Question

好朋友。我正在使用Laravel应用程序，就像我国学校的学生管理系统一样。该申请要求我存储了数十名学生，并进一步生成学生达到该分数的期间或学期的报告。

在我的国家，通常一学年有六（6）个学期。一段时间跨越4周。一个学年还有两（2）个学期，跨越3个半月。每学期有四（4）个学期。

这是一张解释事物外观的图片：

由于我必须在给定学期的某个时期生成每个学生的报告，这就是我的概念模型的样子：

学生表：

students
--------
id *
name
grade_id (fk)

主题表：

subjects
--------
id *
name

成绩表：

grades
--------
id *
name

注意成绩表存储学校中每个班级的名称。恩。 grade 12，grade 11等。

现在，这些是学生在特定时期或学期的科目中得分的表格。

period_one (1st Period)
-----------------------
id *
student_id (fk)
subject_id (fk)
grade_id (fk)
score

first_exam (1st Sem. Exam)
-----------------------
id *
student_id (fk)
subject_id (fk)
grade_id (fk)
score

period_four (4th Period)
-----------------------
id *
student_id (fk)
subject_id (fk)
grade_id (fk)
score

注意同样适用于second period，third period，5th period，six period等

此时我知道有人在这里看到过腥或愚蠢的东西。我在每个表格中重复相同的列，我相信它可以做得更好。

但我有理由这样做，这与生成报告有关。通过上述设置，我可以轻松构建查询以生成报告。

这是我用来生成定期报告的查询

// the function runs a select all query for all the period/term grades tables
// It displays the periodic grades sheet

$student_id = 1; // any id could be passed
$term = "period_one"; // giving the exact table name to fetch record from
$class_id = 2; // 2 may represent grade 12 in this case.

function periodic_report($term, $student_id, $class_id)
{
    global $connection;
    $query = "SELECT $table.student_id, score, subject_name, class_name
                from $term
                INNER JOIN students
                        ON students.student_id = $table.student_id
                    INNER JOIN classes
                        ON classes.class_id = $table.class_id
                    INNER JOIN subjects
                        ON subjects.subject_id = $table.subject_id
              WHERE $table.student_id = $student_id AND $table.class_id = $class_id";

    $result_set = mysqli_query($connection, $query);

    // Test if there was a query/syntax error
    confirm_query($result_set);
    return $result_set;
}

这是结果的样子：

查询以生成第一学期报告：

// the function runs a select all query for semester grades
// It displays the first semester grades sheet
function first_semester_report($student_id) {
    global $connection;
    $query = "SELECT subjects.subject_name, period_one.score, period_two.score, period_three.score, first_exam.score
        FROM subjects

        LEFT JOIN period_one 
            ON period_one.subject_id = subjects.subject_id 
            AND period_one.student_id = {$student_id}

        LEFT JOIN period_two 
            ON period_two.subject_id = subjects.subject_id 
            AND period_two.student_id = {$student_id}

        LEFT JOIN period_three
            ON period_three.subject_id = subjects.subject_id 
            AND period_three.student_id = {$student_id}

        LEFT JOIN first_exam
            ON first_exam.subject_id = subjects.subject_id 
            AND first_exam.student_id = {$student_id}

        LEFT JOIN students
            ON students.student_id = {$student_id}

        WHERE period_one.subject_id = subjects.subject_id OR period_two.subject_id = subjects.subject_id OR period_three.subject_id = subjects.subject_id OR first_exam.subject_id = subjects.subject_id
        ORDER BY subjects.subject_name";      

    $result_set = mysqli_query($connection, $query);

    if(!$result_set){
        die("Database query failed. Error from first semester report: ".mysqli_error($connection)." Error number is: ".mysqli_errno($connection));
    }


    // Test if there was a query/syntax error
    //confirm_query($result_set);
    return $result_set;
}

查询以生成第二学期报告：

// the function runs a select all query for semester grades
// It displays the second semester grades sheet
function second_semester_report($student_id) {
    global $connection;
    $query = "SELECT subjects.subject_name, period_four.score, period_five.score, period_six.score, second_exam.score
        FROM subjects

        LEFT JOIN period_four 
            ON period_four.subject_id = subjects.subject_id 
            AND period_four.student_id = {$student_id}

        LEFT JOIN period_five 
            ON period_five.subject_id = subjects.subject_id 
            AND period_five.student_id = {$student_id}

        LEFT JOIN period_six
            ON period_six.subject_id = subjects.subject_id 
            AND period_six.student_id = {$student_id}

        LEFT JOIN second_exam
            ON second_exam.subject_id = subjects.subject_id 
            AND second_exam.student_id = {$student_id}

        LEFT JOIN students
            ON students.student_id = {$student_id}

        WHERE period_four.subject_id = subjects.subject_id OR period_five.subject_id = subjects.subject_id OR period_six.subject_id = subjects.subject_id OR second_exam.subject_id = subjects.subject_id
        ORDER BY subjects.subject_name";      

    $result_set = mysqli_query($connection, $query);

    if(!$result_set){
        die("Database query failed. Error from second semester report: ".mysqli_error($connection)." Error number is: ".mysqli_errno($connection));
    }


    // Test if there was a query/syntax error
    //confirm_query($result_set);
    return $result_set;
}

这是学期报告根据学期查询的外观：

有时候我很难做到这一点可以做得更好，但遗憾的是我不知道怎么做。

有更好的方法吗？此外，只有70％的学校在其他时期使用不同方式计划学年。有没有办法以学校可以指定学年的期间或期限的方式来定制这个，并为每个期间或期限创建一个表格？

将欣赏实际意义和建议。感谢!!!!

如何让这个概念模型更好？

0 个答案: