I am trying to fetch the output of a variable stored in a file in another shell script.
Example:
cat abc.log
var1=2
var2=2
var3=25
I am writing a script to fetch the value of var3.
Thank you in advance.
awk -F= '$1 ~ /^[[:space:]]*var3/ { print $2 }' abc.log
Set the field delimiter to = and then where the line contains "var3", print the second field.
Alternatively, you could:
source abc.log
and then:
echo $var3
Using sed you can isolate 25 with particularity with:
sed -n '/^[[:space:]]*var3=/s/^[^=]*=//p' file
Explanation
This is the general substitution form s/find/replace/ with a matching expression preceding it. The total form is /match/s/find/replace/. The option -n suppresses the normal printing of pattern-space and the p at the end tells sed to print the line where the match and substitution took place. Specifically,
/match/ locates a line with any number of preceding whitespace characters followed by var3=. The POSIX [:space:] character class matches any whitespace,
the /find/ is all characters anchored from the '^' beginning that are not the [^=] character and then match the literal '=' character, and finally
the /replace/ is the empty-string leaving the 25 alone which is printed.
Example Use/Output
$ sed -n '/^[[:space:]]*var3=/s/^[^=]*=//p' file
25
A grep one-liner, if your grep has support for Perl-compatible regular expressions (the -P option; not all greps support that)
grep -Po '^\s*var3=\K.*' abc.log
or,
grep -Po '^\s*var3=\K.*' abc.log | tail -n1
in order to get the last value of the var3, if multiple var3s is a possibility.
Would like to insert line number at specific location in file
e.g.
apple
ball
should be
(1) apple
(2) ball
Using command
sed '/./=' <FileName>| sed '/./N; s/\n/ /'
It generates
1 Apple
2 Ball
1st solution: This should be an easy task for awk.
awk '{print "("FNR") "$0}' Input_file
2nd solution: With pure sed as per OP's attempt try:
sed '=' Input_file | sed 'N; s/^/(/;s/\n/) /'
Easy to do with perl instead:
perl -ne 'print "($.) $_"' foo.txt
If you want to modify the file in-place instead of just printing out the numbered lines on standard output:
perl -ni -e 'print "($.) $_"' foo.txt
Many ways are there to insert line numbers in a file
some of them are :-
1.Using cat command
cat -n file.txt > newfile.txt
2.Using nl command
nl -b a file.txt
Awk and perl both are very usefull and powerfull. But if, like me, you are reluctant to learn yet another programming language, you can complete this task with the bash commands you probably know already.
With bash you can
increment a sequence number n: $((++n))
read all lines from a file foo into a variable l: while read -r l;do ...;done <foo, where the option -r serves to treat backslashes as just characters.
print formatted output to a line: printf "plain text %i %s\n" number string
Now suppose you want to enclose your sequence number in parentheses, and format them to 8 digits with leading zeroes, then you combine all this to get:
n=0;while read -r l;do printf "(%08i) %s\n" $((++n)) "$l";done <foo >numberedfoo
Note that you do not need to initialize the variable n to use it as a sequence number further on. But if you experiment with this command a few times without reinitializing n, your lines will be numbered from where your previous try stopped incrementing.
Finally, if you don't like the C-like formatting syntax of printf, just use plain echo, and leave the formatting to bash variable expansion. Here is how to format a number like in the command above (do type a space before the -, and a ; before the echo) :
nformat="0000000$n"; echo "(${nformat: -8}) ...";
Fist I am trying to print a file with the word 'Guess' in it and change the word fall to bar.
This what I have tried:
sed -n -e '/Guess/p' -e 's/Fall/bar/' data.txt
The commands work fine alone however, together only the first part is working.
To print line containing 'Guest' and change word 'Fall' in that line,
I would try experimenting with this command>
cat data.txt | sed -n '/Guest/{ s/Fall/bar/p }'
However, this print nothing, if the line with 'Guest' does not contain the word 'Fall'. (Both scenarios - Guest + Fall are required)
If you want to print line containing 'Guest' no matter if substitution finds a word 'Fall', I suggest trying:
cat data.txt | sed -n '/Guest/ { s/Fall/bar/;p }'
I have a file like
key\0value\n
akey\0value\n
key2\0value\n
I have to create a script that take as argument a word. I have to return every lines having a key exactly the same than the argument.
