how to determine the number of instances of a character in string in Oracle

I want to determine how many of a certain letter in a string, e.g., MTWRF10,TR34,TBD

these are class periods, so M=Mon, T=Tues, R=Thurs, etc.

I want to determine how many M's, for example. I can't figure out how to do this in Sql, plus PL/SQL is probably better, no ?

I guess it would just be walking character by character and counting up, I suppose. The logic processes a day at a time, so as I process Monday, I would just walk through the string and count my M's (while ignoring TBD, etc.).

So if there is a Sql trick, please advise. If not, and you have a code snippet to address, that would be excellent.

(oracle is such a broad term, but that looks like the only topic that applies, I don't see PL/SQL).

(I also don't see where to assign points anymore . . . is everything now 500 points ???)
Gadsden ConsultingIT SpecialistAsked:
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slightwv (䄆 Netminder) Commented:
Single character?

length(string) - length(replace(string,'M'))

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Gadsden ConsultingIT SpecialistAuthor Commented:
yes, single character.

very slick ! (I should have thought of that . . . )
Gadsden ConsultingIT SpecialistAuthor Commented:
just what the patient needed !

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slightwv (䄆 Netminder) Commented:
Just thought of something.

You wanted to ignore TBD but what I posted will include the 'T' when looking for Tuesday.

Simple fix.  Just grab everything up to the first comma and do the length as above.

There are a few ways but try this:

If you are dealing with a LOT of data this one might be faster:
slightwv (䄆 Netminder) Commented:
>>(I also don't see where to assign points anymore . . . is everything now 500 points ???)

Looks like I even missed that change.  I was just informed that everything is now 500 points.
Gadsden ConsultingIT SpecialistAuthor Commented:
thanks - I used replace TBD with ''

regexp_substr looks good but I don't understand what is going on there . . .

same with the substr / instr . . . not sure there.

I will try both out tomorrow.

yeah on the 500 points - grade inflation !
slightwv (䄆 Netminder) Commented:
The magic is [^,]+

This looks for 1 or more characters '+'  that is NOT '^'  in this list '[]' of characters.  In this case only one, a comma ','

>>same with the substr / instr . . . not sure there.

instr returns the position of the first comma.  Then I subtract one position to remove the comma (really unnecessary for the regex), then grabs the substring from position 1 to that number.

>>grade inflation !

Changes at corporate.
Gadsden ConsultingIT SpecialistAuthor Commented:
ok, thanks for the expln . . . I'll try tomorrow.
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