I am running the following command using awk on file.txt ,currently its running the command on the ids present in file.txt from top to bottom..i want the commmand to be run in the reverse order for the ids present in file.txt..any inputs on how we can do this?
git command $(awk '{print $1}' file.txt)
file.txt contains.
97a65fd1d1b3b8055edef75e060738fed8b31d3
fb8df67ceff40b4fc078ced31110d7a42e407f16
a0631ce8a9a10391ac4dc377cd79d1adf1f3f3e2
.....
If you aren't bound to using awk then tail with the -r (for reverse) argument will do the trick...
myFile.txt
97a65fd1d1b3b8055edef75e060738fed8b31d3
fb8df67ceff40b4fc078ced31110d7a42e407f16
a0631ce8a9a10391ac4dc377cd79d1adf1f3f3e2
Now to print it in reverse...
$ tail -r myFile.txt
a0631ce8a9a10391ac4dc377cd79d1adf1f3f3e2
fb8df67ceff40b4fc078ced31110d7a42e407f16
97a65fd1d1b3b8055edef75e060738fed8b31d3
EDIT:
To output this to a file simply redirect it out...
$ tail -r myFile.txt > newFile.txt
EDIT:
Want to write to the same file? No problem!
tail -r myFile.txt > temp.txt; cat temp.txt > myFile.txt; rm temp.txt;
For some reason when I redirected tail -r to the same file it came back blank, this workaround avoids that issue by writing to a temporary "buffer" file.
To reverse the lines in a file using awk, use
awk '{a[i++]=$0} END {for (j=i-1; j>=0;) print a[j--] }' file
use $1 instead of $0 above to operate on the first field only instead of the whole line.
Related
I have a awk program add_hashtag.awk
BEGIN{printf("#")}1
and a bash program
for file in *.asc; do awk -f add_hashtag.awk "$file" > "$file"_in; done
that add hashtag into file. It works, however, I would like to get files with same names. When I run
for file in *.asc; do awk -f add_hashtag.awk "$file" > "$file"; done
I get files only with #.
How to do that? Thank you
Could you please try following.
for file in *.asc; do awk -f add_hashtag.awk "$file" > "temp_file" && mv "temp_file" "$file"; done
I am going with approach where creating a temp_file for output and later renaming it to Input_file so that there will not be any danger of losing or truncating actual Input_file. Also it will not rename temp_file to actual Input_file until/unless awk command is a success(with use of &&)
With gawk 4.1.0 version or so try(haven't tested it since no samples were given):
awk -i inplace -f add_hashtag.awk *.asc
OR in case you want to inplace edit files along with taking their backup:
awk -i inplace -v INPLACE_SUFFIX=.backup -f add_hashtag.awk *.asc
I'm writing a shell script which shut down some services and trying to get its pid by using the following awk script.
However, this awk script can't get pid. What's wrong with that?
ps -ef | awk -v port_no=10080 '/[m]ilk.*port=port_no/{print $2}'
The result of ps -ef is like this:
username 13155 27705 0 16:06 pts/2 00:00:00 /home/username/.rbenv/versions/2.3.6/bin/ruby /home/username/.rbenv/versions/2.3.6/bin/milk web --no-browser --host=example.com --port=10080
This process is working with a different port argument as well, so I want to kill the process only working on port=10080.
The awk script below works fine, but when I specify the port no using awk -v like the above, it doesn't work well.
ps -ef | awk '/[m]ilk.*port=10080/{print $2}'
awk version: GNU Awk 4.0.2
The syntax for pattern matching with /../ does not work with variables in the regular expression. You need to use the ~ syntax for it.
awk -v port_no=10080 '$0 ~ "[m]ilk.*port="port_no{print $2}'
If you notice the regex carefully, the regex string on the r.h.s of ~ is under the double-quotes ".." except the variable name holding the port number which shouldn't be under quotes, for the expansion to happen.