I tried
grep -aF "$key\x0"
but grep seems to do not understand the \x0 (\0 same result). Futhermore, I have to check that the line begins with "$key\0"
I only can use sed grep and tr and other no maching commands
To have the \0 taken into account try :
grep -Pa "^key\x0"
it works for me.
Using sed
sed will work:
$ sed -n '/^key1\x00/p' file
key1value
The use of \x00 to represent a hex character is a GNU extension to sed. Since this question is tagged linux, that is not a problem.
Since the null character does not display well, one might (or might not) want to improve the display with something like this:
$ sed -n 's/^\(akey\)\x00/\1-->/p' file
akey-->value
Using sed with keys that contain special characters
If the key itself can contain sed or shell active characters, then we must escape them first and then run sed against the input file:
#!/bin/bash
printf -v script '/^%s\\x00/p' "$(sed 's:[]\[^$.*/]:\\&:g' <<<"$1")"
sed -n "$script" file
To use this script, simply supply the key as the first argument on the command line, enclosed in single-quotes, of course, to prevent shell processing.
To see how it works, let's look at the pieces in turn:
sed 's:[]\[^$.*/]:\\&:g' <<<"$1"
This puts a backslash escape in front of all sed-active characters.
printf -v script '/^%s\\x00/p' "$(sed 's:[]\[^$.*/]:\\&:g' <<<"$1")"
This creates a sed command using the escaped key and stores it in the shell variable script.
sed -n "$script" file
This runs sed using the shell variable script as the sed command.
Using awk
The question states that awk is not an acceptable tool. For completeness, though, here is an awk solution:
$ awk -F'\x00' -v k=key1 '$1 == k' file
key1value
Explanation:
-F'\x00'
awk divides the input up into records (lines) and divides the records up into fields. Here, we set the field separator to the null character. Consequently, the first field, denoted $1, is the key.
-v k=key1
This creates an awk variable, called k, and sets it to the key that we are looking for.
$1 == k
This statement looks for records (lines) for which the first field matches our specified key. If a match is found, the line is printed.
I have a file as below:
line1
line2
line3
And I want to get:
prefixline1
prefixline2
prefixline3
I could write a Ruby script, but it is better if I do not need to.
prefix will contain /. It is a path, /opt/workdir/ for example.
# If you want to edit the file in-place
sed -i -e 's/^/prefix/' file
# If you want to create a new file
sed -e 's/^/prefix/' file > file.new
If prefix contains /, you can use any other character not in prefix, or
escape the /, so the sed command becomes
's#^#/opt/workdir#'
# or
's/^/\/opt\/workdir/'
awk '$0="prefix"$0' file > new_file
In awk the default action is '{print $0}' (i.e. print the whole line), so the above is equivalent to:
awk '{print "prefix"$0}' file > new_file
With Perl (in place replacement):
perl -pi 's/^/prefix/' file
You can use Vim in Ex mode:
ex -sc '%s/^/prefix/|x' file
% select all lines
s replace
x save and close
If your prefix is a bit complicated, just put it in a variable:
prefix=path/to/file/
Then, you pass that variable and let awk deal with it:
awk -v prefix="$prefix" '{print prefix $0}' input_file.txt
Here is a hightly readable oneliner solution using the ts command from moreutils
$ cat file | ts prefix | tr -d ' '
And how it's derived step by step:
# Step 0. create the file
$ cat file
line1
line2
line3
# Step 1. add prefix to the beginning of each line
$ cat file | ts prefix
prefix line1
prefix line2
prefix line3
# Step 2. remove spaces in the middle
$ cat file | ts prefix | tr -d ' '
prefixline1
prefixline2
prefixline3
If you have Perl:
perl -pe 's/^/PREFIX/' input.file
Using & (the whole part of the input that was matched by the pattern”):
cat in.txt | sed -e "s/.*/prefix&/" > out.txt
OR using back references:
cat in.txt | sed -e "s/\(.*\)/prefix\1/" > out.txt
Using the shell:
#!/bin/bash
prefix="something"
file="file"
while read -r line
do
echo "${prefix}$line"
done <$file > newfile
mv newfile $file
While I don't think pierr had this concern, I needed a solution that would not delay output from the live "tail" of a file, since I wanted to monitor several alert logs simultaneously, prefixing each line with the name of its respective log.
