How to create a bash function with variable parameters/arguments to grep several keywords/tags - linux

I'm using the :!grep "tag1" filename | grep "tag2" filename | grep -n "tag3 or more" filename command in vim to search for my code snippets based on their tags (a simple comment at the top of a snippet) in one big file, similar to firefox's tag functionality. I use snippets to remember tricky things.
This is painful to write out each time. I'd like to make an alias, or function to do something like this:
:!greptag tag1 tag2 ... tag39
And it should search the current doc and return the lines with all the tags on them.
Vim is set to interactive shell mode so that it can parse my bashrc for aliases/functions.
set shellcmdflag=-ic "lets vim use bashrc
How can I construct a function that allows for variable arguments like this in bash?

You could also use sed:
sed e '/tag1/!d;/tag2/!d;.../tagN/!d' filename
The /tag1/! command is an address prefix to tell sed to only execute the command if the line contains tag1. The command we execute is "d", to delete the line.
A function (in .bashrc) would then be:
greptags () {
local filename="$1"
shift
sed "$filename" -e "$(echo "$#" | sed -e 's,\([^ ]*\) *,/\1/\!d;,g')"
}
Notice that we use another sed command to parse the arguments into the final sed command list. Basically, for every word (no spaces), we remove the spaces that follows it and put it into the /tagN/!d; command form.
And call it:
greptags filename tag1 tag2 tag3 tag4
Hope this helps =)

How about something like this:
greptags() {
if [[ -z "$1" ]] ; then
cat
else
local t="$1" ; shift
greptags "$#" | grep "$t"
fi
}
greptag() {
local f="$1" ; shift
local t="$2" ; shift
grep "$t" "$f" | greptags "$#"
}
(untested and, probably not exactly what you want, but illustrating the idea).

Related

bash script to replace all occurrences of placeholders in file

I'm trying to write a bash script to replace all occurrences of a placeholder in a file with an environment variable of the same name. As an example, if I have a file like the following...
This is an {{VAR1}} {{VAR2}}.
It should work across multiple lines in this {{VAR2}}.
... and I have the following environment variables set:
VAR1='example'
VAR2='file'
after running the script on my file, I should get the output:
This is an example file.
It should work across multiple lines in this file.
I'm sure there must be a solution using awk/sed, but so far the closest I've come can't handle if there's more than one variable on a line. Here's my attempt so far:
cat example.txt | grep -o '{{.*}}' > temp
while read placeholder; do
varName=$(echo "$placeholder" | tr -d '{}')
value="${!varName}"
sed -i "s/$placeholder/$value/g" "$file"
done < temp
rm -rf temp
I'd use Perl:
perl -pe 's/{{(.*?)}}/$ENV{$1}/g' filename
This assumes that VAR1 and VAR2 are environment variables (i.e., are exported), so that Perl can pick them out of its environment. This would be required of any approach that isn't pure shell; I just mention it to avoid confusion.
This works as follows:
s/pattern/replacement/g is a substitution command; you may recognize it from sed. The difference is that here we can use Perl's more powerful regex engine and variables. The g flag makes it so that all matches are replaced; without it, it would apply only to the first.
In the pattern, .*? matches non-greedily, so that in a line that contains foo {{VAR1}} bar {{VAR2}} baz, the pattern {{.*?}} matches only {{VAR1}} instead of {{VAR1}} bar {{VAR2}}.
The part between {{ and }} is captured because it is between () and can be reused as $1
$ENV{$1} in the replacement uses the special %ENV hash that contains the environment of the Perl process. $ENV{$1} is the value of the environment variable that has the name $1, which is the captured group from before.
Only bash and sed:
$ VAR1='example'
$ VAR2='file'
$ export VAR1 VAR2
$ sed -e '{s/{{\([^{]*\)}}/${\1}/g; s/^/echo "/; s/$/";/}' -e e filename
This is an example file.
It should work across multiple lines in this file.
sed -e '{s/{{\([^{]*\)}}/${\1}/g;}' filename:
This is an ${VAR1} ${VAR2}.
It should work across multiple lines in this ${VAR2}.
{{\([^{]*\)}} - Search for {{..}}
[^{] - Non greedy match
\1 - Access to the bracketed values \(...\).
sed -e '{s/{{\([^{]*\)}}/${\1}/g; s/^/echo "/; s/$/";/}' filename:
echo "This is an ${VAR1} ${VAR2}.";
echo "It should work across multiple lines in this ${VAR2}.";
s/^/echo "/ - Replace the beginning of the line with echo "
s/$/";/ - Replace the end of the line with ";
I was just playing with your original approach. Wouldn't adding another loop on $varName work?
cat example.txt | grep -o '{{.*}}' > temp
while read placeholder; do
varName=$(echo "$placeholder" | tr -d '{}')
for i in $varName; do
value="${!i}"
sed -i "s/{{$i}}/$value/g" example.txt
done
done < temp
rm -rf temp

