確定兩個MySQL數據庫模式之間的差異

Question

我想要獲取數據庫AA中數據庫AA中缺失的任何表或字段。我正在使用INFORMATION_SCHEMA.columns獲取信息。所以，我寫了一個'缺失記錄'查詢來找到它們。在測試中，我使用了2個數據庫，我知道BB在另一個表中有1個缺失的表和1個缺失的字段。
這是我第一次嘗試：

SELECT AA.table_name, 
     AA.column_name, 
     BB.table_name, 
     BB.column_name 
FROM information_schema.columns AS AA 
     LEFT JOIN information_schema.columns AS BB 
       ON (AA.table_name = bb.table_name) 
       AND (AA.column_name = BB.column_name) 
WHERE AA.table_schema = 'wireless-2015-05' 
    AND BB.table_schema = 'wireless-2015-04' 
    AND BB.column_name IS NULL 

這返回0的記錄。所以，然後我嘗試：

SELECT AA.table_name, 
     AA.column_name 
FROM information_schema.columns AS AA 
WHERE AA.table_schema = 'wireless-2015-04' 
    AND NOT EXISTS(SELECT BB.table_name, 
         BB.column_name 
        FROM information_schema.columns AS BB 
        WHERE BB.table_schema = 'wireless-2015-05') 

我再次得到0條記錄。最後我試過這個：

SELECT table_name, 
     column_name 
FROM (SELECT DISTINCT table_name, 
         column_name 
     FROM information_schema.columns 
     WHERE table_schema = 'wireless-2015-04' 
     UNION ALL 
     SELECT DISTINCT table_name, 
         column_name 
     FROM information_schema.columns 
     WHERE table_schema = 'wireless-2015-05') AS tbl 
GROUP BY table_name, 
      column_name 
HAVING Count(*) = 1 

這產生了預期的結果。

雖然我不介意使用第三個查詢，但我無法弄清楚爲什麼前兩個不起作用。我想知道以供將來參考。任何人都可以發現問題嗎？

---

更新：
對於那些感興趣的，這裏有4個查詢的工作，以及運行每一個的時間。按照最快的順序列出，並且在查詢下方列出時間。

SELECT AA.table_name, 
     AA.column_name 
FROM information_schema.columns AS AA 
     LEFT JOIN (SELECT table_name, 
         column_name 
        FROM information_schema.columns 
        WHERE table_schema = 'wireless-2015-04') BB 
       ON AA.table_name = BB.table_name 
       AND AA.column_name = BB.column_name 
WHERE AA.table_schema = 'wireless-2015-05' 
     AND BB.table_name IS NULL; 

0.047秒

SELECT table_name, 
     column_name 
FROM (SELECT DISTINCT table_name, 
         column_name 
     FROM information_schema.columns 
     WHERE table_schema = 'wireless-2015-04' 
     UNION ALL 
     SELECT DISTINCT table_name, 
         column_name 
     FROM information_schema.columns 
     WHERE table_schema = 'wireless-2015-05') AS tbl 
GROUP BY table_name, 
      column_name 
HAVING Count(*) = 1; 

0.078秒

SELECT DISTINCT table_name, 
       column_name, 
       Concat(table_name, '--', column_name) AS tc 
FROM information_schema.columns 
WHERE table_schema = 'wireless-2015-05' 
HAVING tc NOT IN(SELECT DISTINCT Concat(table_name, '--', column_name) 
       FROM information_schema.columns 
       WHERE table_schema = 'wireless-2015-04'); 

0.125秒（一個新的解決方案，我認爲今天上午的）

SELECT aa.table_name, 
     aa.column_name 
FROM information_schema.columns aa 
WHERE table_schema = 'wireless-2015-05' 
     AND NOT EXISTS (SELECT 1 
         FROM information_schema.columns 
         WHERE table_schema = 'wireless-2015-04' 
           AND table_name = aa.table_name 
           AND column_name = aa.column_name); 

