0

在我的源代码中当我按下删除按钮时，内部 isset 代码未执行。任何人都可以帮助我，

<body>
<form method="post" action="<?php  echo $_SERVER['PHP_SELF'];?>">

<?php

$dbc=mysqli_connect("localhost","root","","elvis_store") or die("Error Connecting to Mysql Database");

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


echo "Hello";
foreach($_POST['todelete'] as $delete_id){

$query="DELETE FROM email_list WHERE id=$delete_id";
mysqli_query($dbc,$query) or die("Error Querying Database");

}

echo "Customer(s) Removed";


}



$query="SELECT * FROM email_list";
$result=mysqli_query($dbc,$query)or die("Query Syntaxt is Incorrect");

while($row=mysqli_fetch_array($result)){

echo '<input type="checkbox" value="' . $row['id'] . '" name="todelete[]" />';
echo $row['first_name']." ".$row['last_name']." ".$row['email'];
echo "<br/>";




}



mysqli_close($dbc);

?>

<input type="submit" name"submit" value="Remove"/>
</form>




</body>
4

3 回答 3

3

也许这

 name"submit"

是问题吗？

于 2012-07-24T08:15:22.383 回答
2

尝试：

if(isset($_POST) && !empty($_POST))  {


}
于 2012-07-24T08:19:49.643 回答
0

首先 html 不正确 -> name"submit" 必须是 name="submit"

其次，我建议您首先检查是否设置了 $_POST['submit'] 以及是否设置了 php else 显示表单。

于 2012-07-24T08:17:17.633 回答