I'd like to create a script that greps for a specific string in a log file that is being written to. I'd like to take the first result and put that into a variable for later use. This will be used though an SSH connection like so:
ssh 'user@xxx.xxx.xxx.xxx' 'bash -s' < /usr/local/bin/checklog.sh string
The command in a regular terminal
tail -f /var/log/named.log | grep $1 > $var
echo "${var}"
When I try the above method, there's no output
Thank you all for the input. You helped figure out a better way to do it, using while and tail without -f
This just reads the last line in the file. Since the file is being written to, the last line changes which is the result I was looking for.
Using a
while
loop may work for your situation, but be aware that it's not guaranteed to catch every line of the log file. Consider a situation where the log writer includes one action that writes out two lines:It's very likely that your loop will only see the second line when it performs the
tail -1
action.Not only that, but the while loop implementation means your spinning the CPU in a loop, constantly firing off
tail
commands (take a look attop
while thewhile
implementation runs, versus atail -f
).This question has some good suggestions if you just want to stop monitoring once the pattern is matched. (Note the concerns of the tail process hanging around.)
This monstrosity is probably not optimal, but it catches every line, uses minimal CPU while waiting for new lines, terminates the tail when it's done, and gives you the flexibility to write in some extra logic (like performing actions based on different matched patterns):
> $var
doesn't do what you think it does.It redirects the output of the preceding command to a file with name of what
$var
contains.To capture the output of a command and put it into a variable, use
variableName="$(...)"
.