I have a problem with reading the last line of file in Linux Ubuntu.
I have a file named auth.log and I'm trying to read it last line after new line was added (after file was modified).
I know i need to use
tail -1 /var/log/auth.log to get last line but I don't know how to check the file every time it was modified.
After reading the last line i need to check if it contains "login" word and if yes I need to send this line by email. As far as I know grep needs to be use with mail command.
Any help would be awesome, thanks!
Maybe something like this would work for you, using tail -f to continuously monitor the file:
while read -r; do
echo "$REPLY" | mail your#email.address
done < <(tail -f /var/log/auth.log | grep --line-buffered login)
But this would be running continuously in the background, rather than as a cron job or something.
Related
In a open terminal how could I see all the new content added to a file whenever a process writes data into it?
I've tried combinations with cat and tee but no success
Use tail with -f
tail -f filename
Taken from the man pages for tail:
-f, --follow[={name|descriptor}]
output appended data as the file grows;
an absent option argument means 'descriptor'
You can not do it with cat, you should use tail -f <filename> or less <filename> and push F in order to wait data.
$ man less
...
F Scroll forward, and keep trying to read when the end of file is reached. Normally this
command would be used when already at the end of the file. It is a way to monitor the
tail of a file which is growing while it is being viewed. (The behavior is similar to
the "tail -f" command.)
...
One of my Linux MySQL servers suffered from a crash. So I put back a backup, however this time the MySQL is running local (localhost) instead of remotely (IP-address).
Thanks to Stack Overflow users I found an excellent command to find the IP-address in all .php files in a given directory! The command I am using for this is:
grep -r -l --include="*.php" "100.110.120.130" .
This outputs the necessary files with its location ofcourse. If it were less than 10 results, I would simply change them by hand obviously. However I received over 200 hits/results.
So now I want to know if there is a safe command which replaces the IP-address (example: 100.110.120.130) with the text "localhost" instead for all .php files in the given directory (/var/www/vhosts/) recursively.
And maybe, if only possible and not to much work, also output the changed lines to a file? I don't know if thats even possible.
Maybe someone can provide me with a working solution? To be honest, I dont dare to fool around out of the blue with this. Thats why I created a new thread.
The most standard way of replacing a string in multiple files would be to use a tool such as sed. The list of files you've obtained via grep could be read line by line (when output to a file) using a while loop in combination with sed.
$ grep -r -l --include="*.php" "100.110.120.130" . > list.txt
# this will output all matching files to list.txt
Replacing IP in matched files:
while read -r line ; do echo "$line" >> updated.txt ; sed -i 's/100.110.120.130/localhost/g' "${line}" ; done<list.txt
This will take list.txt and read it line by line to the sed command which should replace all occurrences of the IP to "localhost". The echo command directly before sed outputs all the filenames that will be modified into a file updated.txt (it isn't necessary though as list.txt contains the same exact filenames, although it could be used as a means of verification perhaps).
To do a dry run before modifying all of the matched files remove the
-i from the sed command and it will print the output to stdout
instead of in-place modifying the files.
I need to copy only the last line of a lot of files to another file. How can I do that? Please help me.
I know tail to take the last file and > to put that to other file but I can do the same thing to a lot of files?
Try:
tail -qn 1 inputfile1 inputfile2 ... > outputfile
-n 1 for outputting only the last line, -q for suppressing the header.
See:
man tail
I want to make a simple dmenu command that reads a file of commands and names. Then takes the names and displays them using dmenu then takes dmenu's output and runs the associated command using the file again.
I got to the point where dmenu displays the names, but I don't really know where to go from there. Learning bash is a really daunting task to me and I don't really know where to start with this seemingly simple script/command.
here is the file:
Pushbullet
google-chrome-stable --app=https://www.pushbullet.com
Steam
steam
Chrome
google-chrome-stable
Libre Office
libreoffice
Transmission
transmission-qt
Audio Control Panel
sudo pavucontrol & bluberry
and here is what I have so far for my command:
awk 'NR % 2 != 0' /home/rocco/programlist | dmenu | ??(grep -l "stdout" /home/rocco/programlist....)
It was my thinking that I could somehow pipe into grep or awk with the name of the application then get the line number then add one and pipe that into sh.
Thanks
I have no experience with dmenu but if I understand how it works correctly, this should do what you want. Wrapping a command in $(…) returns the output as a variable, which we can pass on to another command.
#!/bin/bash
plist="/home/rocco/programlist"
# pipe every second line to dmenu
selected=$(awk 'NR % 2 != 0' "$plist" | dmenu)
# search for the selected item, get the command after it
cmd=$(grep -A1 "$selected" "$plist" | tail -n 1)
# run the command
$cmd
Worth mentioning a mistake in your question. dmenu sends to stdout, or standard output, but the next program in line would be reading stdin, or standard input. In any case, grep can't take patterns on standard input, which is why I've saved to a variable instead of trying to pipe it somewhere.
Assuming you have programlist.txt in the working directory you can use:
awk 'NR%2 !=0' programlist.txt |dmenu |awk '{system("grep --no-group-separator -A 1 '"'"'"$0"'"'"' programlist.txt");}' |awk '{if(NR==2){system($0);}}'
Note the quoting of the $0 in the first awk envocation. This is necessary to get names with spaces in them like "Libre Office"
I use screen to run a minecraft server .jar, and I would like to write a bash script to see if the most recent line has changed every five minutes or so. If it has, then the script would start from the beginning and make the check again in another five minutes. If not, it should kill the java process.
How would I go about getting the last line of text from a screen via a bash script?
If I have understand, you can redirect the output of your program in a file and work on it, with the operator >.
Try to run :
ls -l > myoutput.txt
and open the file created.
You want to use the tail command. tail -n 1 will give you the last line of the file or redirected standard output, while tail -f will keep the tail program going until you cancel it yourself.
For example:
echo -e "Jello\nPudding\nSkittles" | tail -n 1 | if grep -q Skittles ; then echo yes; fi
The first section simply prints three lines of text:
Jello
Pudding
Skittles
The tail -n 1 finds the last line of text ("Skittles") and passes that to the next section.
grep -q simply returns TRUE if your pattern was found or FALSE if not, without actually dumping or outputting anything to screen.
So the if grep -q Skittles section will check the result of that grep -q Skittles pattern and, if it found Skittles, prints 'yes' to the screen. If not, nothing gets printed (try replacing Skittles with Pudding, and even though it was in the original input, it never made it out the other end of the tail -n 1 call).
Maybe you can use that logic and output your .jar to standard output, then search that output every 5 minutes?