問題的描述如下:給定資料庫中的兩列,每個列內的所有記錄可以視為一個集合,如何求這兩個集合的交集,差集等。樣本:
table1中欄位firstname
- tom
- kevin
- john
- steven
- marry
- anthony
|
table2中欄位username
- jack
- tom
- william
- tom
- marry
- Thomas
|
兩個列的交集是tom, marry。解決的方法是採用union和group by:
SELECT name
FROM (SELECT firstname as name FROM table1 union SELECT username as name FROM table2) as alltable
group by name having count(*) > 1;
兩個列的交集的補集:
SELECT name
FROM (SELECT firstname as name FROM table1 union SELECT username as name FROM table2) as alltable
group by name having count(*) = 1;
第一個列和第二個列的差集:
SELECT * FROM table1
WHERE firstname not in
(SELECT name
FROM (SELECT firstname as name FROM table1 union SELECT username as name FROM table2) as alltable
group by name having count(*) > 1)
類似的可以求第二個列和第一個列的差集。如果一個集合是另一個集合的子集,情況會簡單一點。如果希望包含重複的記錄,使用union all.
大家可以自己考慮一下。當然這個肯定不是唯一的解決方案了,就算拋磚引玉了。
關於union,可以參考http://www.w3schools.com/sql/sql_union.asp
參考:http://www.mysqltutorial.org/compare-two-tables-to-find-unmatched-records-mysql.aspx