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How to write a SQL query to present values horizontally

开发者 https://www.devze.com 2023-02-14 19:14 出处:网络
Suppose I have one table called Jobs: CREATE TABLE [Jobs] ( [JOBID] [int] IDENTITY(1,1) NOT NULL, [PARTDESC] [nvarchar](64) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,

Suppose I have one table called Jobs:

CREATE TABLE [Jobs]
(
    [JOBID] [int] IDENTITY(1,1) NOT NULL,
    [PARTDESC] [nvarchar](64) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
    [OrderPlacedBy] [nvarchar](64) COLLATE SQL_Latin1_General_CP1_CI_AS NOT NULL,
    [SpecialistName] [nvarchar](64) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
    [Priority] [int] NOT NULL,
    [Symptoms] [ntext] COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
    [CustomerNotes] [ntext] COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
    [ShopNotes] [ntext] COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
    [JobType] [nvarchar](32) COLLATE SQL_Latin1_General_CP1_CI_AS NULL,
    [JobState] [nvarchar](32) COLLATE Latin1_Ge开发者_如何学运维neral_CI_AS NULL,
    [JobAddedDate] [datetime] NOT NULL,
    [JobStartedDate] [datetime] NULL,
    [JobFinishedDate] [datetime] NULL,
    [JobShippedDate] [datetime] NULL,
    [RecievedDate] [datetime] NULL
)

I want to see the specialist name and his jobs IDs horizontally.

ANA 201,502,605,701,774
BEN 102,103,051
JEN 705,401,402,509,409,408

A specialist may have n jobs. Suppose specialist ANA has 10 jobs where BEN has 5 jobs.

In this way I want to show specialist his jobs horizontally where the number of jobs may vary per specialist.

How can I do this in SQL?


Marc is right. Don't do this in SQL, do it at the presentation level. Still, if you want to do this, take a look here: How to return 1 single row data from 2 different tables with dynamic contents in sql

Applied to your situation, the code might be:

     select SpecialistName , LEFT(JobsIds, len(JobsIds)-1) as JobsIds from 
(SELECT j.SpecialistName ,

      ( SELECT cast(j1.JobsId as varchar(10)) + ','

           FROM Jobs j1

          WHERE j1.SpecialistName = j.SpecialistName

          ORDER BY JobId

            FOR XML PATH('') ) AS JobsIds
      FROM Jobs j
      GROUP BY SpecialistName )A;
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