Unfortunately, sed, cut, etc. introduced too much buffering and kept me from seeing the most current lines. Steven Penny's suggestion to use the -s option of nl was intriguing, and testing proved that it did not introduce the unwanted buffering that concerned me.
There were a couple of problems with using nl, though, related to the desire to strip out the unwanted line numbers (even if you don't care about the aesthetics of it, there may be cases where using the extra columns would be undesirable). First, using "cut" to strip out the numbers re-introduces the buffering problem, so it wrecks the solution. Second, using "-w1" doesn't help, since this does NOT restrict the line number to a single column - it just gets wider as more digits are needed.
It isn't pretty if you want to capture this elsewhere, but since that's exactly what I didn't need to do (everything was being written to log files already, I just wanted to watch several at once in real time), the best way to lose the line numbers and have only my prefix was to start the -s string with a carriage return (CR or ^M or Ctrl-M). So for example:
#!/bin/ksh
# Monitor the widget, framas, and dweezil
# log files until the operator hits <enter>
# to end monitoring.
PGRP=$$
for LOGFILE in widget framas dweezil
do
(
tail -f $LOGFILE 2>&1 |
nl -s"^M${LOGFILE}> "
) &
sleep 1
done
read KILLEM
kill -- -${PGRP}
Using ed:
ed infile <<'EOE'
,s/^/prefix/
wq
EOE
This substitutes, for each line (,), the beginning of the line (^) with prefix. wq saves and exits.
If the replacement string contains a slash, we can use a different delimiter for s instead:
ed infile <<'EOE'
,s#^#/opt/workdir/#
wq
EOE
I've quoted the here-doc delimiter EOE ("end of ed") to prevent parameter expansion. In this example, it would work unquoted as well, but it's good practice to prevent surprises if you ever have a $ in your ed script.
Here's a wrapped up example using the sed approach from this answer:
$ cat /path/to/some/file | prefix_lines "WOW: "
WOW: some text
WOW: another line
WOW: more text
prefix_lines
function show_help()
{
IT=$(CAT <<EOF
Usage: PREFIX {FILE}
e.g.
cat /path/to/file | prefix_lines "WOW: "
WOW: some text
WOW: another line
WOW: more text
)
echo "$IT"
exit
}
# Require a prefix
if [ -z "$1" ]
then
show_help
fi
# Check if input is from stdin or a file
FILE=$2
if [ -z "$2" ]
then
# If no stdin exists
if [ -t 0 ]; then
show_help
fi
FILE=/dev/stdin
fi
# Now prefix the output
PREFIX=$1
sed -e "s/^/$PREFIX/" $FILE
You can also achieve this using the backreference technique
sed -i.bak 's/\(.*\)/prefix\1/' foo.txt
You can also use with awk like this
awk '{print "prefix"$0}' foo.txt > tmp && mv tmp foo.txt
Using Pythonize (pz):
pz '"preix"+s' <filename
Simple solution using a for loop on the command line with bash:
for i in $(cat yourfile.txt); do echo "prefix$i"; done
Save the output to a file:
for i in $(cat yourfile.txt); do echo "prefix$i"; done > yourfilewithprefixes.txt
You can do it using AWK
echo example| awk '{print "prefix"$0}'
or
awk '{print "prefix"$0}' file.txt > output.txt
For suffix: awk '{print $0"suffix"}'
For prefix and suffix: awk '{print "prefix"$0"suffix"}'
For people on BSD/OSX systems there's utility called lam, short for laminate. lam -s prefix file will do what you want. I use it in pipelines, eg:
find -type f -exec lam -s "{}: " "{}" \; | fzf
...which will find all files, exec lam on each of them, giving each file a prefix of its own filename. (And pump the output to fzf for searching.)
If you need to prepend a text at the beginning of each line that has a certain string, try following. In the following example, I am adding # at the beginning of each line that has the word "rock" in it.
sed -i -e 's/^.*rock.*/#&/' file_name
SETLOCAL ENABLEDELAYEDEXPANSION
YourPrefix=blabla
YourPath=C:\path
for /f "tokens=*" %%a in (!YourPath!\longfile.csv) do (echo !YourPrefix!%%a) >> !YourPath!\Archive\output.csv