Using a variable to replace lines in a file with backslashes

I want to add the string %%% to the beginning of some specific lines in a text file.
This is my script:
#!/bin/bash
a="c:\Temp"
sed "s/$a/%%%$a/g" <File.txt
And this is my File.txt content:
d:\Temp
c:\Temp
e:\Temp
But nothing changes when I execute it.
I think the 'sed' command is not finding the pattern, possibly due to the \ backslashes in the variable a.
I can find the c:\Temp line if I use grep with -F option (to not interpret strings):
cat File.txt | grep -F "$a"
But sed seems not to implement such '-F` option.
Not working neither:
sed 's/$a/%%%$a/g' <File.txt
sed 's/"$a"/%%%"$a"/g' <File.txt
I have found similar threads about replacing with sed, but they don't refer to variables.
How can I replace the desired lines by using a variable adding them the %%% char string?
EDIT: It would be fine that the $a variable could be entered via parameter when calling the script, so it will be assigned like:
a=$1
Try it like this:
#!/bin/sh
a='c:\\Temp' # single quotes
sed "s/$a/%%%$a/g" <File.txt # double quotes
Output:
Johns-MacBook-Pro:sed jcreasey$ sh x.sh
d:\Temp
e:\Temp
%%%c:\Temp
You need the double slash '\' to escape the '\'.
The single quotes won't expand the variables.
So you escape the slash in single quotes and pass it into the double quotes.
Of course you could also just do this:
#!/bin/sh
sed 's/\(.*Temp\)/%%%&/' <File.txt
If you want to get input from the command line you have to allow for the fact that \ is an escape character there too. So the user needs to type 'c:\\' or the interpreter will just wait for another character. Then once you get it, you will need to escape it again. (printf %q).
#!/bin/sh
b=`printf "%q" $1`
sed "s/\($b\)/%%% &/" < File.txt
The issue you are having has to do with substitution of your variable providing a regular expression looking for a literal c:Temp with the \ interpreted as an escape by the shell. There are a number of workarounds. Seeing the comments and having worked through the possibilities, the following will allow the unquoted entry of the search term:
#!/bin/bash
## validate that needed input is given on the command line
[ -n "$1" -a "$2" ] || {
printf "Error: insufficient input. Usage: %s <term> <file>\n" "${0//*\//}" >&2
exit 1
}
## validate that the filename given is readable
[ -r "$2" ] || {
printf "Error: file not readable '%s'\n" "$2" >&2
exit 1
}
a="$1" # assign a
filenm="$2" # assign filename
## test and fix the search term entered
[[ "$a" =~ '/' ]] || a="${a/:/:\\}" # test if \ removed by shell, if so replace
a="${a/\\/\\\\}" # add second \
sed -e "s/$a/%%%$a/g" "$filenm" # call sed with output to stdout
Usage:
$ bash sedwinpath.sh c:\Temp dat/winpath.txt
d:\Temp
%%%c:\Temp
e:\Temp
Note: This allows both single-quoted or unquoted entry of the dos path search term. To edit in place use sed -i. Additionally, the [[ operator and =~ operator are limited to bash.
I could have sworn the original question said replace, but to append, just as you suggest in the comments. I have updated the code with:
sed -e "s/$a/%%%$a/g" "$filenm"
Which provides the new output:
$ bash sedwinpath.sh c:\Temp dat/winpath.txt
d:\Temp
%%%c:\Temp
e:\Temp
Remember: If you want to edit the file in place use sed -i or sed -i.bak which will edit the actual file (and if -i.bak is given create a backup of the original in originalname.bak). Let me know if that is not what you intended and I'm happy to edit again.
Creating your script with a positional parameter of $1
#!/bin/bash
a="$1"
cat <file path>|sed "s/"$1"/%%%"$1"/g" > "temporary file"
Now whenever you want sed to find "c:\Temp" you need to use your script command line as follows
bash <my executing script> c:\\\\Temp
The first backslash will make bash interpret any backslashes that follows therefore what will be save in variable "a" in your executing script is "c:\\Temp". Now substituting this variable in sed will cause sed to interpret 1 backlash since the first backslash in this variable will cause sed to start interpreting the other backlash.
when you Open your temporary file you will see:
d:\Temp
%%%c:\Temp
e:\Temp

