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auto numbering sub-items
Is there a way to auto number sub-items that have been indented? I know you can auto number all of the rows, but that doesn't change anything from the basic row numbers. You can add a suffix (.1), yet it applies that same suffix to all the rows; defeating the purpose. It would be much more useful (unless it can already be done); to have each sub-item (indented) to break into it's own numbering system; i.e. 1.1, 1.2, 1.3, 2.1, 2.2, 3.1, 3.2, etc... This way you can add/delete items and all of the numbers can be easily updated. This numbering helps to visually connect all of the tasks and sub-items associated with each task.
Any help or ability to make this happen in the future would be fantastic. Thanks!
Meyer
Comments
-
Meyer,
Short(er) answer:
I create three columns for data entry (assuming 3 levels of outline) and then show the outline in another column.
Data entry is just as fast (tab/arrow key instead of typing the "."), solves the sort problem (below).
Longer answer:
What you are trying to build are not numbers, unfortunately.
1.1
1.2
...
1.9
1.10 (not 2.0)
and
1.1.1
1.1.2
1.2
2.0
etc...
There's also the added problem with sorting (if you ever need to do that)
For example 1.2 comes before 1.1.1 when sorting in Smartsheet.
You can test for whether something is text (ISTEXT()) or a number (ISNUMBER()) but can't force something to be a number. You can force it to be text, however
=1 + ""
will a be "1" in text and treated as such.
Data entry would be a pain, though.
So that's why I use the level columns and formatting.
Everyone has an opinion on whether
1
1.
1.0.0
is the correct way to format the first level.
And I don't account for 1.0.1
I don't account for 0's at all, but could.
Hope this helps.
Craig
-
Looks like the formulas show to small.
The first one (with ".") is
'=[Level 1]1 + "."
+ [Level 2]1 + "."
+ [Level 3]1 + ""
The second one (without) is
"=[Level 1]7
+ IF(ISNUMBER([Level 2]7), "." + [Level 2]7 + ".", "")
+ IF(ISNUMBER([Level 3]7), [Level 3]7 + ".", "")"
for rows 1 and 7 respectively.
Craig
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