嵌套SQL查詢

Question

我怎麼能寫只是一個SQL查詢來獲取相同的結果通過執行這兩個查詢（步驟）：

第一步：

SELECT old_id, new_id FROM history WHERE flag = 1; 

結果：

+--------+--------+ 
| old_id | new_id | 
+--------+--------+ 
|  11 |  22 | 
|  33 |  44 | 
|  55 |  66 | 
+--------+--------+ 

，然後使用以前的結果，執行這個查詢：

UPDATE other_tabla SET somefk_id = CASE somefk_id 
    WHEN 11 THEN 22 
    WHEN 33 THEN 44 
    WHEN 55 THEN 66 
END WHERE somefk_id IN (11,33,55) 

Answer 1

我覺得這是你描述：

UPDATE `other_tablea` 
JOIN `history` ON history.old_id = other_tablea.somefk_id AND history.flag = 1 
SET other_tablea.somefk_id = history.new_id 

Answer 2

子查詢似乎做的伎倆：

update other_tabla 
set  somefk_id = coalesce((
      select new_id 
      from history 
      where flag = 1 
        and old_id = other_tabla.somefk_id 
     ), other_tabla.somefk_id) 

Answer 3

您可以使用臨時表來存儲德第一次查詢的結果，然後resuse數據在第二個查詢。

SELECT old_id, new_id 
INTO #tmpTable1 
FROM history 
WHERE flag = 1; 

UPDATE other_tabla SET somefk_id = t.new.id 
FROM other_tabla as o 
INNER JOIN #tmpTable1 t ON o.somefk_id=t.old_id 

DROP TABLE #tmpTable1 

Answer 4

你不需要case

update 
    other_table, history 
set 
    other_table.somefk_id=history.new_id 
where 
    other_table.somefk_id=history.old_id and history.flag=1;