找不到错误:/

时间:2018-01-25 05:48:05

标签: php database mysqli error-handling syntax

有一张包含客户数据的表格。试图合并编辑功能。目前,该表格中包含<a href>“编辑”链接。

        <tr>

            <th><strong>Extras</strong></th>

        </tr>
        <?php while($r = mysqli_fetch_assoc($res)){
        ?>
        <tr> 
            <td><a href="crud/update.php?id='.$r['id'].'">Edit</a></td>
            <td><input type="button" onClick="deleteme(<?php echo $r['u_uid']; ?>)" name="Delete" value="Delete"></td>
             </tr>

   function deleteme(delid)
  { if(confirm("Are you sure you want to Delete?")){
    window.location.href='crud/delete.php';
   }
 } 
 </script>
        <?php } ?>

请注意:

$ReadSql = "SELECT * FROM `contact` WHERE users_id=$uid ORDER BY Name";
$res = mysqli_query($connection, $ReadSql);

我希望将用户带到crud / update.php,它以:

开头
<?php
 error_reporting();
   require_once('connect.php');
   $id = $_GET['id'];
   $SelSql = "SELECT * FROM `contact` WHERE id=$id";
   $res = mysqli_query($connection, $SelSql);
   $r = mysqli_fetch_assoc($res);

if(isset($_POST) & !empty($_POST)){
 $name =  mysqli_real_escape_string($connection,$_POST['name']);
  //and other credentials 

  $UpdateSql = "UPDATE `contact` SET Name='$name', Company='$comp', 
 Title='$title', Phone='$phone', Email='$email', Address='$location' WHERE id=$id"; 

拔出我的头发,因为我无法找到每次按“编辑”链接时出错的原因。

干杯队员。

1 个答案:

答案 0 :(得分:1)

您获得404的原因是您没有正确的链接。

您已在线上混合使用HTML和PHP

<a href="crud/update.php?id='.$r['id'].'">Edit</a>

应该是:

<a href="crud/update.php?id=<?php echo $r['id'] ?>">Edit</a>