Repository.Action.GetAllActionsDictionary: Difference between revisions
Jump to navigation
Jump to search
Alex Ulanov (talk | contribs) |
Kate Racaza (talk | contribs) No edit summary |
||
Line 1: | Line 1: | ||
{{Template: | {{Template:API}}{{Template:APIBreadcrumb|10|.[[Repository Class|Repository]]}} | ||
== | <onlyinclude>Gets a dictionary of all existing actions in the current project. Returns a '''Dictionary(Of Integer, cAction)''' object, where the keys are the ID numbers of the actions.</onlyinclude> | ||
== Syntax == | |||
'''.Action.GetAllActionsDictionary()''' | |||
== Example == | |||
This example assumes that actions exist in the first available project of the repository. | |||
'''VB.NET''' | |||
{{APIComment|' | {{APIPrefix|Dim}} MyRepository {{APIPrefix|As New}} Repository | ||
{{APIComment|...'Add code to connect to a Synthesis repository.}} | |||
{{APIComment|'Get | {{APIComment|'Get a dictionary of all actions in project #1.}} | ||
Dim | {{APIPrefix|Dim}} ActionsDict {{APIPrefix|As}} Dictionary (of Integer, cAction) | ||
MyRepository.Project.SetCurrentProject(1) | |||
ActionsDict = MyRepository.Action.GetAllActionsDictionary() | |||
{{APIComment|'Output sample: Display the number of available actions in the project. }} | |||
MsgBox(ActionsDict.Count) |
Revision as of 17:51, 21 July 2015
Member of: SynthesisAPI10.Repository
Gets a dictionary of all existing actions in the current project. Returns a Dictionary(Of Integer, cAction) object, where the keys are the ID numbers of the actions.
Syntax
.Action.GetAllActionsDictionary()
Example
This example assumes that actions exist in the first available project of the repository.
VB.NET Dim MyRepository As New Repository ...'Add code to connect to a Synthesis repository. 'Get a dictionary of all actions in project #1. Dim ActionsDict As Dictionary (of Integer, cAction) MyRepository.Project.SetCurrentProject(1) ActionsDict = MyRepository.Action.GetAllActionsDictionary() 'Output sample: Display the number of available actions in the project. MsgBox(ActionsDict.Count)