如何使用PHP和MySQL正確顯示學生的總分分數？

Question

到目前爲止，我認爲我做得很對，但我無法正確顯示grade_points這裏是我在下面顯示的內容。我如何解決這個問題，以便我的代碼能夠正確顯示grade_points？

輸出數據

這裏是我的電流輸出：

user_id Array ([1] => 1 [3] => 2) 
rank Array ([0] => 1 [1] => 3) 
grade_points Array ([0] => [1] =>) 
users Array ([0] => 3 [1] => 2) 
the rank of user_id 3 is #2 

，這裏是我想要的輸出：

user_id Array ([1] => 1 [3] => 2) 
rank Array ([0] => 1 [1] => 3) 
grade_points Array ([0] => 8 [1] => 5) 
users Array ([0] => 3 [1] => 2) 
the rank of user_id 3 is #2 

PHP & MySQL代碼

這是我的PHP & MySQL的代碼。

$i = 1; 

$u = array(); 
$user = array(); 
$rank = array(); 
$gp = array(); 

$dbc = mysqli_query($mysqli,"SELECT SUM(grade_points) as p, grades.grade_points, assignment_grades.*, COUNT(*) as u, users_assignment.user_id 
           FROM users_assignment 
           LEFT JOIN grades ON users_assignment.user_id = grades.letter_grade 
           LEFT JOIN assignment_grades ON grades.id = assignment_grades.grade_id 
           GROUP BY users_assignment.user_id 
           ORDER BY p DESC"); 


if (!$dbc) { 
    print mysqli_error($mysqli); 
} else { 
    while($row = mysqli_fetch_array($dbc)){ 
     $u[$row['user_id']] = $i++; 
     $rank[] = $row['user_id']; 
     $user[] = $row['u']; 
     $gp[] = $row['p']; 
    } 
} 

echo 'user_id '; print_r($u); echo '<br />'; 
echo 'rank '; print_r($rank); echo '<br />'; 
echo 'points '; print_r($gp); echo '<br />'; 
echo 'users '; print_r($user); 

echo "the rank of user_id 3 is" . $u[3]; 

MySQL表

CREATE TABLE assignment_grades ( 
id INT UNSIGNED NOT NULL AUTO_INCREMENT, 
grade_id INT UNSIGNED NOT NULL, 
users_assignment_id INT UNSIGNED NOT NULL, 
PRIMARY KEY (id) 
); 



CREATE TABLE grades ( 
id INT UNSIGNED NOT NULL AUTO_INCREMENT, 
letter_grade VARCHAR NOT NULL, 
grade_points FLOAT UNSIGNED NOT NULL DEFAULT 0, 
PRIMARY KEY (id) 
); 



CREATE TABLE users_assignment (
id INT UNSIGNED NOT NULL AUTO_INCREMENT, 
user_id INT UNSIGNED NOT NULL, 
assignment_content LONGTEXT NOT NULL, 
grade_average VARCHAR DEFAULT NULL, 
PRIMARY KEY (id) 
); 

TABLE DATA

個

assignment_grades

id  grade_id  users_assignment_id 
15  15    35 
16  16    35 
17  17    33 

等級

id  letter_grade  grade_points 
15  C+     3 
16  A+     5 
17  A+     5 

users_assignment

id  user_id  assignment_content  grade_average 
32  1   some content   NULL 
33  1   some content   A+ 
34  3   some content   NULL 
35  3   some content   B+ 
36  1   some content   NULL 

Answer 1

我認爲這個問題是在您的查詢。嘗試更換SUM（grade_points）以SUM（grade.grade_points）

Answer 2

你查詢的連接並不匹配字段正確 - 嘗試一下本作查詢：

SELECT SUM(grade_points) as p, grades.grade_points, assignment_grades.*, COUNT(*) as u, users_assignment.user_id 
FROM users_assignment 
LEFT JOIN assignment_grades ON users_assignment.id = assignment_grades.users_assignment_id 
LEFT JOIN grades ON grades.id = assignment_grades.grade_id 
GROUP BY users_assignment.user_id 
ORDER BY p DESC