Find position of last alphabetic letter in a cell

[anglais428]

I am looking for a formula that will identify the position of the last letter in a cell, ignoring punctuation and special characters.
E.g. If the text in a cell is:
This,text,in,a,cell,,,,,,
The result should be 19 as the last "L" is in position 19. It should ignore all the commas at the end.

 

[Fluff]

how about

+Fluff.xlsm
	A	B
1
2	abc123xyz,,,??	9
3	789xyz123abc,,*?@,	12
Main

Cell Formulas
Range	Formula
B2:B3	B2	=AGGREGATE(14,6,SEARCH(CHAR(ROW($65:$90)),A2),1)

 

[anglais428]

Thanks for your reply. It works only if there are no repeated characters (and I have repeated characters). So, for example, if you change your cell A2 value to "abc123abc,,,??" the result is 3, whereas I would still like it to be 9.
Any ideas?

 

[Fluff]

In that case, I'm afraid I don't know.

 

[sandy666]

is that what you want?

Raw	Text	Position
abc123xyz,,,??	z	9
789xyz123abc,,*?@,	c	12
abc,,,abc…abc	c	13

 

[anglais428]

In your third example, I would expect the position number to be 15 (the last C in the cell), rows 1 and 2 are perfect.

 

[sandy666]

three dots together are treated as special "single character" (sometimes)
I can do nothing with that, sorry
but you can try replace . (dot) with any other special character, eg.#

Source
Raw
abc123xyz,,,??
789xyz123abc,,*?@,
abc,,,abc...abc
l;kjri2678%^&*(!!M@:Kk|)*&}{

Raw	Text	Position
abc123xyz,,,??	z	9
789xyz123abc,,*?@,	c	12
abc,,,abc###abc	c	15
l;kjri2678%^&*(!!M@:Kk|)*&}{	k	22

 

[jimrward]

Doing a search of the forum I found the following

Formula to Find the Position of the Last Alpha Character in a Text String?

I am looking to see if there is a simple formula that will return the position of the last alpha character in a text string. The text strings will be of varying lengths and will contain alphanumeric characters and spaces only. So one string might be: 123 45 Abcde 678 So I just want to...

www.mrexcel.com

 

[sandy666]

continue from post#7
so end result is:

Source
Raw		Text	Position
abc123xyz,,,??		z	9
789xyz123abc,,*?@,		c	12
abc,,,abc...abc		c	15
l;kjri2678%^&*(!!M@:Kk|)*&}{		k	22

 

[anglais428]

Thanks Jim, I can get something from that thread. Sandy, what formula are you using?