如何列出所有與它在postgresql中的引用像mysql中的列？

Question

所以，我一直在通過互聯網搜索，似乎無法找到這個解決方案...

我試圖列出所有與它在postgresql中的參考列。在MySQL中，查詢是這樣的：

select table_schema, table_name, column_name, referenced_table_schema, referenced_table_name, referenced_column_name from informatioN_schema.key_column_usage where table_schema = '(Table Schema)'; 

，其結果是：

setandlog | access  | kode_privilege | NULL     | NULL     | NULL 
setandlog | access  | kode_feature | NULL     | NULL     | NULL 
setandlog | access  | kode_app  | NULL     | NULL     | NULL 
setandlog | access  | kode_credential | NULL     | NULL     | NULL 
setandlog | access  | username  | NULL     | NULL     | NULL 
setandlog | access  | kode_credential | setandlog    | cred_access   | kode_credential 
setandlog | access  | kode_privilege | setandlog    | cred_access   | kode_privilege 
setandlog | access  | kode_feature | setandlog    | cred_access   | kode_feature 
setandlog | access  | kode_app  | setandlog    | cred_access   | kode_app 
setandlog | access  | username  | setandlog    | login     | username 

但是當IM使用此查詢在PostgreSQL中嘗試它：

select r.table_schema as table_schema, r.table_name as table_name, r.column_name as column_name, u.table_schema as referenced_table_schema, u.table_name as referenced_table_name, u.column_name as referenced_column_name 
from INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE as u 
inner join INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS as FK 
    on U.CONSTRAINT_CATALOG = FK.UNIQUE_CONSTRAINT_CATALOG 
    and U.CONSTRAINT_SCHEMA = FK.UNIQUE_CONSTRAINT_SCHEMA 
    and U.CONSTRAINT_NAME = FK.UNIQUE_CONSTRAINT_NAME 
inner join INFORMATION_SCHEMA.KEY_COLUMN_USAGE as R 
    ON R.CONSTRAINT_CATALOG = FK.CONSTRAINT_CATALOG 
    AND R.CONSTRAINT_SCHEMA = FK.CONSTRAINT_SCHEMA 
    AND R.CONSTRAINT_NAME = FK.CONSTRAINT_NAME 

結果不正如我在MySQL查詢中所期望的那樣... 結果如下：

setandlog | access   | kode_credential   | setandlog    | cred_access   | kode_credential 
setandlog | access   | kode_privilege   | setandlog    | cred_access   | kode_credential 
setandlog | access   | kode_feature    | setandlog    | cred_access   | kode_credential 
setandlog | access   | kode_app     | setandlog    | cred_access   | kode_credential 
setandlog | access   | kode_credential   | setandlog    | cred_access   | kode_privilege 
setandlog | access   | kode_privilege   | setandlog    | cred_access   | kode_privilege 
setandlog | access   | kode_feature    | setandlog    | cred_access   | kode_privilege 
setandlog | access   | kode_app     | setandlog    | cred_access   | kode_privilege 
setandlog | access   | kode_credential   | setandlog    | cred_access   | kode_feature 
setandlog | access   | kode_privilege   | setandlog    | cred_access   | kode_feature 
setandlog | access   | kode_feature    | setandlog    | cred_access   | kode_feature 
setandlog | access   | kode_app     | setandlog    | cred_access   | kode_feature 
setandlog | access   | kode_credential   | setandlog    | cred_access   | kode_app 
setandlog | access   | kode_privilege   | setandlog    | cred_access   | kode_app 
setandlog | access   | kode_feature    | setandlog    | cred_access   | kode_app 
setandlog | access   | kode_app     | setandlog    | cred_access   | kode_app 
setandlog | access   | username     | setandlog    | login     | username 

好像有很多redudant數據... 當我檢查查詢，似乎是在UNIQUE_CONSTRAINT_NAME沒有diferrence在INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS表...

有人可以幫助我這個問題......？

Answer 1

你這裏的問題是，所引用的表中有一個是由多個列的PRIMARY KEY：

PRIMARY KEY(kode_credential, kode_privilege, kode_feature, kode_app) 

這4列連接到其他表，然後你會得到一個笛卡爾乘積：4行爲每個外鍵中的每一列。您需要使用ordinal_position（您的案例中引用表中的約束列 - access）和position_in_unique_constraint（引用表中的約束列 - cred_access），兩者均在key_column_usage視圖中。爲了退出這個功能，你需要一個自聯接，如下所示：

SELECT k1.table_schema, 
     k1.table_name, 
     k1.column_name, 
     k2.table_schema AS referenced_table_schema, 
     k2.table_name AS referenced_table_name, 
     k2.column_name AS referenced_column_name 
FROM information_schema.key_column_usage k1 
JOIN information_schema.referential_constraints fk USING (constraint_schema, constraint_name) 
JOIN information_schema.key_column_usage k2 
    ON k2.constraint_schema = fk.unique_constraint_schema 
AND k2.constraint_name = fk.unique_constraint_name 
AND k2.ordinal_position = k1.position_in_unique_constraint; 

這是相當「神奇」是，MySQL可以做到這一點沒有具體加入像上面。讓你想知道還有什麼其他的兜帽裏的東西可以繼續......