创建删除 php 表单
我在尝试使用PHP中创建删除表单以使用提供给客人的ID删除数据时遇到问题。我是这种东西的新手。
这是错误：
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
这是我的删除.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>" . $conn->error;
}
?>
<!DOCTYPE html>
<html>
<body>
<a href='index.php'>Home</a><br>
<h2>Delete User</h2>
<form action="/delete.php" method="post">
Guest ID:<br>
<input type="text" name="id" >
<br><br>
<input type="submit" value="Submit">
</form>
</body>
</html>
慕田峪4524236浏览 86回答 2
2回答
-
蝴蝶刀刀
作为最简单的解决方法：<?phpinclude_once('mysql.php');// make sure that form is submitted if (isset($_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>" . $conn->error; }}?>另请注意，将原始 /值传递到查询文本中是不安全的。由于您没有提到您使用的API（PDO或mysqli），因此我在查询中串联了值，但您不得这样做。尽快移至预准备语句。$_POST$_GET$_POST预准备语句版本（我假设您根据比较使用）：mysqli$ins = "DELETE FROM MyGuests WHERE id=? LIMIT 1";$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;} -
慕工程0101907
$id未定义。您正在提交表单，因此它可能应该是：DELETE FROM MyGuests WHERE id={$_POST['id']}你也不应该需要 LIMIT 1。我想它总是只删除1条记录。
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