Question

也许我已经累了，也许我只是在犯这个错误，但不管怎么说都不对。

勾选复选框后，数据库会更新，但忽略$ id，只更新具有空pu_time列的第一条记录。应该发生的是当勾选复选框时，当前时间被插入到提供了id的相应记录中。我可以回复$ id，但它仍然被忽略。

我已成功使用文本框进行此工作....

$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

只需进行两项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，以避免不必要的数据库调用，因为我们正在重定向到新页面（或同一页面）：

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;

错误的DB记录正在更新

1 个答案: