创建删除php表单_Php_Mysql - Fatal编程技术网

创建删除php表单

php mysql

创建删除php表单,php,mysql,Php,Mysql,我在尝试用PHP创建一个delete表单以使用给来宾的ID删除数据时遇到了问题。我是这方面的新手以下是错误： Error: DELETE FROM MyGuests WHERE id= LIMIT 1 You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'LIMIT 1' at

我在尝试用PHP创建一个delete表单以使用给来宾的ID删除数据时遇到了问题。我是这方面的新手

以下是错误：

Error: DELETE FROM MyGuests WHERE id= LIMIT 1
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'LIMIT 1' at line 1

以下是我的delete.php：

<?php

include_once('mysql.php');

$ins= "DELETE FROM MyGuests WHERE id=$id LIMIT 1";

if ($conn->query($ins) === TRUE) {
    echo "Record deleted successfully";
} else {
    echo "Error: " . $ins . "<br>" . $conn->error;
}

?>

<!DOCTYPE html>
<html>
<body>
<a href='index.php'>Home</a><br>

<h2>Delete User</h2>

<form action="/delete.php" method="post">
  Guest ID:<br>
  <input type="text" name="id" >
  <br><br>
  <input type="submit" value="Submit">
</form> 

</body>
</html>




删除用户
来宾ID:

作为最简单的修复：

<?php

include_once('mysql.php');

// make sure that form is submitted    
if (isset($_POST['id'])) {

    $ins = "DELETE FROM MyGuests WHERE id=" . $_POST['id'] . " LIMIT 1";
    if ($conn->query($ins) === TRUE) {
        echo "Record deleted successfully";
    } else {
        echo "Error: " . $ins . "<br>" . $conn->error;
    }
}?>

$id

未定义。您正在提交一个表单，因此它可能是：

DELETE FROM MyGuests，其中id={$\u POST['id']}

你也不需要限制1。我想它总是只删除一条记录。

您缺少定义$id的实际内容，所以添加$id=。。。请仔细阅读并使用准备好的声明

$ins = "DELETE FROM MyGuests WHERE id=? LIMIT 1";
$st = $conn->prepare($ins);
$st->bind_param('i', $_POST['id']);

if ($st->execute() === TRUE) {
    echo "Record deleted successfully";
} else {
    echo "Error: " . $ins . "<br>" . $conn->error;
}