If this line is really "working in Linux", then Linux must be wrong. Compare this line to what i wrote:

Code:

iNrOfAttempts=$(grep -c "$iCurrDay" "$fLog")

"iNrOfAttempts" is a variable, which gets assigned with whatever comes out of the command:

Code:

grep -c "$iCurrDay" "$fLog"

Now, "iCurrDay" is another variable: an integer holding the day of month. This is used as an expression to search a file located at "$fLog", where "fLog" is another variable, supposed to contain a path name pointing to a file. So, filling the line with possible example values, what finally gets executed would read:

Code:

iNrOfAttempts=$(grep -c "28" "/path/to/blabla")

This counts the lines in /path/to/blabla, which contain the string "28" - for the example let us suppose there are 2 - and print that to stdout. The shell now executes whatever is inside "$(...)" and replaces that with the outcome, hence the line finally executed is:

Code:

iNrOfAttempts=2

What you did was to let grep search for:

Code:

grep -c "$ i CurrDay"

which is, because "$" is the regexp for "line end": the end of the line, followed by a space, an "i", another space and the string "CurrDay", which is obviously nonsense. This is why it isn't working - and i'd bet my daily dose of espresso against a single american coffee it isn't working in Lunix either. Because you let it search in a file without a fully qualified path chances are it won't search in the right file either, but that only as an aside.

I hope this helps.

bakunin