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I wanted to capture to a file the errors being returned on the command line from grep. For example,

grep foo.lookup
No such file in directory

I want to output that to a log file. This is my shell script:

var1=$(grep $lookUpVal foo.lookup) >>lookup.log 2>$1

It creates the file lookup.log but doesn't write the error on it.

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up vote 1 down vote accepted

If I understand it correct, you want to capture the output of grep into a variable and append any error to the logfile.

You could say:

var1=$(grep $lookUpVal foo.lookup 2>>lookup.log)

The $(...) syntax denotes command substitution, i.e. outputs the result of the command into a variable. By default it would capture the STDOUT of the command into the variable and the STDERR is printed to the console. In order to redirect the STDERR to a file, you would need to perform the redirection within the command itself, i.e. within $(...).

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devnull, thank you! you're the best. thank you very much! :) – user60216 Mar 1 '14 at 6:20

grep foo.lookup isn't a correct example. It's missing an argument, either a search parameter or a filename.

To capture just the error messages that grep returns to a file you could do this:

$ grep $lookUpVal foo.lookup 2> errors.log

All errors are sent to STDERR (2>) while all output is sent to STDOUT (1>). They're combined like this:

$ grep $lookUpVal foo.lookup > output.log 2>&1
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grep foo.lookup is the error message. I want that to be printed in the file. The grep is part of the shell script. It has to return a value to a variable. – user60216 Mar 1 '14 at 5:52

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