As part of this script i need to be able to check if the first argument given matches the first word of the file If it does, exit with an error message; if it doesn't, append the arguments to the file. I understand how to write the
if statement, but not how to use
grep within a script. I understand that
grep will look something like this
grep ^$1 schemas.txt
I feel like this should be much easier than I am making it.
I'm getting an error "too many arguments" on the
if statement. I got rid of the space between
grep -q and then got an error binary operator expected.
if [ grep -q ^$1 schemas.txt ] then echo "Schema already exists. Please try again" exit 1 else echo "$@" >> schemas.txt fi
grep returns a different exit code if it found something (zero) vs. if it hasn't found anything (non-zero). In an
if statement, a zero exit code is mapped to "true" and a non-zero exit code is mapped to false. In addition, grep has a
-q argument to not output the matched text (but only return the exit status code)
So, you can use grep like this.
if grep -q PATTERN file.txt; then echo found else echo not found fi
As a quick note, when you do something like
if [ -z "$var" ]… , it turns out that
[ is actually a command you're running, just like grep. On my system, it's
/usr/bin/[ . (Well, technically, your shell probably has it built-in, but that's an optimization. It behaves as if it were a command). It works the same way,
[ returns a zero exit code for true, a non-zero exit code for false. (
test is the same thing as
[ , except for the closing