I have the following command using Grep in a complex shell script. Can you please advise what does the below expression mean?
# grep ^${File1}\| $Textfile > /deve/null 2>&1
Questions,
1) Wanted to know the significance of "^" in this line
2)What does a backslash"\" before the pipeline does
3)In the last part I guess they are trying to write the output of the search in the path /deve/null and i dont understand the meaning of 2>&1
Apart from explaining the above questions, please give me an overall idea from your knowledge that what this line tries to do in the script. Thanks in advance for your help..
The carrot (^) signifies the start of a record. Thus the pattern would be matched only if it's at the beginning of the record.
The backslash escapes the character and prevents the shell from interpreting it as a pipe symbol. In this instance it is passed as the end of the pattern to grep rather than causing the shell to treat the next token as a command.
Yes, they are trying to redirect output. The 2>&1 redirects standard error to the same file as standard output. I'd also say that the file they want to redirect to is wrong. The file /dev/null (no in deve) is a special file that drops the output on the floor so to speak.
In this case my guess is that the programmer was interested in having the result of the grep (0/success) if the pattern was found in the file and thus the output could be ignored and the exit code ($?) tested after the command. If your version of grep supports it, use the -q option rather than redirecting the output. This option (quiet) causes grep to not write any output to stdout and will stop on the first match -- much more efficient when a huge file is involved. Something like this:
if grep -q "^$file1|" $textfile
then
echo "$textfile references $file1"
else
echo "$textfile doesn't reference $file1"
fi
You'll also notice that I put the pattern in quotes and that eliminates the need to escape the pipe symbol -- makes the pattern much easier to read.
