Question

PHP多个更新记录无效这个程序可以更新多个记录但只有一行，是否有必要修复或添加以使其工作？我需要帮助... TNX

这是代码：

<?php require_once('Connections/tlsc_conn.php'); ?>
<?php
$maxRows_Recordset1 = 10;
$pageNum_Recordset1 = 0;
if (isset($_GET['pageNum_Recordset1'])) {
  $pageNum_Recordset1 = $_GET['pageNum_Recordset1'];
}
$startRow_Recordset1 = $pageNum_Recordset1 * $maxRows_Recordset1;

mysql_select_db($database_tlsc_conn, $tlsc_conn);
$query_Recordset1 = "SELECT * FROM tb_exam";
$query_limit_Recordset1 = sprintf("%s LIMIT %d, %d", $query_Recordset1, $startRow_Recordset1, $maxRows_Recordset1);
$Recordset1 = mysql_query($query_limit_Recordset1, $tlsc_conn) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);

?>

Answer 1

删除它并查看它是否更好

if($row_Recordset1){
header("location:mulupdate.php");
exit;
}

并在表单操作中将其设置为

action="mulupdate.php"

Answer 2

$submit在哪里设置？如果未设置，则您的更新将永远不会运行，$row_Recordset1的值将始终是您的SELECT查询的结果。这意味着header()语句将立即重定向。

如果重定向到同一页面，您将获得循环。

您可能需要在代码顶部使用$submit = $_GET['submit']，然后再使用

if($submit){
  for($i=0;$i<$count;$i++){
    $sql1="UPDATE $tbl_name SET name='$name[$i]', lastname='$lastname[$i]',   email='$email[$i]' WHERE id='$id[$i]'";
    $row_Recordset1=mysql_query($sql1);
  }

  // note: this moved inside if($submit) block
  if($row_Recordset1){ 
    header("location:mulupdate.php");
    exit;
  }
}

（我没有对此进行测试，因此逻辑可能需要进行调整）

Answer 3

也未设置变量：

$name = $_POST['name'];
$lastname= $_POST['lastname'];
...

Answer 4

试试这个，它应该适合你

<?php 
 require_once('Connections/tlsc_conn.php');
 mysql_select_db($database_tlsc_conn, $tlsc_conn);

 if(isset($_POST['submit'])) {

    $count = count($_POST['name']);

    for($i=0;$i<$count;$i++){
          $sql1="UPDATE $tbl_name SET name='$name[$i]',lastname='$lastname[$i]', email='$email[$i]' WHERE id='$id[$i]'";
        $row_Recordset1=mysql_query($sql1);
    }

    if($row_Recordset1){
            header("location:mulupdate.php");
            exit;
    }   
 }

 $query_Recordset1 = "SELECT * FROM tbl_name";
 $Recordset1 = mysql_query($query_Recordset1, $tlsc_conn) or die(mysql_error());
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org    /TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
 <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
 <title>Untitled Document</title>
</head>

<body>
<form name="form2" method="post" action="">
  <p>&nbsp;</p>
  <table width="634" border="1">
    <tr>
       <td>id</td>
       <td>name</td>
       <td>lastname</td>
       <td>email</td>
    </tr>
    <?php while ($row_Recordset1 = mysql_fetch_assoc($Recordset1) { ?>
    <tr>
      <td><?php $id[]=$row_Recordset1['id']; ?><?php echo $row_Recordset1['id'];      ?></td>
      <td><input name="name[]" type="text" value="<?php echo $row_Recordset1['name'];   ?>"></td>
       <td><input name="lastname[]" type="text" value="<?php echo   $row_Recordset1['lastname']; ?>"></td>
       <td><input name="email[]" type="text" value="<?php echo  $row_Recordset1['email']; ?>">
       </td>
    </tr>
     <?php } ?>

   </table>
   <p>
    <label>
    <input type="submit" name="submit" value="Submit" />
     </label>
   </p>
</form>
 </body>
</html>

PHP多个更新记录

4 个答案: