Question

在我的源代码当我按下删除按钮时，内部的代码没有执行。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>

Answer 1

也许这个

 name"submit"

是问题吗？

Answer 2

尝试：

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


}

Answer 3

首先，html不正确 - ＆gt;名称“提交”必须是name =“submit”

其次，我建议你首先检查是否设置了$ _POST ['submit']，如果是，那么php会显示表单。

PHP ISSET函数未执行

3 个答案: