在这部分代码中，分数表和个人资料页面都显示了我想要的所有内容，但是它显示了每个人的详细信息，例如
名字 姓 电子邮件 类别 用户名 但是，我希望用户登录时只能看到自己的详细信息
这是分数.php
<?php
error_reporting(E_ALL);
ini_set('display_errors', TRUE);
ini_set('display_startup_errors', TRUE);
require("db_connect.php");
session_start();
?>
<!doctype html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">
<meta name="description" content="">
<meta name="author" content="">
<link rel="icon" href="../../../../favicon.ico">
<?php
$con=mysqli_connect("localhost","username","Password","Database");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Score where 'Username' LIKE _['Username']");
echo "<table border='1'>
<tr>
<th>ID</th>
<th>Username</th>
<th>Score</th>
<th>Gamedate</th>
<th>QuizTitle</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['ID'] . "</td>";
echo "<td>" . $row['Username'] . "</td>";
echo "<td>" . $row['Score'] . "</td>";
echo "<td>" . $row['Gamedate'] . "</td>";
echo "<td>" . $row['QuizTitle'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
这是 profile.php
<?php
include_once 'db_connect.php';
?>
<!DOCTYPE HTML>
<html lang="en">
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">
<meta name="description" content="">
<meta name="author" content="">
<link rel="icon" href="../../../../favicon.ico">
<title>Profile page </title>
<?php
$con=mysqli_connect("localhost","Username","password","database");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT `FirstName`,`Surname`,`Email`,`Username`,`Date_Creation`FROM Users ");
while($row = mysqli_fetch_array($result))
{
echo "<br />Your <b><i>Profile</i></b> is as follows:<br />";
echo "<b>First name:</b> ". $row['FirstName'];
echo "<br /><b>Last name:</b> ".$row['Surname'];
echo "<br /><b>Email:</b> ".$row['Email'];
echo "<br /><b>Year:</b> ".$row['Username'];
echo "<br /><b>Date created :</b> ".$row['Date_Creation'];
}
mysqli_close($con);
?>
</main>
</html>
这些是我尝试运行该页面时收到的错误。在我更改选择查询之前，它选择了当前登录的用户，但它显示了每个人的信息
最佳答案
SQL 中的单引号用于引用特定字符串，如 'Bob'。如果您尝试检查列的值，请勿将列名称放在引号中，如 'Username'。相反，请将列名称不带引号: where Username like
关于php - 使用 PHP 和 MySQL 仅显示当前用户的详细信息，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50416027/