Here's what I've been using for years, same idea, but more general:
function howmany tg,ctr
if ctr = empty then return 0
repeat for each item i in "8,7,6,5,4"
put numtochar(i) into divChar
if divChar is not in ctr then exit repeat
if i = 4 then return "error in howmany()"
end repeat
replace tg with divChar in ctr
set the itemdelimiter to divChar
put the number of items of ctr into h
if char -1 of ctr <> divChar then put h-1 into h
return h
end howmany
This works for all strings. Note that there is an initial check to ensure that
the replacement character is not contained in the string. The last if-then
covers the anomaly that a trailing itemdelimiter is ignored in "the number of
items of...".
-- Peter
Peter M. Brigham
[email protected]
http://home.comcast.net/~pmbrig
On Feb 22, 2012, at 10:47 PM, Bob Sneidar wrote:
> I am everyone's Uncle. Still, no repeat loops needed. See my solution.
>
> Bob
>
>
> On Feb 22, 2012, at 7:44 PM, Jerry Jensen wrote:
>
>> The offset() function has a third parameter: chars to skip. So if you call
>> offset in a loop with the third param set to the previous result, and count
>> loops until it returns 0, Bob's your uncle.
>>
>> On Feb 22, 2012, at 7:37 PM, Andrew Kluthe wrote:
>>
>>> function substringCount pChar, pString
>>> put 0 into theCount
>>> repeat for each char theChar in pString
>>> if theChar is pChar then
>>> add 1 to theCount
>>> end if
>>> end repeat
>>> return theCount
>>> end substringCount
>>>
>>>
>>> Would work out nicely if there is not built in function for this.
>>>
>>> --
>>> View this message in context:
>>> http://runtime-revolution.278305.n4.nabble.com/Get-number-of-occurrences-of-one-string-in-another-tp4412675p4412707.html
>>> Sent from the Revolution - User mailing list archive at Nabble.com.
>>>
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