在这部分代码中，score表和profile页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，比如
名字姓氏电子邮件类别用户名，但是，我希望它这样，当用户登录时，他们只能看到自己的详细信息
这是score.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>这些是我尝试运行页面时遇到的错误。在我更改select查询之前，它选择了当前登录的用户，但显示了每个人的信息
3条答案
按热度按时间ttisahbt1#
在你的
score.php文件，在这行：看看你的where语句，似乎是错的。
在你的
profile.php，在这行：你不能传递任何where语句。所以，问题在sql中，特别是在where上。
z18hc3ub2#
sql中的单引号用于引用特定字符串，如
'Bob'. 如果要检查列的值，请不要将列名作为'Username'. 相反，请将列名不带引号：where Username likeyc0p9oo03#
关于错误。您可以使用错误处理来显示查询中的问题。其次，可以使用id获取登录用户的数据信息。您可以通过会话保存用户的id，当用户登录时保存id，然后将其用于where子句。