Метод Job.Start

Статья
04/26/2013

Runs the referenced job.

Пространство имен: Microsoft.SqlServer.Management.Smo.Agent
Сборка: Microsoft.SqlServer.Smo (в Microsoft.SqlServer.Smo.dll)

Синтаксис

'Декларация
Public Sub Start
'Применение
Dim instance As Job

instance.Start()

public void Start()

public:
void Start()

member Start : unit -> unit

public function Start()

Примеры

The following code example creates a job and starts it.

Server srv = new Server("(local)");
Job jb = new Job(srv.JobServer, "Test Job");
jb.Create();
JobStep jbstp = new JobStep(jb, "Test Job Step");
jbstp.OnSuccessAction = StepCompletionAction.QuitWithSuccess;
jbstp.OnFailAction = StepCompletionAction.QuitWithFailure;
jbstp.Create();
jb.ApplyToTargetServer(srv.Name);
jb.IsEnabled = true;
jb.Start();

PowerShell

$srv = new-object Microsoft.SqlServer.Management.Smo.Server("(local)")
$jb = new-object Microsoft.SqlServer.Management.Smo.Agent.Job($srv.JobServer, "Testrrt Job")
$jb.Create()
$jbstp = new-object Microsoft.SqlServer.Management.Smo.Agent.JobStep($jb, "Test Job Step")
$jbstp.OnSuccessAction = [Microsoft.SqlServer.Management.Smo.Agent.StepCompletionAction]::QuitWithSuccess
$jbstp.OnFailAction = [Microsoft.SqlServer.Management.Smo.Agent.StepCompletionAction]::QuitWithFailure
$jbstp.Create()
$jb.ApplyToTargetServer($srv.Name)
$jb.IsEnabled = $TRUE
$jb.Start()