<?php
$connection = mysqli_connect("localhost","root","","loginapp");
if(!$connection){
echo "Connection Failed.";
}
$query = "SELECT * FROM users";
$result = mysqli_query($connection,$query);
if(!$result){
die("Query dismissed".mysqli_error($connection));
}
if(isset($_POST['submit'])){
$username = $_POST['name'];
$password = $_POST['password'];
$id = $_POST['id'];
$query = "UPDATE users SET username = '$username' password = '$password' WHERE id = $id ";
$result = mysqli_query($connection, $query);
if(!$result){
die("Query FAILED" . mysqli_error($connection));
}
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Document</title>
</head>
<body>
<form action = "Update.php" method = "post">
<input type = "name" name = "name" placeholder = "Type your name">
<input type = "password" name = "password" placeholder = "Type your password"><br>
<select name="id" id="">
<?php
while($row = mysqli_fetch_assoc($result)){
$id = $row["id"];
echo "<option value='$id'>$id</option>";
}
?>
</select>
<input type = "submit" name = "submit" value = "Update">
</form>
</body>
</html>
Output result: Failed
输出结果:失败
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'password = 'ffffffffffff' WHERE id = 1' at line 1
您的SQL语法有错误;检查与MariaDB服务器版本对应的手册,找到正确的语法,在第1行使用“password =”附近的“ffffffffffffffffff”,其中id = 1
1 个解决方案
#1
4
It's missing a comma after username = '$username'
.
它在username = '$username'后面缺少一个逗号。
The query should look like: UPDATE users SET username = '$username', password = '$password' WHERE id = $id
查询应该是这样的:更新用户设置用户名= '$username',密码= '$password', id = $id。
#1
4
It's missing a comma after username = '$username'
.
它在username = '$username'后面缺少一个逗号。
The query should look like: UPDATE users SET username = '$username', password = '$password' WHERE id = $id
查询应该是这样的:更新用户设置用户名= '$username',密码= '$password', id = $id。