up}' WHERE id = '{$id}'";
                                    mysql_query($update_p,$con)
                                    or die;
                                    unset($_GET['pick_up_time']);
                                    echo '<script type="text/JavaScript">
                                            window.location.href = "new_index.php";
                                          </script>';
                                    }
                        }
                        else {
                            echo "<td>".$row['pu_time']."</td>";
                            }

Answer 1

只需两个更改即可解决您的问题:

将行 ID 添加到隐藏字段中，然后提交:

<?php echo '
<td><form id = "pu_time">
    <input name="pick_up_time" type="checkbox" value="" onchange="if(this.checked)this.form.submit()" />
    <input name="rowId" type="hidden" value="'.$id.'" />
</form></td>';

当然，数据库更新部分将在循环之外，以避免在我们重定向到新页面(或同一页面)时不必要的数据库调用:

if (isset($_GET['pick_up_time']) && !empty($_GET['rowId'])){ // must have id;
    $pick_up = date('g:ia');
    $rowId = $_GET['rowId']; //mark this!
    $update_p = "UPDATE dispatch SET pu_time = '{$pick_up}' WHERE id = '{$rowId}'";
    mysql_query($update_p,$con)
    or die;
    unset($_GET['pick_up_time']);
    echo '<script type="text/JavaScript">
        window.location.href = "new_index.php";
      </script>';
}

 //and then write your code what is currently in the first line: $sql = "SELECT * FROM  dispatch WHERE dr ...... and so;