Propriedade do RDL Job.StartStepID

Artigo
04/26/2013

Gets or sets the ID value of the first job step in the sequence to be executed.

Namespace: Microsoft.SqlServer.Management.Smo.Agent
Assembly: Microsoft.SqlServer.Smo (em Microsoft.SqlServer.Smo.dll)

Sintaxe

'Declaração
<SfcPropertyAttribute(SfcPropertyFlags.Standalone)> _
Public Property StartStepID As Integer 
    Get 
    Set
'Uso
Dim instance As Job 
Dim value As Integer 

value = instance.StartStepID

instance.StartStepID = value

[SfcPropertyAttribute(SfcPropertyFlags.Standalone)]
public int StartStepID { get; set; }

[SfcPropertyAttribute(SfcPropertyFlags::Standalone)]
public:
property int StartStepID {
    int get ();
    void set (int value);
}

[<SfcPropertyAttribute(SfcPropertyFlags.Standalone)>]
member StartStepID : int with get, set

function get StartStepID () : int 
function set StartStepID (value : int)

Valor da propriedade

Tipo: System.Int32
An Int32 value that specifies the ID value of the first job step in the sequence.

Comentários

Microsoft SQL Server Agent job steps are identified by a user-defined, integer value. If no value is specified when you use SQL Server Management Objects (SMO) to create a job, job steps are given an identifier value when the job is added to the Jobs collection of a JobServer object.

By default, the StartStepID value is the value of ID property of the JobStep at the first ordinal position of the JobSteps collection property of the Job object.

Exemplos

The following code example creates a job and specifies that the job begins execution on step 4.

Server srv = new Server("(local)");
Job jb = new Job(srv.JobServer, "Test Job");
jb.Create();
jb.StartStepID = 4;

PowerShell

$srv = new-object Microsoft.SqlServer.Management.Smo.Server("(local)")
$jb = new-object Microsoft.SqlServer.Management.Smo.Agent.Job($srv.JobServer, "Test Job")
$jb.Create()
$jb.StartStepID = 4