How to insert a text at the beginning of a file?

So far I've been able to find out how to add a line at the beginning of a file but that's not exactly what I want. I'll show it with an example:
File content
some text at the beginning
Result
<added text> some text at the beginning
It's similar but I don't want to create any new line with it...
I would like to do this with sed if possible.
sed can operate on an address:
$ sed -i '1s/^/<added text> /' file
What is this magical 1s you see on every answer here? Line addressing!.
Want to add <added text> on the first 10 lines?
$ sed -i '1,10s/^/<added text> /' file
Or you can use Command Grouping:
$ { echo -n '<added text> '; cat file; } >file.new
$ mv file{.new,}
If you want to add a line at the beginning of a file, you need to add \n at the end of the string in the best solution above.
The best solution will add the string, but with the string, it will not add a line at the end of a file.
sed -i '1s/^/your text\n/' file
If the file is only one line, you can use:
sed 's/^/insert this /' oldfile > newfile
If it's more than one line. one of:
sed '1s/^/insert this /' oldfile > newfile
sed '1,1s/^/insert this /' oldfile > newfile
I've included the latter so that you know how to do ranges of lines. Both of these "replace" the start line marker on their affected lines with the text you want to insert. You can also (assuming your sed is modern enough) use:
sed -i 'whatever command you choose' filename
to do in-place editing.
Use subshell:
echo "$(echo -n 'hello'; cat filename)" > filename
Unfortunately, command substitution will remove newlines at the end of file. So as to keep them one can use:
echo -n "hello" | cat - filename > /tmp/filename.tmp
mv /tmp/filename.tmp filename
Neither grouping nor command substitution is needed.
To insert just a newline:
sed '1i\\'
You can use cat -
printf '%s' "some text at the beginning" | cat - filename
To add a line to the top of the file:
sed -i '1iText to add\'
my two cents:
sed -i '1i /path/of/file.sh' filename
This will work even is the string containing forward slash "/"
Hi with carriage return:
sed -i '1s/^/your text\n/' file
Note that on OS X, sed -i <pattern> file, fails. However, if you provide a backup extension, sed -i old <pattern> file, then file is modified in place while file.old is created. You can then delete file.old in your script.
There is a very easy way:
echo "your header" > headerFile.txt
cat yourFile >> headerFile.txt
PROBLEM: tag a file, at the top of the file, with the base name of the parent directory.
I.e., for
/mnt/Vancouver/Programming/file1
tag the top of file1 with Programming.
SOLUTION 1 -- non-empty files:
bn=${PWD##*/} ## bn: basename
sed -i '1s/^/'"$bn"'\n/' <file>
1s places the text at line 1 of the file.
SOLUTION 2 -- empty or non-empty files:
The sed command, above, fails on empty files. Here is a solution, based on https://superuser.com/questions/246837/how-do-i-add-text-to-the-beginning-of-a-file-in-bash/246841#246841
printf "${PWD##*/}\n" | cat - <file> > temp && mv -f temp <file>
Note that the - in the cat command is required (reads standard input: see man cat for more information). Here, I believe, it's needed to take the output of the printf statement (to STDIN), and cat that and the file to temp ... See also the explanation at the bottom of http://www.linfo.org/cat.html.
I also added -f to the mv command, to avoid being asked for confirmations when overwriting files.
To recurse over a directory:
for file in *; do printf "${PWD##*/}\n" | cat - $file > temp && mv -f temp $file; done
Note also that this will break over paths with spaces; there are solutions, elsewhere (e.g. file globbing, or find . -type f ... -type solutions) for those.
ADDENDUM: Re: my last comment, this script will allow you to recurse over directories with spaces in the paths:
#!/bin/bash
## https://stackoverflow.com/questions/4638874/how-to-loop-through-a-directory-recursively-to-delete-files-with-certain-extensi
## To allow spaces in filenames,
## at the top of the script include: IFS=$'\n'; set -f
## at the end of the script include: unset IFS; set +f
IFS=$'\n'; set -f
# ----------------------------------------------------------------------------
# SET PATHS:
IN="/mnt/Vancouver/Programming/data/claws-test/corpus test/"
# https://superuser.com/questions/716001/how-can-i-get-files-with-numeric-names-using-ls-command
# FILES=$(find $IN -type f -regex ".*/[0-9]*") ## recursive; numeric filenames only
FILES=$(find $IN -type f -regex ".*/[0-9 ]*") ## recursive; numeric filenames only (may include spaces)
