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10 notes have parent tag 'x' and 9 of them have one or another of x's 3 child tags. What is search formula to identify which note is the odd one out?

Go to solution Solved by JP MTP,

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I'd be grateful for instruction on how to use search (presumably Boolean search) to solve a problem like the following example:

    I have parent tag 'x' with a note count of 10.  All 10 are in the same notebook.  
    This tag 'x' has 3 child tags each with a note count as follows:
     child tag: 'xa' (3)
     child tag: 'xb' (3)
     child tag:  'xc' (3)
    To all nine notes have the parent tag 'x' applied to them as well.   
    So one tag with 'x' does not have this and 'xa', or this and 'xb', or this and 'cx'.
    Now I need to determine which of the ten notes that were tagged with 'x' it is that does not have one or other of those three child tags applied to it.
For that expample involving few notes I can of course identify the odd note out by checking each one individually, but I've failed with every attempted formula for getting Evernote to do the work.

Surely there's a way...?


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Thank you for your post.

The odd note out, in the 'ten' of the given example, does have the parent tag on it so will not show with -tag*.

Meanwhile the EN article does not show how to use -tag:xa, -tag:xb and -tag:ac to show which note with tag:x is missing one of these. 



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Glad it worked  - although possibly a bit cheeky to say that you solved the problem!

If that is something you need to do alot, with alot more subtags, it is worth thinking about the naming of your tags. So in your example, if the first tag was called y you could just do

tag:y -tag:x*

That's alot less typing and still works if you add further sub-tags, provided they follow your naming convention.

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