Bash – Insert word between each matching line

bashlinuxshell

I am wondering how I can check the end of the file in while loop.
What I am doing here is if there is some word, such as, "pineapple", "apple", "onion", "orange", or etc, I want to find lines including specific words by using grep command and make some comment "window". For example, if I use grep 'a' 'file', then it will find "pineapple", "apple", and "orange". Then, finally I want to make it printed like "pineapple, window, apple, window, orange, window", something like this. So, I would like to make some condition in while or for loop. Any helps will be really appreciated.

edit:
Sample inputs

apple
banana
pineapple
orange
mandu
ricecake
meat
juice
milk
onion
green onion

Expected outputs when using grep command –> grep 'a' 'file name'

apple
window
banana
window
pineapple
window
orange
window
mandu
window
ricecake
window
meat
window

Best Answer

Using awk

To print window after every line that matches a:

$ awk '/a/{print; print "window"}' filename
apple
window
banana
window
pineapple
window
orange
window
mandu
window
ricecake
window
meat
window

How it works

/a/{...} selects lines that match the regex a. For each such line, the commands in curly braces are executed.

print prints the line containing the match

print "window" prints window.

Using sed

$ sed -n '/a/{s/$/\nwindow/; p}' filename
apple
window
banana
window
pineapple
window
orange
window
mandu
window
ricecake
window
meat
window

How it works

-n tells sed not to print unless we explicitly as it to.

/a/{...} selects lines that match the regex a. For those lines, the commands in curly braces are executed.

s/$/\nwindow/ adds a newline and window after the end of the current line.

p prints.

Related Solutions

Shell – `read` command not working in a Makefile

You're mixing shell and make variables in there. Both make and the shell use $ for their variables.

In Makefile, variables are $(var) or $v for single-letter variables, and $var or ${var} in shells.

But if you write $var in a Makefile, make will understand it as $(v)ar. If you want to pass a literal $ to the shell, you need to enter it as $$, as in $$var or $${var} so that it becomes $var or ${var} for the shell.

Also, make runs sh, not bash to interpret that code (Edit, sorry missed your SHELL := /bin/bash above, note that many systems don't have bash in /bin if they have bash at all and := is GNU specific), so you need to use sh syntax there. echo -n, read -s are zsh/bash syntax, not sh syntax.

Best here would be to add a zsh/bash script to do that (and add a build dependency on bash). Something like:

#! /usr/bin/env bash
printf 'Welcome\nPlease enter the MySQL host (default: localhost): '
read host || exit
host=${host:-localhost}
printf "Please enter the MySQL username: "
read username || exit
printf "Please enter the MySQL password:"
IFS= read -rs password || exit
printf '\n'
mv includes/config.php.example includes/config.php 2>/dev/null
repl=${password//\\/\\\\}
repl=${repl//&/\\&}
repl=${repl//:/\\:}
{ rm includes/config.php && 
  sed 's/"USER", ""/"USER", "'"$username"'"/g
       s:"PASSWORD", "":"PASSWORD", "'"$repl"'"/g' > includes/config.php
} < includes/config.php || exit
mysql -u "$username" -p"$password" codeday-team < ./codeday-team.sql || exit

echo "Configuration complete. For further configuration options, check the config file"

It addresses a few more issues:

you need -r when reading the password if you want to allow the user to use backslashes in their password.
you need IFS= if you want to allow the user to have a password that starts or ends in blanks
same would apply for the username, but here we're making the assumption that the user won't use anything silly for the username, and the blank stripping and backslash handling could be seen as a /feature/.
your ;true after mv doesn't do what you think it does. It doesn't cancel the effect of the errexit option (for make implementations that call the shell with -e). You'd need ||true instead. Here, we're not using errexit but doing the error handling ourselves with || exit where needed.
You need to escape backslash, & and the s:pattern:repl: separator (here :) or it won't work (and could have nasty side effects).
You can't do sed ... < file > file, as file would be truncated before sed is even started. Some sed implementations support a -i or -i '' option for that. Alternatively you could use perl -pi. Here we're doing the equivalent of perl -pi manually (delete and recreate the input file after it has been open for reading) but without taking care of the file's metadata.

Here, it would be better to use the example one as input and the final one as output.

It still doesn't address a few more issues:

the new config.php is created with permissions derived from the current umask which is likely to be world readable and owned by the user running make. You may need to adapt the umask and/or change ownership if the includes dir is not otherwise restricted as that file contains sensitive information.
if the password contains " characters, that will likely break (this time for php). You'd want to either forbid those (by returning an error) or add another layer of escaping for them in the right syntax for that php file. You're likely to have similar problems with backslash and you may want to exclude non-ascii or control characters as well.
Passing the password on the command-line of mysql is generally a bad idea as that means it shows in the output of ps -f. It would be better to use:
```
 mysql --defaults-file=<(
   printf '[client]\nuser=%s\npassword="%s"\n' "$username" "$password") ...
```
printf being built-in, it wouldn't show up in ps output.
the $host variable is not used.
prompting the user like that means that your script can not easily be automated. You could take the input from arguments or (better for the password) with environment variables instead.

Bash while condition

You're giving sleep two arguments, 3 and $(( i++ )). With GNU tools, this means that the loop would sleep an increasing number of seconds in each iteration. On a non-GNU system, you would get an error from sleep.
Your loop will always run at least 10 times due to the way you have written your test. Your test says "Iterate while $i is less or equal to 10". Once $i reaches 10, the other part of the test will come into effect. You probably want a logical AND test here rather than an OR.
You should not use ls | wc -l to count files in a directory. This will give the wrong result in some (albeit pathological) cases.

Instead:

i=1
while [ "$i" -ne 10 ]; do
    set -- /opt/hosts/*
    if [ "$#" -ge 2 ]; then
        break
    fi

    sleep 3         # or, to sleep longer and longer: sleep "$(( 3 + i ))"
    i=$(( i + 1 ))
done

This properly increments i in each iteration, and also uses a safer way to count the number of names in the /opt/hosts directory (by expanding the * glob in the directory and counting the number of names that it expands to). The loop exits whenever the number of names is two or more, or whenever $i reaches the value 10.

After the loop, if $i is 10, then the files failed to materialize.

If you need to preserve your positional parameters (these would be overwritten by the set command), then expand the glob into an array with names=( /opt/hosts/* ) and then use "${#names[@]}" to get the length of that array instead of using "$#".

You could also write it as

i=1
while [ "$i" -ne 10 ] && ( set -- /opt/hosts/*; [ "$#" -lt 2 ] )
do
    sleep 3
    i=$(( i + 1 ))
done

This would not clobber your existing positional parameters as set is running in a subshell. It would also be closer to the type of thing you attempted yourself.

As a "one-liner":

i=1; while [ "$i" -ne 10 ]; do set -- /opt/hosts/*; [ "$#" -ge 2 ] && break; sleep 3; i=$(( i + 1 )); done

Or, with an "arithmetic for loop" in bash:

for (( i=1; i<=10; i++ )); do set -- /opt/hosts/*; [ "$#" -ge 2 ] && break; sleep 3; done

Or, with that last piece of code above the divider:

i=1; while [ "$i" -ne 10 ] && ( set -- /opt/hosts/*; [ "$#" -lt 2 ] ); do sleep 3; i=$(( i + 1 )); done