テーブルのプライマリキーに基づいて2つのテーブルから選択

Question

2種類のレコードの共通の値を表すテーブルがあり、互いに異なるデータを保持する別の2つのテーブルがあります。

バンキング

+-----------------+--------------+------+-----+---------+----------------+ 
| Field   | Type   | Null | Key | Default | Extra   | 
+-----------------+--------------+------+-----+---------+----------------+ 
| id    | int(11)  | NO | PRI | NULL | auto_increment | 
| officer   | varchar(32) | NO |  | NULL |    | 
| bank   | varchar(64) | NO |  | NULL |    | 
| branch   | varchar(64) | NO |  | NULL |    | 
| amount   | int(11)  | NO |  | NULL |    | 
| date   | date   | NO |  | NULL |    | 
| sys_date  | date   | NO |  | NULL |    | 
| source_document | varchar(360) | NO |  | NULL |    | 
+-----------------+--------------+------+-----+---------+----------------+ 

machinery_bankingをspare_parts_banking

+----------------+-------------+------+-----+---------+-------+ 
| Field   | Type  | Null | Key | Default | Extra | 
+----------------+-------------+------+-----+---------+-------+ 
| banking_id  | int(11)  | NO |  | NULL |  | 
| engine_no  | varchar(64) | NO |  | NULL |  | 
| chassis_no  | varchar(64) | NO |  | NULL |  | 
| receipt_number | varchar(64) | NO |  | NULL |  | 
| payment_type | int(11)  | NO |  | NULL |  | 
+----------------+-------------+------+-----+---------+-------+ 

を以下のように表1である

+----------------+-------------+------+-----+---------+-------+ 
| Field   | Type  | Null | Key | Default | Extra | 
+----------------+-------------+------+-----+---------+-------+ 
| banking_id  | int(11)  | NO |  | NULL |  | 
| invoice_number | varchar(32) | NO |  | NULL |  | 
+----------------+-------------+------+-----+---------+-------+ 

バンキングテーブルに入力された各データは、各テーブルmachinery_bankingまたはspare_parts_bankingのいずれかに他のデータがあります。

私はで、次のデータが

を次のようにがあるspare_parts_banking に

+------------+-----------+------------+----------------+--------------+ 
| banking_id | engine_no | chassis_no | receipt_number | payment_type | 
+------------+-----------+------------+----------------+--------------+ 
|   1 | 2324234 | NL234234 | RAN234   |   1 | 
+------------+-----------+------------+----------------+--------------+ 

データをmachinery_bankingいる

+----+---------+------------+------------+-----------+------------+------------+------------------------------------------------------------------------------------------+ 
| id | officer | bank  | branch  | amount | date  | sys_date | source_document                   | 
+----+---------+------------+------------+-----------+------------+------------+------------------------------------------------------------------------------------------+ 
| 1 | prasad | Sampath | Kaduruwela | 234234234 | 2017-06-28 | 2017-06-28 | image | 
| 2 | prasad | Commercial | Colombo | 234234234 | 2017-05-28 | 2017-05-28 | image | 
+----+---------+------------+------------+-----------+------------+------------+------------------------------------------------------------------------------------------+ 

銀行テーブルに次のデータを持っています

+------------+----------------+ 
| banking_id | invoice_number | 
+------------+----------------+ 
|   2 | INVRAN1  | 
+------------+----------------+ 

私は

、私はどのようなクエリを使用することができます

+----+---------+------------+------------+-----------+------------+------------+-----------+------------+------------------------------------------------------------------------------------------+----------------+ 
| id | officer | bank  | branch  | amount | date  | sys_date | engine_no | chassis_no | source_document                   | invoice_number | 
+----+---------+------------+------------+-----------+------------+------------+-----------+------------+------------------------------------------------------------------------------------------+----------------+ 
| 1 | prasad | Sampath | Kaduruwela | 234234234 | 2017-06-28 | 2017-06-28 | 2324234 | NL234234 | http://res.cloudinary.com/randeepa-com/image/upload/v1498455394/dmhxqal8hjcthhgav0n9.jpg |    | 
| 2 | prasad | Commercial | Colombo | 234234234 | 2017-05-28 | 2017-05-28 |   |   | http://res.cloudinary.com/randeepa-com/image/upload/v1498455394/dmhxqal8hjcthhgav0n9.jpg | INVRAN1  | 
+----+---------+------------+------------+-----------+------------+------------+-----------+------------+------------------------------------------------------------------------------------------+----------------+ 

を次のように私が取得を希望する結果が

+----+---------+------------+------------+-----------+------------+------------+-----------+------------+------------------------------------------------------------------------------------------+----------------+ 
| id | officer | bank  | branch  | amount | date  | sys_date | engine_no | chassis_no | source_document                   | invoice_number | 
+----+---------+------------+------------+-----------+------------+------------+-----------+------------+------------------------------------------------------------------------------------------+----------------+ 
| 1 | prasad | Sampath | Kaduruwela | 234234234 | 2017-06-28 | 2017-06-28 | 2324234 | NL234234 | http://res.cloudinary.com/randeepa-com/image/upload/v1498455394/dmhxqal8hjcthhgav0n9.jpg | INVRAN1  | 
| 2 | prasad | Commercial | Colombo | 234234234 | 2017-05-28 | 2017-05-28 | 2324234 | NL234234 | http://res.cloudinary.com/randeepa-com/image/upload/v1498455394/dmhxqal8hjcthhgav0n9.jpg | INVRAN1  | 
+----+---------+------------+------------+-----------+------------+------------+-----------+------------+------------------------------------------------------------------------------------------+----------------+ 

を次のクエリ

SELECT a.id, a.officer, a.bank, a.branch, a.amount, a.date, a.sys_date, b.engine_no, b.chassis_no, a.source_document, c.invoice_number 
FROM banking a,machinery_banking b, spare_parts_banking c 
WHERE a.id = b.banking_id OR a.id = c.banking_id; 

を試してみましたが、次の結果を終了していますこの結果を得るには

Answer 1

理由は、あなたはデータを持っている理由重複が、私はこれが

SELECT a.id, 
    a.officer, 
    a.bank, 
    a.branch, 
    a.amount, 
    a.date, 
    a.sys_date, 
    b.engine_no, 
    b.chassis_no, 
    a.source_document, 
    c.invoice_number 
FROM banking a 
LEFT JOIN machinery_banking b 
ON a.id = b.banking_id 
LEFT JOIN spare_parts_banking c 
ON a.id = c.banking_id; 

はまた新しいのJOIN構文に移動動作するはずと信じているため OR

条件の間のものであり、理解する方がはるかに簡単です。

SQL構文に参加 - Reference

Answer 2

はこれを試してみてください：私はPAYMENT_TYPEある外部キーである列を持っていると私はどのようにすることができ、テーブルPAYMENT_TYPEからpaymnent_nameを取得する必要がありmachinery_bankingで

SELECT a.id, a.officer, a.bank, a.branch, a.amount, a.date, a.sys_date, b.engine_no, b.chassis_no, a.source_document, c.invoice_number 
FROM banking a,machinery_banking b, spare_parts_banking c 
WHERE a.id = b.banking_id(+) 
AND a.id = c.banking_id(+);