Find nth occurance of character in Rexx - mainframe

Question

I want to find a nth occurance of character in string in Rexx. Is there any built in function to do this? If no, how can I do this?

Answer 1

Use the POS function n number of times.
E.g to find the 3rd 'a' in a string, use pos("a",string,currentPos) 3 times with the initial start point being 1 and the subsequent start points being the result of the previous pos.
/* REXX */
string = "The quick brown fox jumps over the lazy dog"
srchchar = "e"
count = 2
say "The position of the '"count"' occurrence of '"srchChar"'"||,
" in '"string"' is at position : "
say findChar(srchChar, string, count)
exit
findChar: procedure
/* Find the nth (count) position of the string needle in haystack */
arg needle, haystack, count
currentPos = 0
do count
currentPos = currentPos + 1
currentPos = pos(needle, haystack, currentPos)
end
return currentPos
If you run this, you get :
The position of the '2' occurrence of 'e' in 'The quick brown fox jumps over the lazy dog' is at position :
29
***
This will accept a string as the search argument, not just a single character. If you wanted to enforce the search argument being a single character you could, but I can't see any value in doing so as it adds code and reduces functionality.
E.g.
string = "The quick brown fox jumps over the lazy dog"
srchchar = "the"
count = 2
say "The position of the '"count"' occurrence of '"srchChar"'"||,
" in '"string"' is at position : "
say findChar(srchChar, string, count)
returns:
The position of the '2' occurrence of 'the' in 'The quick brown fox jumps over the lazy dog' is at position :
32
***
EDIT - This is my old (incorrect as I mis-read the quest) answer, that counts the number of times the search character appears in the string.
No built-in function that I know of, but you could write a subroutine to step through the search string character-by-character, comparing the current character with the one you are looking for:
/* REXX */
string = "The quick brown fox jumps over the lazy dog"
srchchar = "q"
say "There is/are "count(srchchar, string)" occurrances of "||,
"'"srchchar"' in '"string"'"
exit
count: procedure
arg needle, haystack
count = 0
do i = 1 to length(haystack)
currentChar = substr(haystack,i,1)
if currentChar = needle then count = count + 1
end
return count
I'm sure there are other ways too.

Answer 2

Here is a sample implementation of a function to return the position of the nth occurrence of a needle within a haystack. The function is called NthPositionOfNeedle.
/*- REXX --------------------------------------------------------------------------
This function returns the position of the nth occurrence of a given character
in a given string. A negative return value indicates a parameter error, or
the fact that the nth occurrence was not found.
Parameters:
Haystack A non-empty string of characters to be search.
Needle A single character to be searched for in the haystack.
Count A whole number indicating the nth occurrence to be found.
Return values:
n where n > 0 indicates the position of the nth occurrence
of the needle within the haystack.
-1 The haystack is an empty string.
-2 The needle is not a single character.
-3 The count is not a whole number > 0.
-4 The needle was not found at all within the haystack.
-5 The nth occurrence was not found, but the needle was
found at least once.
---------------------------------------------------------------------------------*/
NthPositionOfNeedle: procedure
Haystack = arg(1)
Needle = arg(2)
Count = arg(3)
/* The haystack must not be an empty string. Return -1 if it is */
if length( Haystack ) = 0
then return -1
/* The Needle must be a single character. Return -2 if it is not */
if length( Needle ) <> 1
then return -2
/* The count must be a whole number > 0. Return -3 if it is not */
if datatype( Count, "W" ) = 0 | Count < 1
then return -3
/* Does needle exist at least once in the haystack? Return -4 if not */
if pos( Needle, Haystack ) = 0
then return -4
CurrCnt = 0
Start = 1
do while ( CurrCnt < Count & Start <= length( Haystack ) )
NeedlePos = pos( Needle, Haystack, Start )
/* No more occurrences? */
if NeedlePos = 0
then leave
CurrCnt = CurrCnt + 1
Start = NeedlePos + 1
end
/* Return the position of the nth occurrence, if so many exist, else return zero */
if CurrCnt = Count
then return NeedlePos
else return -5
This is the code to test above funtion
/*- REXX --------------------------------------------------------------------------
Code to test the function "NthPositionOfNeedle".
---------------------------------------------------------------------------------*/
say "1st occurrence of 't' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( 'this is a string with multiple characters i.', 't', 1 )
/* ....+....1....+....2....+....3....+....4.... */
say "1st occurrence of 'i' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( 'this is a string with multiple characters i.', 'i', 1 )
say "2nd occurrence of 'i' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( 'this is a string with multiple characters i.', 'i', 2 )
say "3rd occurrence of 'i' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( 'this is a string with multiple characters i.', 'i', 3 )
say "6th occurrence of 'i' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( 'this is a string with multiple characters i.', 'i', 6 )
say "7th occurrence of 'i' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( 'this is a string with multiple characters i.', 'i', 7 )
say "0th occurrence of 'i' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( 'this is a string with multiple characters i.', 'i', 0 )
say "2nd occurrence of 'x' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( 'this is a string with multiple characters i.', 'x', 2 )
say "1st occurrence of 'i' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( '', 'i', 1 )
say "1st occurrence of 'i' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( 'this is a string with multiple characters i.', '', 1 )
say "1st occurrence of 'i' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( 'this is a string with multiple characters i.', 'i', 0 )
say "1st occurrence of 'i' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( 'this is a string with multiple characters i.', 'i', "a" )
say "1st occurrence of 'i' in 'this is a string with multiple characters i.' is" ,
NthPositionOfNeedle( 'this is a string with multiple characters i.', 'i', 1.1 )
exit