# echo '$FILES:' ## single-quoted, (literally) prints: $FILES:
# echo "$FILES" ## double-quoted, prints path/, filename (one per line)
# ----------------------------------------------------------------------------
# MAIN LOOP:
for f in $FILES
do
# Tag top of file with basename of current dir:
printf "[top] Tag: ${PWD##*/}\n\n" | cat - $f > temp && mv -f temp $f
# Tag bottom of file with basename of current dir:
printf "\n[bottom] Tag: ${PWD##*/}\n" >> $f
done
unset IFS; set +f
Just for fun, here is a solution using ed which does not have the problem of not working on an empty file. You can put it into a shell script just like any other answer to this question.
ed Test <<EOF
a
.
0i
<added text>
.
1,+1 j
$ g/^$/d
wq
EOF
The above script adds the text to insert to the first line, and then joins the first and second line. To avoid ed exiting on error with an invalid join, it first creates a blank line at the end of the file and remove it later if it still exists.
Limitations: This script does not work if <added text> is exactly equal to a single period.
echo -n "text to insert " ;tac filename.txt| tac > newfilename.txt
The first tac pipes the file backwards (last line first) so the "text to insert" appears last. The 2nd tac wraps it once again so the inserted line is at the beginning and the original file is in its original order.
The simplest solution I found is:
echo -n "<text to add>" | cat - myFile.txt | tee myFile.txt
Notes:
Remove | tee myFile.txt if you don't want to change the file contents.
Remove the -n parameter if you want to append a full line.
Add &> /dev/null to the end if you don't want to see the output (the generated file).
This can be used to append a shebang to the file. Example:
# make it executable (use u+x to allow only current user)
chmod +x cropImage.ts
# append the shebang
echo '#''!'/usr/bin/env ts-node | cat - cropImage.ts | tee cropImage.ts &> /dev/null
# execute it
./cropImage.ts myImage.png
Another solution with aliases. Add to your init rc/ env file:
addtail () { find . -type f ! -path "./.git/*" -exec sh -c "echo $# >> {}" \; }
addhead () { find . -type f ! -path "./.git/*" -exec sh -c "sed -i '1s/^/$#\n/' {}" \; }
Usage:
addtail "string to add at the beginning of file"
addtail "string to add at the end of file"
With the echo approach, if you are on macOS/BSD like me, lose the -n switch that other people suggest. And I like to define a variable for the text.
So it would be like this:
Header="my complex header that may have difficult chars \"like these quotes\" and line breaks \n\n "
{ echo "$Header"; cat "old.txt"; } > "new.txt"
mv new.txt old.txt
TL;dr -
Consider using ex. Since you want the front of a given line, then the syntax is basically the same as what you might find for sed but the option of "in place editing" is built-in.
I cannot imagine an environment where you have sed but not ex/vi, unless it is a MS Windows box with some special "sed.exe", maybe.
sed & grep sort of evolved from ex / vi, so it might be better to say sed syntax is the same as ex.
You can change the line number to something besides #1 or search for a line and change that one.
source=myFile.txt
Front="This goes IN FRONT "
man true > $source
ex -s ${source} <<EOF
1s/^/$Front/
wq
EOF
$ head -n 3 $source
This goes IN FRONT TRUE(1) User Commands TRUE(1)
NAME
Long version, I recommend ex (or ed if you are one of the cool kids).
I like ex because it is portable, extremely powerful, allows me to write in-place, and/or make backups all without needing GNU (or even BSD) extensions.
Additionally, if you know the ex way, then you know how to do it in vi - and probably vim if that is your jam.
Notice that EOF is not quoted when we use "i"nsert and using echo:
str="+++ TOP +++" && ex -s <<EOF
r!man true
1i
`echo "$str"`
.
"0r!echo "${str}"
wq! true.txt
EOF
0r!echo "${str}" might also be used as shorthand for :0read! or :0r! that you have likely used in vi mode (it is literally the same thing) but the : is optional here and some implementations do not support "r"ead address of zero.
"r"eading directly to the special line #0 (or from line 1) would automatically push everything "down", and then you just :wq to save your changes.
$ head -n 3 true.txt | nl -ba
1 +++ TOP +++
2 TRUE(1) User Commands TRUE(1)
3
Also, most classic sed implementations do not have extensions (like \U&) that ex should have by default.
cat concatenates multiple files. <() sends output of a command as a file. Combining these two, we can insert lines at the beginning and end of a file by,
cat <(echo "line before the file") file.txt <(echo "line after the file")

