DYN-3844 Open Dynamo Template as new workspace #14871
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UI Smoke TestsTest: success. 2 passed, 0 failed. |
RobertGlobant20
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LGTM with some comments
| /// Present the open dialog and open the template that is selected. | ||
| /// </summary> | ||
| /// <param name="parameter"></param> | ||
| private void ShowOpenTemplateDialog(object parameter) |
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Any way we can combine this with ShowOpenDialogAndOpenResult ?
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yeah I thought about this but we need to call Open function below with different tuple and use TemplatesDirectory as start directory, currently there is nothing we can use to differentiate two workflows other than use different command for each menu item. I can try think about this more
| <MenuItem Header="{x:Static p:Resources.DynamoViewFileMenuOpenTemplate}" | ||
| Name="openTemplateMenuItem" | ||
| Command="{Binding ShowOpenTemplateDialogCommand}" | ||
| CommandParameter="Template" |
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@zeusongit I decided to use command param so we can distinguish if Dynamo is opening a template so we can reuse more file open code. We cant really use the same command because I think command and shortcut keys are 1:1 relationship and Open Template need its own shortcut keys.
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Merging this PR in to create the base of template function. |
Purpose
Introduce open by template function in Dynamo.
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