Answer 3

Parse is nice. Usually keeps things short. Here's one way of using it in context of this Q:
/* REXX: PosOfNth: Position of nth character(string) in haystack */
/* tests */
s='Foo Bar Baz'; c='F'; n=1; Say n'-th 'c' in 's"="PosOfNth(s,c,n)
s='Foo Bar Baz'; c='a'; n=1; Say n'-th 'c' in 's"="PosOfNth(s,c,n)
s='Foo Bar Baz'; c='a'; n=2; Say n'-th 'c' in 's"="PosOfNth(s,c,n)
s='Foo Bar Baz'; c='z'; n=1; Say n'-th 'c' in 's"="PosOfNth(s,c,n)
s='Foo Bar Baz'; c='z'; n=2; Say n'-th 'c' in 's"="PosOfNth(s,c,n)
s='Lorem Ipsum'; c='m'; n=1; Say n'-th 'c' in 's"="PosOfNth(s,c,n)
s='Lorem Ipsum'; c='z'; n=1; Say n'-th 'c' in 's"="PosOfNth(s,c,n)
s='Lorem Ipsum'; c='Ips'; n=1; Say n'-th 'c' in 's"="PosOfNth(s,c,n)
/* "visual" test */
Say
s='Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod...'
c='a'; n=2
Say s; Say Left('',PosOfNth(s,c,n)-1,' ')||'^ - 'n'-th 'c' is here'
Exit
/* ===== PROCS ===== */
PosOfNth: Procedure ;
Parse Arg haystack, needle, n
posN = 0
Do n
oHaystack = haystack
Parse Value haystack With head (needle) haystack
If haystack='' & oHaystack<>head||needle Then Do
posN = 0
Leave
End; Else posN = posN + Length(head) + 1
End
posN = posN + (Length(needle)-1) * (n-1)
Return posN
EDIT
As #phunsoft pointed out, there was an issue with multi-char needle (search string). I added line posN = posN + (Length(needle)-1) * (n-1) to adjust for it.

Answer 4

public class character_occurance {
public static void main(String[] args) {
String str1 = "bcdabghbghjbjkib";
int count = 0;
for (int i = 0; i <str1.length(); i++) {
if (str1.charAt(i) == 'j') {
count=count+1;
}
}System.out.println(count);
}
}