Question

我正在尝试使用id从50个表格中选择FIND_IN_SET。是否可以使用LEFT JOIN以外的其他方式？ 我只给了3张桌子，因为50张桌子不合适。

注意：每个表中都有about_education列。

下面的代码无法正常工作，那么我还要怎么做才能避免使用一千行代码？

nose_1

|id|class|about_education|
| 1|    1|            a01|
| 2|    0|            a02|


coins_2

|id|class|about_education|
| 1|    1|            a01|
| 2|    0|            a02|


money_3

|id|class|about_education|
| 1|    1|            a01|
| 2|    1|            a10|


SELECT COUNT(id) AS id FROM nose_1, coins_2, money_3 WHERE class=1 AND FIND_IN_SET('a01', about_education)");
SELECT COUNT(id) AS id FROM nose_1, coins_2, money_3 WHERE class=1 AND FIND_IN_SET('a02', about_education)");
SELECT COUNT(id) AS id FROM nose_1, coins_2, money_3 WHERE class=1 AND FIND_IN_SET('a02', about_education)");
SELECT COUNT(id) AS id FROM nose_1, coins_2, money_3 WHERE class=1 AND FIND_IN_SET('a03', about_education)");
SELECT COUNT(id) AS id FROM nose_1, coins_2, money_3 WHERE class=1 AND FIND_IN_SET('a04', about_education)");
SELECT COUNT(id) AS id FROM nose_1, coins_2, money_3 WHERE class=1 AND FIND_IN_SET('a05', about_education)");
SELECT COUNT(id) AS id FROM nose_1, coins_2, money_3 WHERE class=1 AND FIND_IN_SET('a06', about_education)");
SELECT COUNT(id) AS id FROM nose_1, coins_2, money_3 WHERE class=1 AND FIND_IN_SET('a07', about_education)");
SELECT COUNT(id) AS id FROM nose_1, coins_2, money_3 WHERE class=1 AND FIND_IN_SET('a08', about_education)");
SELECT COUNT(id) AS id FROM nose_1, coins_2, money_3 WHERE class=1 AND FIND_IN_SET('a09', about_education)");
SELECT COUNT(id) AS id FROM nose_1, coins_2, money_3 WHERE class=1 AND FIND_IN_SET('a10', about_education)");

输出：3,0,0,0,0,0,0,0,0,1

Answer 1

您需要编写所有表的子查询UNION ALL，然后使用count的{{1}}函数从ID中获取计数。

CASE WHEN

sqlfiddle：http://sqlfiddle.com/#!9/6a87fa/4

结果

SELECT 
     COUNT(CASE WHEN about_education = 'a01' and class = 1 then 1 else null end) a01,
     COUNT(CASE WHEN about_education = 'a02' and class = 1 then 1 else null end) a02, 
     COUNT(CASE WHEN about_education = 'a03' and class = 1 then 1 else null end) a03, 
     COUNT(CASE WHEN about_education = 'a04' and class = 1 then 1 else null end) a04, 
     COUNT(CASE WHEN about_education = 'a05' and class = 1 then 1 else null end) a05, 
     COUNT(CASE WHEN about_education = 'a06' and class = 1 then 1 else null end) a06, 
     COUNT(CASE WHEN about_education = 'a07' and class = 1 then 1 else null end) a07, 
     COUNT(CASE WHEN about_education = 'a08' and class = 1 then 1 else null end) a08, 
     COUNT(CASE WHEN about_education = 'a09' and class = 1 then 1 else null end) a09,
     COUNT(CASE WHEN about_education = 'a10' and class = 1 then 1 else null end) a10
FROM 
(
  SELECT id,class,about_education from nose_1
  UNION ALL
  SELECT id,class,about_education FROM coins_2
  UNION ALL
  SELECT id,class,about_education FROM money_3
) t

Answer 2

union all是正确的方法，但是我认为除非您确实想要列中的值，否则我建议进行定期聚合：

select about_education, count(*)
from (select id, class, about_education from nose_1
      union all
      select id, class, about_education from coins_2
      union all
      select id, class, about_education from money_3
     ) t
where class = 1
group by about_education;

在没有LEFT JOIN的情况下从多个表中选择

2 个答案: