Split out fields after specfic characters and lengths

[lbanham]

Hi
I have been trying to sovle this for a day now, and its frustated me.
I have data in a column (Name) and dependant on where the data is in the string and the number of characters, depends on what column it should sit in. I have pasted a table below showing the possible combinations and how it should split out. The fields have a set number of characters with exception of Set 4 which can be 1 -8 characters in lenth.

Set 1 & 2 are fine, using Mid/left etc that works fine. its getting Set 3 and Set4 to work for each scenario i cant get to work. I have tried various combinations of Mid/LEN/FIND etc with no joy.
Any help would be appreciated


Thanks
Lynsey

	6 Char	5 Char	5 Char	max 8 Char
Name	Set 1	Set 2	Set 3	Set 4
115113.52010.1ABCD	115113	52010	1ABCD
115113.52010.1ABCD.5TEN56	115113	52010	1ABCD	5TEN56
115113.52010.5TEN5678	115113	52010		5TEN5678
115113.52010	115113	52010

<TBODY>
</TBODY>

 

[s hal]

Are these the only possible combinations and exact data? If so, you could actually write a pretty simple formula that spits out the elements of the string in one of 4 different ways with the above 4 options (I'm guessing its not that simple, but if it is, I'd probably go that route for this).

The next option would be to search and then search against the next string for the periods like a nested search/mid formula

 

[lbanham]

There is an error in my message above, Set 1 can range from 4 -6 characters. Thought this may make a difference as any reference to data length cannot be fixed.

 

[lbanham]

Hi
There are another few combinations, but I have tried various formulas but the problem is the information when it has the four sections, as it cant reference a fixed number of cells. Or if the third section of data i.e. 5TEN5678 in row 3 below, actually belongs in SET4 not, Set 3.

The actual data will change (its about 3000 rows of data). I have pasted a new table below.

Hope that makes sense

	6 Char</SPAN>	5 Char</SPAN>	5 Char</SPAN>	max 8 Char</SPAN>
Name</SPAN>	Set 1</SPAN>	Set 2</SPAN>	Set 3</SPAN>	Set 4</SPAN>
115113.52010.1ABCD</SPAN>	115113</SPAN>	52010</SPAN>	1ABCD</SPAN>
115113.52010.1ABCD.5TEN56</SPAN>	115113</SPAN>	52010</SPAN>	1ABCD</SPAN>	5TEN56</SPAN>
115113.52010.5TEN5678</SPAN>	115113</SPAN>	52010</SPAN>		5TEN5678</SPAN>
115113.52010</SPAN>	115113</SPAN>	52010</SPAN>
1234.52010.1ABCD</SPAN>	1234</SPAN>	52010</SPAN>	1ABCD</SPAN>
1234.52010</SPAN>	1234</SPAN>	52010</SPAN>
1234.52010.1ABCD</SPAN>	1234</SPAN>	52010</SPAN>	1ABCD</SPAN>
1234.52010.1ABCD.5TEN56</SPAN>	1234</SPAN>	52010</SPAN>	1ABCD</SPAN>	5TEN56</SPAN>
1234.52010.56238942</SPAN>	1234</SPAN>	52010</SPAN>		56238942</SPAN>

<TBODY>
</TBODY><COLGROUP><COL><COL span=4></COLGROUP>

 

[s hal]

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Does set 1 always begin with a 1, set 2 always with a 5, set 3 always with a 1 and set 4 always with a 5?

There are ways to split the pieces into the right buckets, but only if you can clearly define what causes an item to be in one bucket vs another, if that makes sense...?

 

[lbanham]

No the numbers will always vary. What i was doing for Set 1 was, left of the first ".", Set 2 was 5 characters after the second decimel. Set 3 and Set 4 are then defined by if/where the decimal point is and the characters lengths. IF the number of characters after the second "." are more than 5 (and there is not a third ".") then it is Set 4 data, if it is max 5 characters after the second "." it is Set 3, if there is a third "." it should always split it to Set 3 (5 Char) and Set 4 (1-8 Char)

Thanks

 

[lbanham]

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This is a copy of the formula I am currently trying - apologies for the format it wont allow attachements. Set 3</SPAN>	Set 4</SPAN>
IFERROR(IF(LEN(MID(E1,FIND("^^",SUBSTITUTE(Pivot!E1,".","^^",2))+1,30))=5,MID(E1,FIND("^^",SUBSTITUTE(Pivot!E1,".","^^",2))+1,5),""),"")</SPAN>	IFERROR(IF(H1="",MID(Pivot!E1,FIND("^^",SUBSTITUTE(Pivot!E1,".","^^",2))+1,8),MID(Pivot!E1,FIND("^^",SUBSTITUTE(Pivot!E1,".","^^",3))+1,8)),"")</SPAN>

<TBODY>
</TBODY><COLGROUP><COL><COL></COLGROUP>

 

[s hal]

I was working on a formula and got a headache

Essentially you need to check in set 3 that the data exists at all (since you have some that stop at set 2), probably best accomplished by checking the total len() vs the len() of set 1 + set 2 + a period

Then you need to do a find of the next period starting at the end of the first two strings together and check the length to see if its 5, otherwise that value goes to set 4 (in set 4's own formula)..

and that's about the point I got a headache

I hope its enough to get you back on track, however...

 

[lbanham]

Thanks. I eventually got it to work by using your suggsestion of calculating the differing lengths and by adding two helper columns to calculate if there was 2 or 3 "." and then the subsequent position number of these. It was just Set 3 formula that needed adjusted. It is below incase it helps anyone else FYI . helper columns are J4 (".") and K4 (Position number)

Helper Columns
Firstly calculate how many "." are in a cell - IF($E4="",$J3,LEN($E4)-LEN(SUBSTITUTE($E4,".","")))
Second is the position of the last "." - IF($E4="",$K3,FIND("^^",SUBSTITUTE(Pivot!$E4,".","^^",$J4)))

Main Formula
IF($E4="",$H3,IF($J4=1,"",IF(IF(AND((LEN($E4)-FIND("^^",SUBSTITUTE(Pivot!$E4,".","^^",$J4))=5),$J4=2),MID($E4,$K4+1,5),MID($E4,$K4-5,5))=$G4,"",IF(AND((LEN($E4)-FIND("^^",SUBSTITUTE(Pivot!$E4,".","^^",$J4))=5),$J4=2),MID($E4,$K4+1,5),MID($E4,$K4-5,5)))))

This now works for all length variations and combinations of data sets. Phew! No doubt there is a cleaner way to do this, but it works for me just now.

Thanks again