S
scabHead
I am using VB to create a calendar type "report" to overcome the
much-discussed weaknesses of Project's built-in calendar view. As part of
this effort, I filter the list of tasks according to task start and task
finish and use the data to draw a calendar in Excel. Things are basically
working pretty well but there are a few headaches. I check each task for its
calendar and if none, use the project's calendar for purposes of determining
work days so I don't show a task working on a weekend when it really only
occurs on say a friday and the following monday. I use the weekday
properties to help me with this. Where I have trouble is when a task is
entered as an eday type task. In such a case the calendars are overridden. But
what vb code can I use to see if eday approach has been used on a given task?
I haven't stumbled on the property and the duration property is of no use
for this purpose as far as I can see. My only workaround right now is to
set all eday tasks to 24-hour calendars before I set them to eday tasks.
This is less than an elegant solution.
much-discussed weaknesses of Project's built-in calendar view. As part of
this effort, I filter the list of tasks according to task start and task
finish and use the data to draw a calendar in Excel. Things are basically
working pretty well but there are a few headaches. I check each task for its
calendar and if none, use the project's calendar for purposes of determining
work days so I don't show a task working on a weekend when it really only
occurs on say a friday and the following monday. I use the weekday
properties to help me with this. Where I have trouble is when a task is
entered as an eday type task. In such a case the calendars are overridden. But
what vb code can I use to see if eday approach has been used on a given task?
I haven't stumbled on the property and the duration property is of no use
for this purpose as far as I can see. My only workaround right now is to
set all eday tasks to 24-hour calendars before I set them to eday tasks.
This is less than an elegant solution.