SQL-Anweisung mit zwei Tabellen verbinden und eine andere Tabelle

Question

Ok, hier ist die Info nach Ihrem Leitfaden. Dies ist das Tabellenskript + einige Tabellen.

USE [tempdb] 
GO 

CREATE TABLE [dbo].[table1](
    [cn] [nvarchar](1024) NULL, 
    [member] [nvarchar](1024) NULL, 
    [description] [nvarchar](1024) NULL, 
    [Date] [datetime] NULL, 
     [IsArchived] [bit] null 
) 

GO 


CREATE TABLE [dbo].[table2](
    [cn] [nvarchar](1024) NULL, 
    [member] [nvarchar](1024) NULL, 
    [description] [nvarchar](1024) NULL, 
    [Date] [datetime] NULL, 
     [IsArchived] [bit] null 
) 

GO 


CREATE TABLE [dbo].[table4](
    [cn] [nvarchar](1024) NULL, 
    [member] [nvarchar](1024) NULL, 
    [description] [nvarchar](1024) NULL, 
    [Date] [datetime] NULL, 
     [IsArchived] [bit] null 
) 

GO 

CREATE TABLE [dbo].[table3](
    [ID] [int] not NULL, 
    [ProductID] [nvarchar](1024) NULL, 
    [ProductOmschrijving] [nvarchar](1024) NULL, 
    [ProductPrijs] [money] NULL, 
) 

GO 

INSERT INTO table1 VALUES ('grp-sec-spla-office','john; chris; jack; marc;','112-112','', ''); 
INSERT INTO table1 VALUES ('grp-sec-spla-office-prof','jack; marc;','114-114','', ''); 
INSERT INTO table2 VALUES ('grp-sec-spla-office','cees; klaas','112-112','', ''); 
INSERT INTO table2 VALUES ('grp-sec-spla-office-prof','jan; piet','114-114','', ''); 
INSERT INTO table4 VALUES ('grp-sec-spla-office-prof','jack; marc;','114-114','', ''); 
INSERT INTO table4 VALUES ('grp-sec-spla-office','piet; ellen','112-112','', ''); 
INSERT INTO table4 VALUES ('grp-sec-spla-visio','henk; alwin','112-116','', ''); 
INSERT INTO table3 VALUES (1,'112-112','grp-sec-spla-office-eng','10.12'); 
INSERT INTO table3 VALUES (2,'114-114','grp-sec-spla-office-prof-2016','5.45'); 
INSERT INTO table3 VALUES (3,'112-116','grp-sec-spla-visio-blabla','7.12'); 
INSERT INTO table3 VALUES (4,'112-118','grp-sec-ac-office-sta-eng','2.45'); 
INSERT INTO table3 VALUES (5,'112-120','grp-sec-ac-office-pro-eng','2,50'); 
GO 

Mein Ergebnis mit der Abfrage ist:

+-------------+-------------------------------+--------------+-------+ 
| description |  ProductOmschrijving  | ProductPrijs | Total | 
+-------------+-------------------------------+--------------+-------+ 
| 112-112  | grp-sec-spla-office-eng  | 10.12  |  9 | 
| 114-114  | grp-sec-spla-office-prof-2016 | 5.45   |  8 | 
+-------------+-------------------------------+--------------+-------+ 

Aber ich brauche zu bekommen:

+-------------+-------------------------------+--------------+-------+ 
| description |  ProductOmschrijving  | ProductPrijs | Total | 
+-------------+-------------------------------+--------------+-------+ 
| 112-112  | grp-sec-spla-office-eng  | 10.12  |  9 | 
| 114-114  | grp-sec-spla-office-prof-2016 | 5.45   |  8 | 
| 112-116  | grp-sec-spla-visio-blabla  | 7.12   |  2 | 
| 112-118  | grp-sec-ac-office-sta-eng  | 2.45   |  0 | 
| 112-120  | grp-sec-ac-office-pro-eng  | 250.00  |  0 | 
+-------------+-------------------------------+--------------+-------+ 

Dies ist die Abfrage i jetzt verwenden:

