Php 从多个没有左联接的表中选择

我正试图从50个表中选择

id

，其中

FIND\u IN\u SET

。是否有一种方法可以使用除左连接以外的其他方法？我只给了3张桌子，因为50张不合适

注意：每个表中都有关于教育的

列

下面的代码无法工作，那么我还需要做些什么来避免代码中出现上千行

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
您需要编写一个子查询
UNION ALL
ALL表格，然后
count
函数使用
CASE，当
您可以从ID获取计数时

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

sqlfiddle：

结果

| a01 | a02 | a03 | a04 | a05 | a06 | a07 | a08 | a09 | a10 |
|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|
|   3 |   0 |   0 |   0 |   0 |   0 |   0 |   0 |   0 |   1 |

union all
是正确的方法，但我认为我建议定期聚合，除非您真的想要列中的值：

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;

在你的例子中，左join在哪里？有交叉连接，笛卡尔乘法。你能提供一些样本数据并期望结果吗？@D-Shih现在检查一下。你需要
union
有趣的方法。也会帮助我。