sql-server - sp_start_job 等待作业完成

Question

可能重复：
从存储过程执行 SQL Server 代理作业并返回作业结果

使用 sp_start_job 启动 sql 代理作业后，有没有办法确定它何时完成？

Answer 1

本文介绍了一个 SP 启动一个 sql 代理作业并等待。

-- output from stored procedure xp_sqlagent_enum_jobs is captured in the following table
    declare @xp_results TABLE ( job_id                UNIQUEIDENTIFIER NOT NULL,
                                last_run_date         INT              NOT NULL,
                                last_run_time         INT              NOT NULL,
                                next_run_date         INT              NOT NULL,
                                next_run_time         INT              NOT NULL,
                                next_run_schedule_id  INT              NOT NULL,
                                requested_to_run      INT              NOT NULL, -- BOOL
                                request_source        INT              NOT NULL,
                                request_source_id     sysname          COLLATE database_default NULL,
                                running               INT              NOT NULL, -- BOOL
                                current_step          INT              NOT NULL,
                                current_retry_attempt INT              NOT NULL,
                                job_state             INT              NOT NULL)

    -- start the job
    declare @r as int
    exec @r = msdb..sp_start_job @job

    -- quit if unable to start
    if @r<>0
        RAISERROR (N'Could not start job: %s.', 16, 2, @job)

    -- start with an initial delay to allow the job to appear in the job list (maybe I am missing something ?)
    WAITFOR DELAY '0:0:01';
    set @seccount = 1

    -- check job run state
    insert into @xp_results
    execute master.dbo.xp_sqlagent_enum_jobs 1, @job_owner, @job_id

    set @running= (SELECT top 1 running from @xp_results)

    while @running<>0
    begin
        WAITFOR DELAY '0:0:01';
        set @seccount = @seccount + 1

        delete from @xp_results

        insert into @xp_results
        execute master.dbo.xp_sqlagent_enum_jobs 1, @job_owner, @job_id

        set @running= (SELECT top 1 running from @xp_results)
    end

    -- result: not ok (=1) if still running

    if @running <> 0 begin
        -- still running
        return 0
    end
    else begin

        -- did it finish ok ?
        set @run_status = 0

        select @run_status=run_status
        from msdb.dbo.sysjobhistory
        where job_id=@job_id
          and cast(run_date as bigint) * 1000000 + run_time >= @start_job

        if @run_status=1
            return 1  --finished ok
        else  --error
            RAISERROR (N'job %s did not finish successfully.', 16, 2, @job)

    end

    END TRY

Answer 2

XP_SQLAGENT_ENUM_JOBS可以使用，但没有记录。它通常用于检测长时间运行的作业。

当然，也有sp_help_jobs或只是监控作业历史表

Answer 3

SELECT TOP 1 1 AS FinishedRunning
FROM msdb..sysjobactivity aj
JOIN msdb..sysjobs sj on sj.job_id = aj.job_id
WHERE aj.stop_execution_date IS NOT NULL
AND aj.start_execution_date IS NOT NULL
AND sj.name = 'YourJobNameHere'
AND NOT EXISTS
(
    SELECT TOP 1 1
    FROM msdb..sysjobactivity New
    WHERE New.job_id = aj.job_id
    AND new.start_execution_date > aj.start_execution_date
)

Answer 4

实际上我最近不得不这样做，这就是我正在考虑实施它的方式。我正在通过 sp_add_job 和 sp_add_jobstep 创建一个临时作业，并将 @delete_level 设置为 3（总是在运行后删除）。

我不是 100% 相信这种方法的，正如您可能从存储过程的标题中看出的那样。但是，它确实有效：

CREATE PROCEDURE spWorkaround_checkJobExists

@job_id UNIQUEIDENTIFIER   
, @thisIteration tinyint  
, @maxRecurse tinyint

AS

IF (@thisIteration <= @maxRecurse)
BEGIN
    IF EXISTS(
    select * FROM msdb.dbo.sysjobs where job_id = @job_id
    ) 
    BEGIN
        WAITFOR DELAY '00:00:01'
        DECLARE @nextIteration int
        SET @nextIteration = @thisIteration + 1
        EXEC dbo.spWorkaround_checkJobExists  @job_id, @nextIteration, @maxRecurse
    END
END

当然，您需要输入一些代码来确保递归的最大次数，但您明白了。您还可以传入一个参数来控制递归发生的频率。就我而言，十秒钟后，结果毫无意义。

我在这里所做的可以通过更改选择标准以检查作业的执行状态来修改不打算在执行后立即删除的作业，例如，使用 sp_help_job 传递 @job_name 或 @job_id 和 @execution_status = 0。

Answer 5

 sp_help_job   @job_name   @execution_status = 0