首先,我使用PHP Dreamweaver制作一个带有phpMyAdmin数据库的程序。当我提交按钮时,该程序将我的第一行记录设为空白,并且行中的剩余记录仍然存在,即第二行及以下,但第一行中的记录仍在数据库中,为什么会发生这种情况?我有任何解决这个程序的想法,它也不会更新我的记录。这让我感到恶心,我需要帮助! :( tnx...
这是代码:
<?php require_once('Connections/tlsc_conn.php');
mysql_select_db($database_tlsc_conn, $tlsc_conn);
$query_Recordset1 = "SELECT * FROM tbl_name";
$Recordset1 = mysql_query($query_Recordset1, $tlsc_conn) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);
if(isset($_POST['submit'])) {
$count = count($_POST['id']);
$submit = $_GET['id'];
for($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:lulu.php");
exit;
}
}
?>
<!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="">
<table width="634" border="1">
<tr>
<td>id</td>
<td>name</td>
<td>lastname</td>
<td>email</td>
</tr>
<?php do { ?>
<tr>
<td><?php $id[]=$row_Recordset1['id']; ?><?php echo $row_Recordset1['id']; ?>
<input name="id[]" type="hidden" value="<?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 $row_Recordset1['email']; ?>"> </td>
</tr>
<?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); ?>
</table>
<p>
<input type="submit" name="submit" value="Submit" />
</p>
</form>
<p>
</p>
</body>
</html>
最佳答案
您需要输入:
$name = array_map('mysql_real_escape_string', $_POST['name']);
$lastname = array_map('mysql_real_escape_string', $_POST['lastname']);
$email = array_map('mysql_real_escape_string', $_POST['email']);
$id = array_map('mysql_real_escape_string', $_POST['id']);
在 for
循环之前。您正在使用这些变量,但从未从表单输入中填写它们。
关于PHP 多重更新函数未更新并获取第一行空白,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18885632/