44.382秒。顯然不是一個好的現實世界的解決方案。

Answer 1

假設記錄看起來像這樣：

schema    table column 
    ---------------- ----- ------ 
1. wireless-2015-05 T1  F1 
2. wireless-2015-05 T1  F2 
3. wireless-2015-05 T2  F1 
4. wireless-2015-04 T1  F1 

注意，無線-2015-04缺少表T2和列T1.F2。我們將在描述和SQL Fiddle示例中使用此示例。你在前兩次嘗試中相當接近。只需稍作修改（下面包含）就可以確定它。

查詢1

讓我們打破第一個查詢。我們將離開where子句，因爲上面的例子只有where子句中提到的那兩個模式。

SELECT ... 
FROM information_schema.columns AS AA 
LEFT JOIN information_schema.columns AS BB 
    on aa.table_name = bb.table_name 
    and aa.column_name = bb.column_name 

的wireless-2015-05 + T1 + F1第一個記錄是匹配的（基於表和列名），在同一個表中的所有記錄。所以，

• AA的記錄＃1將匹配BB的記錄＃1和＃4
• AA的記錄＃2將匹配BB的記錄＃2
• AA的記錄＃3將匹配BB的記錄＃3
• AA的記錄＃4將匹配BB的記錄＃1和＃4

例子：http://sqlfiddle.com/#!9/6b704/4

會有與沒有記錄BB.column_name。所以沒有記錄被提取。但是，這不是你正在尋找的。

查詢1改進

，你可以重新編寫查詢1使用這樣的事情，給你正確的結果：

SELECT AA.table_name, 
     AA.column_name 
FROM information_schema.columns AS AA 
LEFT JOIN 
( 
    select table_name, column_name from 
    information_schema.columns 
    where table_schema = 'wireless-2015-04' 
) BB 
    on AA.table_name = BB.table_name 
    and AA.column_name = BB.column_name 
WHERE 
    AA.table_schema = 'wireless-2015-05' 
    and BB.table_name is null 

例子：http://sqlfiddle.com/#!9/6b704/10

查詢2

基本上，查詢2的NOT EXISTS子查詢缺少與AA列匹配的子句。這樣就不會產生你的結果

查詢2改進

該查詢可以做這樣的事情正確改進：

select aa.table_name, aa.column_name 
from information_schema.columns aa 
where table_schema = 'wireless-2015-05' 
and not exists (
    select 1 
    from information_schema.columns 
    where table_schema = 'wireless-2015-04' 
    and table_name = aa.table_name 
    and column_name = aa.column_name 
); 

例子：http://sqlfiddle.com/#!9/6b704/9

希望這有助於。

Answer 2

你的第一個查詢應該是這樣的，

Select AA.* 
(
    SELECT table_name, 
      column_name 
    From information_schema.columns 
    Where table_schema = 'wireless-2015-05' 
) AA 
LEFT JOIN 
(
    SELECT table_name, 
      column_name 
    From information_schema.columns 
    Where table_schema = 'wireless-2015-04' 
)BB 
on AA.table_name = BB.table_name 
AND AA.column_name = BB.column_name 

WHERE BB.table_name is null or BB.column_name is null 

您的查詢問題：

你放在哪裏查詢與錯誤條件

WHERE AA.table_schema = 'wireless-2015-05' 
    AND BB.table_schema = 'wireless-2015-04' 
    AND BB.column_name IS NULL 

當記錄中不存在的BB然後BB.table_schema = 'wireless-2015-04'這種情況變得錯誤，因此整個結果將是錯誤的，所以你沒有重新獲得SULT。

而對於第二個查詢，我認爲@zedfoxus是正確的。

你也可以使用EXCEPT的概念，它給你的願望結果。

以下查詢從查詢​​中返回EXCEPT運算符左邊的所有不同值，這些值在正確的查詢中也找不到。

SELECT DISTINCT table_name, 
       column_name 
FROM information_schema.columns 
WHERE table_schema = 'wireless-2015-05' 

EXCEPT 

SELECT DISTINCT table_name, 
       column_name 
FROM information_schema.columns 
WHERE table_schema = 'wireless-2015-04'