PHP 从数据库中获取用户名，在页面上显示用户名
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PHP Get username from Database, display username on page
提问by NiceDevelopmentq
I am currently trying to get the username from my database into my webpage and display it. I am not having much luck, this is the code that I have so far. Should work, but the $username variable is undefined, not sure why.
我目前正在尝试从我的数据库中获取用户名到我的网页中并显示它。我运气不好，这是我迄今为止的代码。应该可以工作，但 $username 变量未定义，不知道为什么。
<?php
/* Database Information */
$server = '127.0.0.1';
$database = 'nicedevelopment';
$dbuser = 'root';
/* Database Connection */
$connect = mysql_connect("$server", "$dbuser")
OR die(mysql_error());
mysql_select_db("$database", $connect);
/* Escape String */
$username = mysql_real_escape_string($connect);
/* Find and get row*/
$getit = mysql_query("SELECT * FROM users WHERE username='$username'",$connect);
$row = mysql_fetch_array($getit);
$username = $row['username']
?>
<nav>
<div class="nav-wrapper">
<a href="#" class="brand-logo center"><?php echo $row['username']; ?></a>
This is my login page!
这是我的登录页面！
<?php
require('connect.php');
session_start();
// If form submitted, insert values into the database.
if (isset($_POST['email'])){
$email = $_POST['email'];
$password = $_POST['password'];
$email = stripslashes($email);
$email = mysql_real_escape_string($email);
$password = stripslashes($password);
$password = mysql_real_escape_string($password);
//Checking is user existing in the database or not
$query = "SELECT * FROM `users` WHERE email='$email' and password='".md5($password)."'";
$result = mysql_query($query) or die(mysql_error());
$rows = mysql_num_rows($result);
if($rows==1){
$_SESSION['email'] = $email;
header("Location: dashboard.php"); // Redirect user to index.php
}else{
echo "<div class='form'><h4>Email/password is incorrect.</h4><br/>Click here to <a href='index.php'>Login</a></div>";
}
}else{
?>
回答by The Codesee
Updated Answer
更新答案
Since the user is only able to login with their email, you won't be able to use a sessionto display their username (unless you want to display their email!). Your best bet would be to use a query which selects the username from a table where the email equals the email they used to login with:
由于用户只能使用他们的电子邮件登录，您将无法使用 asession显示他们的用户名（除非您想显示他们的电子邮件！）。最好的办法是使用查询从表中选择用户名，其中电子邮件等于他们用来登录的电子邮件：
<?php
$email = $_POST['email'];
$query = "SELECT * FROM users WHERE email='$email'"
$row = mysql_fetch_array($query);
$username = $row['username'];
echo $username;
?>
Original Answer
原答案
According to your query, you're trying to select a username from the database which is equal to $connect = mysql_connect("$server", "$dbuser")- this doesn't make sense at-all.
根据您的查询，您试图从数据库中选择一个用户名，该用户名等于$connect = mysql_connect("$server", "$dbuser")- 这根本没有意义。
You first want to log the user in, like this:
您首先要让用户登录，如下所示：
<?php
session_start();
if(isset($_POST['submit'])) {
$_SESSION['username'] = $_POST['username'];
}
<form method="post">
<input type="text" name="username">
<input type="password>
<input type="submit" name="submit">
</form>
To then display the name of the user, simply use these two lines of code:
要显示用户名，只需使用以下两行代码：
session_start();
echo $_SESSION['username];
Note: this is a very basic script. It doesn't prevent any forms of SQL Injection Attacks and doesn't validate the username and password - it's just an example to get you on the right track.
注意：这是一个非常基本的脚本。它不会阻止任何形式的 SQL 注入攻击，也不会验证用户名和密码——这只是让您走上正轨的一个例子。
回答by shen
Update
更新
Try this code, fill the password field and let me know about the output. Although it is not clear where is your $username variable coming from.
试试这个代码，填写密码字段并让我知道输出。虽然不清楚您的 $username 变量来自哪里。
Edit: I have seen that you updated your question, so I modified the code based on that. I also edited the code based on the feedback of the questioner.
编辑：我已经看到您更新了您的问题，因此我基于此修改了代码。我还根据提问者的反馈编辑了代码。
<?php
$dbservername = "127.0.0.1";
$dbusername = "root";
$dbpassword = "password";
$dbname = "nicedevelopment";
// Create connection
$conn = new mysqli($dbservername, $dbusername, $dbpassword, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$post_email = mysqli_real_escape_string($conn, $_POST['email']);
$sql = "SELECT * FROM users WHERE email='$post_email'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
$row = $result->fetch_assoc();
echo "username: " . $row['username'] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
