如何写mysql的innerquery时，你得到的记录来自同一个表计数

Question

查询到从2个表得到的结果：

SELECT path.* FROM (SELECT tbc.course_name 
slp.course_id,slp.student_type,slp.stu_reference_id, 
count(slp.course_id) as counttotalWatchedStudents 
from tbl_student_learning_path slp LEFT JOIN tbl_courses tbc 
on tbc.course_pid = slp.course_id WHERE slp.stu_reference_id =34 
and slp.student_type='institute' GROUP BY slp.course_id) as path 

查询结果表：

| course_id      | totalCollegeStudents | fullwatchedStudentsCount | counttotalWatchedStudents | sumOfWatchPointsForWatchedStudents | totalStudentsAvg | fullwatchedAvg | 
|-------------------------------|----------------------|--------------------------|---------------------------|------------------------------------|------------------|----------------| 
| Number Systems    | 9     | 0      | 3       | 60         | 20.0000   | 0    | 
| Percentages     | 9     | 0      | 3       | 30         | 10.0000   | 0    | 
| Blood Relations    | 9     | 3      | 3       | 300        | 100.0000   | 300   | 
| Calandar      | 9     | 3      | 3       | 300        | 100.0000   | 300   | 
| Percentages     | 9     | 3      | 3       | 300        | 100.0000   | 300   | 
| Permutation & Combination | 9     | 3      | 3       | 300        | 100.0000   | 300   | 
| Probability     | 9     | 0      | 3       | 90         | 30.0000   | 0    | 
| Ratios      | 9     | 0      | 3       | 120        | 40.0000   | 0    | 
| Time and Work     | 9     | 0      | 3       | 150        | 50.0000   | 0    | 
| Time Speed & Distance  | 9     | 1      | 3       | 140        | 46.6667   | 100   | 
| Averages      | 9     | 3      | 3       | 300        | 100.0000   | 300   | 
| Coding and Decoding   | 9     | 3      | 3       | 300        | 100.0000   | 300   | 

从上面的表格，我想添加内部查询到主查询 查询应该是这样的：

(select count(t1.watched_percentage) from tbl_student_learning_path t1 
WHERE t1.stu_reference_id =34 and t1.student_type='institute' 
AND t1.watched_percentage >= 90 group by t1.course_id ) 
fullwatchedStudentsCount, 

而且将R （这不过是如果第一桌counttotalWatchedStudents值是3，这意味着有2种类型的人 1.学生观看完整（watched_percentage> = 90） 2.学生仍在观看（watched_percentage 1-89） ）

| fullwatchedStudentsCount | fullwatchedStudentsSum | 
|--------------------------|------------------------| 
| 2      | 200     | 
| 1      | 100     | 
| 0      | 0      | 
| 2      | 200     | 
| 1      | 100     | 
| 1      | 100     | 
| 0      | 0      | 
| 2      | 200     | 
| 2      | 200     | 
| 1      | 100     | 
| 2      | 200     | 
| 1      | 100     | 

Answer 1

如果我理解正确的话，你要总结其在另一列看见超过90％的学生：

我会做这样的：

SELECT tbc.course_name, 
     slp.course_id, 
     slp.student_type, 
     slp.stu_reference_id, 
     count(slp.course_id) AS counttotalWatchedStudents, 
     sum(if(slp.watched_percentage>=90, 1, 0)) AS fullwatchedStudentsCount 
     sum(if(slp.watched_percentage>=90, watched_percentage, 0)) AS fullwatchedStudentsSum 
FROM tbl_student_learning_path slp 
LEFT JOIN tbl_courses tbc ON tbc.course_pid = slp.course_id 
WHERE slp.stu_reference_id =34 
    AND slp.student_type='institute' 
GROUP BY slp.course_id 
enter code here 

希望这可以帮助

Answer 2

所以你想要计算学生与watched_percentage >= 90？使用条件汇聚了这一点：

SELECT 
    tbc.course_name, 
    slp.course_id, 
    slp.student_type, 
    slp.stu_reference_id, 
    COUNT(*) as counttotalWatchedStudents, 
    SUM(slp.watched_percentage >= 90) as fullwatchedStudentsCount, 
    SUM(slp.watched_percentage < 90) as stillwatchedStudentsCount 
FROM tbl_student_learning_path slp 
LEFT JOIN tbl_courses tbc ON tbc.course_pid = slp.course_id 
WHERE slp.stu_reference_id = 34 
    AND slp.student_type= 'institute' 
GROUP BY slp.course_id; 

MySQL的对待真= 1，假= 0，这样你就可以简单地总结了trues :-)