mysql選擇是否存在一對多

Question

如何檢查多對一關係是否已經存在？

SampleTable 
╔════╦═══════╗ 
║ ID ║ OB_ID ║ 
╠════╬═══════╣ 
║ 1 ║  1 ║ 
║ 1 ║  2 ║ 
║ 2 ║  3 ║ 
║ 2 ║  1 ║ 
║ 2 ║  2 ║ 
║ 3 ║  1 ║ 
║ 3 ║  3 ║ 
╚════╩═══════╝ 

我想搜索SampleTable.ob_id並選擇SampleTable.id，如果存在。 （1）
return 1;

╔════╗ 
║ ID ║ 
╠════╣ 
║ 1 ║ 
╚════╝ 

搜索（1,2,3）
返回2;

╔════╗ 
║ ID ║ 
╠════╣ 
║ 2 ║ 
╚════╝ 

搜索（2,3）
回報空的記錄（或空）

╔════╗ 
║ ID ║ 
╚════╝ 

Answer 1

SELECT ID 
FROM tableName a 
WHERE ob_id IN (1,2,3) 
GROUP BY ID 
HAVING COUNT(*) = 
      (
       SELECT COUNT(*) 
       FROM TableName b 
       WHERE b.ID = a.ID 
      ) AND 
     COUNT(*) = 3 -- the value of this depend on the number of values 
        -- you have supplied on you where clause. 

• SQLFiddle Demo for values [1, 2]
• SQLFiddle Demo for values [1, 2, 3]
• SQLFiddle Demo for values [2, 3]

如果唯一約束不上ob_ID爲每ID定義，你需要使用DISTINCT

SELECT ID 
FROM tableName a 
WHERE ob_id IN (1,2) 
GROUP BY ID 
HAVING COUNT(DISTINCT ob_id) = 
      (
       SELECT COUNT(DISTINCT ob_id) 
       FROM TableName b 
       WHERE b.ID = a.ID 
      ) AND 
     COUNT(DISTINCT ob_id) = 2