This task is easily accomplished using pgrep:
$ pgrep -f '[m]ilk.*port=10080'
Have a look at man pgrep for details.
i have this file content in my sample file "haproxy-monitoring.conf"
[[inputs.haproxy]]
servers = ["http://localhost:31330/haproxy?stats" ]
Can you please help me, how I can extract just the port number '31330' from the file haproxy-monitoring.conf in a bash script.
with sed
$ sed -rn '/servers/s/.*:([0-9]+).*/\1/p' file
or similarly with awk
$ awk '/servers/{print gensub(/.*:([0-9]+).*/,"\\1",1)}' file
awk -F'[:/]' '{print $5}' file
31330
Or something like
grep -Eo '[0-9]+' file
Questions that state what the output should be without explaining why it should be that leave you open to all sorts of answers unrelated to what you really are trying to do. idk if this is what you want or not since you haven't told us:
$ tr -cd '0-9' < file
31330
I am new to awk and shell based programming. I have a bunch of files name file_0001.dat, file_0002.dat......file_1000.dat. I want to change the file names such as the number after file_ will be a multiple of 4 in comparison to previous file name. SO i want to change
file_0001.dat to file_0004.dat
file_0002.dat to file_0008.dat
and so on.
Can anyone suggest a simple script to do it. I have tried the following but without any success.
#!/bin/bash
a=$(echo $1 sed -e 's:file_::g' -e 's:.dat::g')
b=$(echo "${a}*4" | bc)
shuf file_${a}.dat > file_${b}.dat
This script will do that trick for you:
#!/bin/bash
for i in `ls -r *.dat`; do
a=`echo $i | sed 's/file_//g' | sed 's/\.dat//g'`
almost_b=`bc -l <<< "$a*4"`
b=`printf "%04d" $almost_b`
rename "s/$a/$b/g" $i
done
Files before:
file_0001.dat file_0002.dat
Files after first execution:
file_0004.dat file_0008.dat
Files after second execution:
file_0016.dat file_0032.dat
Here's a pure bash way of doing it (without bc, rename or sed).
#!/bin/bash
for i in $(ls -r *.dat); do
prefix="${i%%_*}_"
oldnum="${i//[^0-9]/}"
newnum="$(printf "%04d" $(( 10#$oldnum * 4 )))"
mv "$i" "${prefix}${newnum}.dat"
done
To test it you can do
mkdir tmp && cd $_
touch file_{0001..1000}.dat
(paste code into convert.sh)
chmod +x convert.sh
./convert.sh
Using bash/sed/find:
files=$(find -name 'file_*.dat' | sort -r)
for file in $files; do
n=$(sed 's/[^_]*_0*\([^.]*\).*/\1/' <<< "$file")
let n*=4
nfile=$(printf "file_%04d.dat" "$n")
mv "$file" "$nfile"
done
ls -r1 | awk -F '[_.]' '{printf "%s %s_%04d.%s\n", $0, $1, 4*$2, $3}' | xargs -n2 mv
ls -r1 list file in reverse order to avoid conflict
the second part will generate new filename. For example: file_0002.dat will become file_0002.dat file_0008.dat
xargs -n2 will pass two arguments every time to mv
This might work for you:
paste <(seq -f'mv file_%04g.dat' 1000) <(seq -f'file_%04g.dat' 4 4 4000) |
sort -r |
sh
This can help:
#!/bin/bash
for i in `cat /path/to/requestedfiles |grep -o '[0-9]*'`; do
count=`bc -l <<< "$i*4"`
echo $count
done
Given this script
#!/bin/awk -f
{
print $1
}
It can be called like so
foo.awk foo.txt
However I would like the script to always call foo.txt. So I would like to modify the script so that it can be called without the input file, like this
foo.awk
#!/bin/awk -f
BEGIN {ARGV[ARGC++] = "foo.txt"}
{print $1}
This will add foo.txt to the end of the arguments list, as if you had put it there on the command line. This has the added bonus of allowing you to extend your script to do more than just print, without having to put everything in the BEGIN block.
I would use a shell wrapper for this:
#!/bin/bash
foo.awk foo.txt
ok, do this trick maybe? (cheat?)
#!/bin/sh
awk '{print $1}' foo.txt
you could name it as foo.awk
and then
chmod +x foo.awk
now try ./foo.awk under same directory.
EDIT
#!/bin/awk -f
BEGIN{while(getline < "/path/to/foo.txt")print $1 }
strange requirement. why bash wrapper doesn't fit your requirement? You did tag the question with shell
anyway, the above script should be what you need.