Simplify a BASH scripting design

I have need to execute a command in a script an arbitrary number of times with associated arbitrary parameters.
I've decided the script will take its cue from a parameter file (parameter.txt) where lines are of the form:
label param1 param2
For each line in parameter.txt, I'll call the command with the specified parameters.
So far, my tinkering is moving along the lines of the following, but it's looking messy:
while read line; do
echo $line | sed -r 's/[^ ]+ ([^ ]+).+/\1/' &&
echo $line | sed -r 's/[^ ]+ [^ ]+ ([^ ]+)/\1/'
done < parameter.txt
My command is of the form:
mycmd -a param1 -b param2 > label
Could I get some suggestions how I might simplify this?
I'm doing this for a small embedded system whose 'helper' commands are in short supply (xargs for example isn't available, and things like awk are hobbled busybox implementations), and I'm using version 2 (2.04g I think) of BASH.
while read label param1 param2; do
mycmd -a "$param1" -b "$param2" > "$label"
done < parameter.txt
I'd suggest a function, as long as there aren't any embedded spaces.
function x()
{
mycmd -a $2 -b $3 >$1
}
while read line; do x $line ; done <parameter.txt
Try this:
while read line ; do
set -- $line
dest="$1"
shift
mycmd "$#" > "$dest"
done < parameter.txt
should work. If the parameters in the file have spaces, you will have to quote them properly.
I suggest to add the -a, -b to the file parameter.txt because generating them on the fly is probably brittle.
If you don't like this solution, then I suggest to create a new script from this one which contains the actual commands. That way, you can easily debug any problems.
When the script looks okay, you can source it with source ./generated.sh (yes, you have to specify the path).

