从php MySQL数据库获取数据不工作

我让我的表单工作得很好，并从数据库中获取数据。但是最近我修改了登录页面中的一些内容，以便将不同的页面分配给不同的用户，一旦他们成功登录，现在我就不会在studentloggedin.php中显示他们的数据（结果）。提前感谢

我认为问题在于我假设的变量

studentloggedin.php

<?php
//code for display marks

$con=mysqli_connect("localhost","kurtfarrugia","1234","kurt_farrugia");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM tblexamresults WHERE admin_usr_name ='$username'");

while($row = mysqli_fetch_array($result))
{
echo "English: " . $row['english'] . " marks";
echo "<br />";
echo "<br />";
echo "Maltese: " . $row['maltese'] . " marks";
echo "<br />";
echo "<br />";
echo "Italian: " . $row['italian'] . " marks";
echo "<br />";
echo "<br />";
echo "PSE: " . $row['pse'] . " marks";
echo "<br />";
echo "<br />";
echo "Social Studies: " . $row['social_studies'] . " marks";
echo "<br />";
echo "<br />";
echo "Physical Education: " . $row['physical_education'] . " marks";
echo "<br />";
echo "<br />";
echo "Drama: " . $row['drama'] . " marks";
echo "<br />";
echo "<br />";
echo "Graphical Communication: " . $row['graphical_communication'] . " marks";
echo "<br />";
echo "<br />";
echo "Computer Studies: " . $row['computer_studies'] . " marks";
echo "<br />";
echo "<br />";
echo "Mathematics: " . $row['mathematics'] . " marks";
echo "<br />";
echo "<br />";
echo "Integrated Science: " . $row['integrated_science'] . " marks";
echo "<br />";
echo "<br />";
echo "Physics: " . $row['physics'] . " marks";
echo "<br />";
echo "<br />";
echo "<br />";

$total = $row['english'] + $row['maltese'] + $row['italian'] + $row['pse'] +       $row['social_studies'] + $row['physical_education'] + $row['drama'] + $row['graphical_communication'] + $row['computer_studies'] + $row['mathematics'] + $row['integrated_science'] + $row['physics'];

$average= $total / 11;

echo "Statistics:"; 
echo "<br />";
echo "<br />";  
echo "Your average mark is: " . round($average, PHP_ROUND_HALF_UP) . " marks";
}

mysqli_close($con);
?>  

---

您的SQL查询中的变量
$username
从未设置

$result = mysqli_query($con,"SELECT * FROM tblexamresults WHERE admin_usr_name ='$username'");

也没有在URL中传递

header( "Location: studentloggedin.php" );

您应该将用户名输入URL，并获取执行查询所需的值。
不要试图猜测问题出在哪里。调试代码以清楚地确定问题的确切位置。检查PHP错误日志，并在每次调用
mysql\u query（）
后检查
mysql\u error（）
。如果仍然注入SQL，则没有理由使用
mysqli
。仔细研究，这样人们就不能将他们想要的任何东西注入到你的代码中。你指的是什么url？这是一个标题（“location:studentloggedin.php”）；但我认为，正如您已经做的那样，将用户名放入会话中，在studentloggedin.php中调用会话_start（），并从会话（而不是从帖子）中获取学生用户名，解决了问题。将变量username分配给my studentloggedin.php页面。谢谢大家！
$result = mysqli_query($con,"SELECT * FROM tblexamresults WHERE admin_usr_name ='$username'");

header( "Location: studentloggedin.php" );