Bash: grep result of command using pattern from a different command?

Question

I'm trying to grep the command df -h with a pattern that comes from a mysql command. Right now I have something like this:

df -hP | grep $(mysql -uroot -e "select statement")

Right now this is trying to grep the result of the mysql query instead of using the result as the pattern to grep the df result

The result of the mysql statement is "raid_48"

I will then want to pipe this into mailx. Maybe I shouldn't be trying to do this with a one liner

Can you add the output of $(mysql -uroot -e "select statement") depending on what it returns it may break grep syntax. — Tiago Lopo
– Tiago Lopo, Commented Jul 14, 2014 at 20:14
If grep is trying to process raid_48 as a file instead of an expression, try adding -e as also suggested in my answer. — konsolebox
– konsolebox, Commented Jul 14, 2014 at 20:24

konsolebox · Accepted Answer · 2014-07-14 20:14:57Z

2

If there may be something wrong with your command you can consider these:

# Add `-e` before the argument to explicitly tell grep that the argument that follows is an expression not an option.
# Quote your argument to prevent word splitting with spaces.
df -hP | grep -e "$(mysql -uroot -e "select statement")"

# Perhaps try using fgrep as well to prevent reading argument as regex.
df -hP | fgrep -e "$(mysql -uroot -e "select statement")"

answered Jul 14, 2014 at 20:14

konsolebox

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