我想在sql中联接三个表，但无法获得所需的结果

这意味着它可以将条目相乘

您可以在下面尝试使用distinct count

SELECT COUNT(TAB1.ID),
       COUNT(TAB2.ID) Results,
       COUNT(TAB3.ID) Dropout 
  FROM TAB1 INNER JOIN  
       TAB2 on TAB1.ID = TAB2.ID INNER JOIN  
       TAB3 on TAB1.ID = TAB3.ID 
 WHERE TAB1.ID='405'

The answer i expect is 2,3,4 but it returns 24,24,24.

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我将切换到相关子查询：

SELECT COUNT(distinct TAB1.ID),COUNT(distinct TAB2.ID) Results,COUNT(distinct 
TAB3.ID) Dropout FROM TAB1 
INNER JOIN  TAB2 on TAB1.ID=TAB2.ID 
INNER JOIN  TAB3 on TAB1.ID=TAB3.ID 
WHERE TAB1.ID='405'

甚至会返回结果为零或退出的行。

我猜COUNTDISTINCT会满足您的要求。没有样本数据，很难提出更具体的建议：

SELECT TAB1.ID, COUNT(TAB1.ID),
       (select COUNT(TAB2.ID) from TAB2 where TAB1.ID=TAB2.ID) Results,
       (select COUNT(TAB3.ID) from TAB3 where TAB1.ID=TAB3.ID) Dropout
FROM TAB1 
WHERE TAB1.ID = '405'  -- to be removed when GROUP BY is added
GROUP BY TAB1.ID  -- more general version

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欢迎使用stack exchange，请添加示例数据和所需输出，并添加为标记什么是您的DBMS2*3*4=24。。。巧合？没用。结果0,0,0但第一次选择必须返回一些结果，否则全部为零

SELECT COUNT(DISTINCT TAB1.ID), 
       COUNT(DISTINCT TAB2.ID) as Results,
       COUNT(DISTINCT TAB3.ID) as Dropout
FROM TAB1 INNER JOIN
     TAB2 
     ON TAB1.ID = TAB2.ID INNER JOIN
     TAB3 
     ON TAB1.ID = TAB3.ID
WHERE TAB1.ID = '405';  -- Are the single quotes really necessary?