Hi,

Im using a find function to find a specific file and print it.
eg.
find /path -name filename -type f -print -exec more '{}' \;

The problem I have is that I need to know if the file has not been fount (ie it does not exist).
I had a look at the find function and could not find anything relevant. I also tried adding another -exec statement that would increment a counter, giving me a number of instances but -exec doesnt seem to accept most shell commands.
Huh, this should be so simple and its been driving me nuts

Also, while at it, is there any way to only process one result... eg. only print the first instance of the file found?

Any ideas?

Thanks!!!

You can try something like this :

find /path -name filename -type f > /tmp/list
if [ $? -ne 0 ]; then
   echo "File not found"
else
   echo "Number of occurence of filename : \c"
   cat /tmp/list | wc -l
   echo "First occurence is : \c"
   head -n1 /tmp/lst
fi

I should mention that I do not want to use any external files... It all needs to be executed within a single script...

Thanks for the idea though!

Its a very simple function the Im writing that finds and prints a specific file...but the location of the file can vary which is why I'm using find.
Also, I want to be able to print an error if the file does not exist.

If you dont want an external file you can do this like that :

find /path -name filename -type f >/dev/null
if [ $? -ne 0 ]; then
   echo "File not found"
else
   i=1
   for file in `find /path -name filename -type f`
   do
      if [ $i -eq 1 ]; then
         echo "First occurence is : $file"
      fi
      i=$(($i + 1))
   done
   echo "Number of occurence of filename : $(($i - 1))
fi

Excellent...

Thanks, that will work.
I tried using $? previously but it tends to return 0 even if the file does not exist.
the "for file in" idea works though.

Thanks very much!!!