Quick unix command to display specific lines in the middle of a file?

Trying to debug an issue with a server and my only log file is a 20GB log file (with no timestamps even! Why do people use System.out.println() as logging? In production?!)
Using grep, I've found an area of the file that I'd like to take a look at, line 347340107.
Other than doing something like
head -<$LINENUM + 10> filename | tail -20
... which would require head to read through the first 347 million lines of the log file, is there a quick and easy command that would dump lines 347340100 - 347340200 (for example) to the console?
update I totally forgot that grep can print the context around a match ... this works well. Thanks!
I found two other solutions if you know the line number but nothing else (no grep possible):
Assuming you need lines 20 to 40,
sed -n '20,40p;41q' file_name
or
awk 'FNR>=20 && FNR<=40' file_name
When using sed it is more efficient to quit processing after having printed the last line than continue processing until the end of the file. This is especially important in the case of large files and printing lines at the beginning. In order to do so, the sed command above introduces the instruction 41q in order to stop processing after line 41 because in the example we are interested in lines 20-40 only. You will need to change the 41 to whatever the last line you are interested in is, plus one.
# print line number 52
sed -n '52p' # method 1
sed '52!d' # method 2
sed '52q;d' # method 3, efficient on large files
method 3 efficient on large files
fastest way to display specific lines
with GNU-grep you could just say
grep --context=10 ...
No there isn't, files are not line-addressable.
There is no constant-time way to find the start of line n in a text file. You must stream through the file and count newlines.
Use the simplest/fastest tool you have to do the job. To me, using head makes much more sense than grep, since the latter is way more complicated. I'm not saying "grep is slow", it really isn't, but I would be surprised if it's faster than head for this case. That'd be a bug in head, basically.
What about:
tail -n +347340107 filename | head -n 100
I didn't test it, but I think that would work.
I prefer just going into less and
typing 50% to goto halfway the file,
43210G to go to line 43210
:43210 to do the same
and stuff like that.
Even better: hit v to start editing (in vim, of course!), at that location. Now, note that vim has the same key bindings!
You can use the ex command, a standard Unix editor (part of Vim now), e.g.
display a single line (e.g. 2nd one):
ex +2p -scq file.txt
corresponding sed syntax: sed -n '2p' file.txt
range of lines (e.g. 2-5 lines):
ex +2,5p -scq file.txt
sed syntax: sed -n '2,5p' file.txt
from the given line till the end (e.g. 5th to the end of the file):
ex +5,p -scq file.txt
sed syntax: sed -n '2,$p' file.txt
multiple line ranges (e.g. 2-4 and 6-8 lines):
ex +2,4p +6,8p -scq file.txt
sed syntax: sed -n '2,4p;6,8p' file.txt
Above commands can be tested with the following test file:
seq 1 20 > file.txt
Explanation:
+ or -c followed by the command - execute the (vi/vim) command after file has been read,
-s - silent mode, also uses current terminal as a default output,
q followed by -c is the command to quit editor (add ! to do force quit, e.g. -scq!).
I'd first split the file into few smaller ones like this
$ split --lines=50000 /path/to/large/file /path/to/output/file/prefix
and then grep on the resulting files.
If your line number is 100 to read
head -100 filename | tail -1
Get ack
Ubuntu/Debian install:
$ sudo apt-get install ack-grep
Then run:
$ ack --lines=$START-$END filename
Example:
$ ack --lines=10-20 filename
From $ man ack:
--lines=NUM
Only print line NUM of each file. Multiple lines can be given with multiple --lines options or as a comma separated list (--lines=3,5,7). --lines=4-7 also works.
The lines are always output in ascending order, no matter the order given on the command line.
