我的表有两列，用户id和用户名。用户以如下用户名登录：
<?php
require('dbConnect.php');
$username = $_POST['username'];
//need to keep this in a session, for other pages later on
session_start();
$_SESSION['username'] = $username;
$sql = "SELECT * FROM user WHERE username = '$username'";
$result = mysqli_query($con,$sql);
$check = mysqli_fetch_array($result);
if(isset($check)) :
//if the username exists in the database, then show a html submit button
$con->close();
?>
<html>
<body>
<form action="UserDetails.php" method="post">
<input type="submit">
</form>
</html>
<?php else :{
//if user is not in db, show this message
echo 'Sorry about that, you can't come in.';
}
$con->close();
?>
<?php endif; ?>
如何获取对应于用户名的用户id，以便以后在页面中使用？
最佳答案
这里有几个问题：
1）如果您使用的是"SELECT * FROM user WHERE username = $1，那么username列应该是唯一的，并且不要忘记转义您的输入：
<?php
$username = mysqli_real_escape_string($con, $_POST['username']);
$sql_query = mysqli_query($con,"SELECT * FROM user WHERE username = '$username'");
2）您可以检查是否有任何结果b4任何东西，并直接从数据库分配这些变量。。。
if (mysqli_num_rows($sql) == 0) {
$error = "This user doesn't exists here...";
}else{
$user_info = mysqli_fetch_assoc($sql_query);
$_SESSION['username'] = $user_info['username'];
$_SESSION['user_id'] = $user_info['user_id'];
}
$con->close();
?>
然后，如果
$_SESSION['user_id']确实存在，则可以为登录用户输出所需的任何内容。。。<html>
<body>
<form action="UserDetails.php" method="post">
<input type="submit">
</form>
</body>
<?php
echo 'Well, here\'s the output: <b>'. (isset($_SESSION['user_id']) ? $_SESSION['username'] : $error).'</b>';
关于php - 如何获取数据库中特定行的相应user_id(自动增量)值？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/39778858/