SELECT 
    table1.description, 
    table3.ProductOmschrijving, 
    table3.ProductPrijs, 
    sum((isnull(LEN(table1.member) - LEN(REPLACE(table1.member, ';', '')),-1) + 1) + 
     isnull(LEN(table2.member) - LEN(REPLACE(table2.member, ';', '')),-1) + 1 + 
     isnull(LEN(table4.member) - LEN(REPLACE(table4.member, ';', '')),-1) + 1) AS Total 
FROM table1 
inner join table3 ON table1.description = table3.ProductID 
left outer join table2 ON table1.description = table2.description 
AND table2.IsArchived = 0 
left outer join table4 on table1.description = table4.description 
and table4.IsArchived = 0 
where table1.IsArchived = 0 
GROUP BY table1.description 
    , ProductOmschrijving 
    , ProductPrijs 

Die Tabelle1, Tabelle2, Tabelle4 kann sich ändern (insbesondere das Element fi eld). Wenn ich später (in der Zukunft) eine neue table5 zur Abfrage hinzufüge, muss das Ergebnis gleich sein, aber nur das Mitglied muss zum Ergebnis hinzugefügt werden. Auch wenn in der neuen Tabelle keine Mitglieder vorhanden sind5.

Ist das klarer? Danke für deine Zeit und Hilfe.

---

Ahhghh, wenn ich den ganzen Beitrag machte mir klar, dass die table3 (die Produkttabelle) muss natürlich die führende Tabelle sein. Also habe ich die Abfrage bearbeitet, um von table3 statt table1 zu gehen. Nein, es funktioniert perfekt !!!

SELECT 
    table3.ProductID, 
    table3.ProductOmschrijving, 
    table3.ProductPrijs, 
    sum((isnull(LEN(table1.member) - LEN(REPLACE(table1.member, ';', '')),-1) + 1) + 
     isnull(LEN(table2.member) - LEN(REPLACE(table2.member, ';', '')),-1) + 1 + 
     isnull(LEN(table4.member) - LEN(REPLACE(table4.member, ';', '')),-1) + 1) 
     AS Total 
FROM table3 
left join table1 ON table3.productid = table1.description 
and table1.IsArchived = 0 
left join table2 ON table3.productid = table2.description 
AND table2.IsArchived = 0 
left join table4 on table3.productid = table4.description 
and table4.IsArchived = 0 
GROUP BY table3.productid 
    , ProductOmschrijving 
    , ProductPrijs 

und das Ergebnis:

+-----------+-------------------------------+--------------+-------+ 
| ProductID |  ProductOmschrijving  | ProductPrijs | Total | 
+-----------+-------------------------------+--------------+-------+ 
| 112-112 | grp-sec-spla-office-eng  | 10.12  |  9 | 
| 112-116 | grp-sec-spla-visio-blabla  | 7.12   |  2 | 
| 112-118 | grp-sec-ac-office-sta-eng  | 2.45   |  0 | 
| 112-120 | grp-sec-ac-office-pro-eng  | 250.00  |  0 | 
| 114-114 | grp-sec-spla-office-prof-2016 | 5.45   |  8 | 
+-----------+-------------------------------+--------------+-------+ 

Answer 1

Die zu Ihrer vorhandenen Abfrage Sie, indem Sie sie in der Liste der Spalten und die Gruppe aufgenommen werden müssen hinzufügen.

SELECT SUM(table1_total + table2_total) AS Total, 
     description 
     , ProductOmschrijving 
     , ProductPrijs 
FROM (
    SELECT 
     table1.description, 
     table3.ProductOmschrijving, 
     table3.ProductPrijs, 
     table1.IsArchived, 
     LEN(table1.member) - LEN(REPLACE(table1.member, ';', '')) + 1 AS table1_total, 
     LEN(table2.member) - LEN(REPLACE(table2.member, ';', '')) + 1 AS table2_total 
    FROM table1 
    INNER JOIN table3 ON table1.description = table3.ProductID 
    LEFT OUTER JOIN table2 ON table1.description = table2.description 
) a 
GROUP BY description 
    , ProductOmschrijving 
    , ProductPrijs 

ODER Sie könnten diese Abfrage wie folgt erheblich vereinfachen.

SELECT 
    table1.description, 
    table3.ProductOmschrijving, 
    table3.ProductPrijs, 
    --table1.IsArchived, --don't think this is required??? 
    sum(LEN(table1.member) - LEN(REPLACE(table1.member, ';', '')) + 1 + LEN(table2.member) - LEN(REPLACE(table2.member, ';', '')) + 1) AS Total 
FROM table1 
INNER JOIN table3 ON table1.description = table3.ProductID 
LEFT OUTER JOIN table2 ON table1.description = table2.description 
GROUP BY description 
    , ProductOmschrijving 
    , ProductPrijs