sed will need to read the data too to count the lines.
The only way a shortcut would be possible would there to be context/order in the file to operate on. For example if there were log lines prepended with a fixed width time/date etc.
you could use the look unix utility to binary search through the files for particular dates/times
Use
x=`cat -n <file> | grep <match> | awk '{print $1}'`
Here you will get the line number where the match occurred.
Now you can use the following command to print 100 lines
awk -v var="$x" 'NR>=var && NR<=var+100{print}' <file>
or you can use "sed" as well
sed -n "${x},${x+100}p" <file>
With sed -e '1,N d; M q' you'll print lines N+1 through M. This is probably a bit better then grep -C as it doesn't try to match lines to a pattern.
Building on Sklivvz' answer, here's a nice function one can put in a .bash_aliases file. It is efficient on huge files when printing stuff from the front of the file.
function middle()
{
startidx=$1
len=$2
endidx=$(($startidx+$len))
filename=$3
awk "FNR>=${startidx} && FNR<=${endidx} { print NR\" \"\$0 }; FNR>${endidx} { print \"END HERE\"; exit }" $filename
}
To display a line from a <textfile> by its <line#>, just do this:
perl -wne 'print if $. == <line#>' <textfile>
If you want a more powerful way to show a range of lines with regular expressions -- I won't say why grep is a bad idea for doing this, it should be fairly obvious -- this simple expression will show you your range in a single pass which is what you want when dealing with ~20GB text files:
perl -wne 'print if m/<regex1>/ .. m/<regex2>/' <filename>
(tip: if your regex has / in it, use something like m!<regex>! instead)
This would print out <filename> starting with the line that matches <regex1> up until (and including) the line that matches <regex2>.
It doesn't take a wizard to see how a few tweaks can make it even more powerful.
Last thing: perl, since it is a mature language, has many hidden enhancements to favor speed and performance. With this in mind, it makes it the obvious choice for such an operation since it was originally developed for handling large log files, text, databases, etc.
print line 5
sed -n '5p' file.txt
sed '5q' file.txt
print everything else than line 5
`sed '5d' file.txt
and my creation using google
#!/bin/bash
#removeline.sh
#remove deleting it comes move line xD
usage() { # Function: Print a help message.
echo "Usage: $0 -l LINENUMBER -i INPUTFILE [ -o OUTPUTFILE ]"
echo "line is removed from INPUTFILE"
echo "line is appended to OUTPUTFILE"
}
exit_abnormal() { # Function: Exit with error.
usage
exit 1
}
while getopts l:i:o:b flag
do
case "${flag}" in
l) line=${OPTARG};;
i) input=${OPTARG};;
o) output=${OPTARG};;
esac
done
if [ -f tmp ]; then
echo "Temp file:tmp exist. delete it yourself :)"
exit
fi
if [ -f "$input" ]; then
re_isanum='^[0-9]+$'
if ! [[ $line =~ $re_isanum ]] ; then
echo "Error: LINENUMBER must be a positive, whole number."
exit 1
elif [ $line -eq "0" ]; then
echo "Error: LINENUMBER must be greater than zero."
exit_abnormal
fi
if [ ! -z $output ]; then
sed -n "${line}p" $input >> $output
fi
if [ ! -z $input ]; then
# remove this sed command and this comes move line to other file
sed "${line}d" $input > tmp && cp tmp $input
fi
fi
if [ -f tmp ]; then
rm tmp
fi
You could try this command:
egrep -n "*" <filename> | egrep "<line number>"
Easy with perl! If you want to get line 1, 3 and 5 from a file, say /etc/passwd:
perl -e 'while(<>){if(++$l~~[1,3,5]){print}}' < /etc/passwd
I am surprised only one other answer (by Ramana Reddy) suggested to add line numbers to the output. The following searches for the required line number and colours the output.
file=FILE
lineno=LINENO
wb="107"; bf="30;1"; rb="101"; yb="103"
cat -n ${file} | { GREP_COLORS="se=${wb};${bf}:cx=${wb};${bf}:ms=${rb};${bf}:sl=${yb};${bf}" grep --color -C 10 "^[[:space:]]\\+${lineno}[[:space:]]"; }

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