Suppress output to StdOut when piping echo

Question

I'm making a bash script that crawls through a directory and outputs all files of a certain type into a text file. I've got that working, it just also writes out a bunch of output to console I don't want (the names of the files) Here's the relevant code so far, tmpFile is the file I'm writing to: ``` for DIR in `find . -type d` # Find problem directories do for FILE in `ls "$DIR"` # Loop through problems in directory do if [[ `echo ${FILE} | grep -e prob[0-9]*_` ]]; then `echo ${FILE} >> ${tmpFile}` fi done done ``` The files I'm putting into the text file are in the format described by the regex prob[0-9]*_ (something like prob12345_01) Where I pipe the output from echo ${FILE} into grep, it still outputs to stdout, something I want to avoid. I think it's a simple fix, but it's escaping me.

Accepted Answer

All this can be done in one single find command. Consider this:

```
find . -type f -name "prob[0-9]*_*" -exec echo {} >> ${tmpFile} \;
```

EDIT:

Even simpler: (Thanks to @GlennJackman)

```
find . -type f -name "prob[0-9]*_*" >> $tmpFile
```

Suppress output to StdOut when piping echo

Solution

Problem