Remove beginning and ending of a string, keep a portion 6

[NewToThis2]

Using CV9.
I have a field called Key-String in which I only need digits 5, 6, 7 and 8th from the right. The field will contain various different numbers of digits but I will always need only the 5th, 6th, 7th and 8th digits from the right. So for this example, I only want to see 3535. Here's a sample string:
E00062904MGRCHGSTAT0000035350001. Could someone please help me with the formula? Thanks so much!

 

[Madawc]

For Crystal 8.5, one can use

mid({Key-String}, 24, 4)

This should also work in higher version, but not if the string itself is variable. If it is, try a pair of formulae:

Right({Key-String}, 8)
   
Left(@Right-key, 4)

Madawc Williams (East Anglia)

 

[lbass]

Or, try the following which allows for different lengths, as long as the characters of interest are always the four digits starting from the 8th position from the right:

strreverse(mid(strreverse({table.key-string}),5,4))

-LB

 

[synapsevampire]

Or:

Mid({table.field},len(trim({table.field}))-5)

This will also allow for padded fields, so if there are spaces it will still work.

-k

 

[lbass]

SV,

I think you mean:

Mid({table.field},len(trim({table.field}))-7,4)

-LB

 

[synapsevampire]

Ahhh, right you are, LB, I should have read it more closely, too busy these days, thanks.

-k

 

[NewToThis2]

Thanks so much everyone. I tried them all. The only one I could get to work was this one:
strreverse(mid(strreverse({table.key-string}),5,4))

A couple of the others brought back this message:
Start position is less than 1 or not an integer.

Thanks again everyone - I got what I needed as I always do whenever I ask a question in this forum. Thanks so much to all of you experts out there who are so willing to help.

 

[mbarron]

I know it's 5 days later, but another option:

left(right({Key-string},8),4)

It's a single formula version of Madawc's second suggestion.

Mike

 

[NewToThis2]

Thanks so much mbarron. It worked like a charm!