Shell Programming and Scripting

I misstype the code the first time.now this is the code and its printing
Command not found
if I put single quotes ' instead of ` gives me No match

Code:

 
 
#!/bin/csh
set res=`find . -name "*.bash" | wc -l`
eval "$res" > result.txt

,

You're using backquotes around the assignment. It should be single quotes.

Code:

#!/bin/csh
set res=`find . -name "*.bash" | wc -l`
eval "$res" > result.txt

Code:

#!/bin/csh
set res='find . -name "*.bash" | wc -l'
eval "$res" > result.txt

Code:

$ rm -f result.txt

$ ./MyTest          
Hello from .cshrc

$ cat result.txt
       5

Thanks,if I put single quotes gives me no mach ,if I put back quotes and print $res,gives me the right result in my case 25
but the next line eval "$res" > result.txt
print Command not found

---------- Post updated at 07:16 AM ---------- Previous update was at 06:55 AM ----------

I found the solution ,I simply had to echho the var

echo $res > result.txt


Thanks for the help !!


I have another question_____how can I cut sample@yahoo.com up to the @ simbol ______ to print just yahoo.com ,the command has to work for a column of words with different length
Thanks again

pipe the column of words to sed:

if it is in a file:

Code:

cat mailfile | sed 's/.*@//g' > outputfile

Do NOT redirect to the same file you read from because you will end up with a blank file.
if it is coming from some other command:

Code:

somecoommand | sed 's/.*@//g' > outputfile

Thank you very mach,could you show me how to print the first part of the email before @ sample@yahoo.com to print just sample
Thanks again

Code:

 
cat file | awk -F\@ '{print $1}' > newfile

The above command is the first thing that comes to mind. There are other ways to do it, its just the way I find easiest.

Thanks a lot ,and another question,how can I replace certain word in text with diferent one