This question already has answers here:
How do I set a variable to the output of a command in Bash?
(15 answers)
Closed 1 year ago.
Got a bit of a problem. I'm new to bash and I'm trying my hardest, but I can't figure out how to assign the desired output to the variable. Running this command in the prompt
wc -l data.txt | awk '{ print $1 }'
yields the result 12, which is desired. However if I put it in the test.sh file, it won't work. I've tried different quotations, but all I've managed to get is the entire line as a string...
Test.sh
#! /bin/bash
# Count lines for data.txt files
data1=wc -l data.txt | awk '{ print $1 }'
echo "Lines in data.txt: $data1"
exit
I think you want:
data1="$(wc -l data.txt | awk '{ print $1 }')"
The $() syntax causes bash to execute that expression and replace it with the results.
Actually, powershell does allow you to do a straight = assignment like you did...
Related
This question already has answers here:
How can I remove the extension of a filename in a shell script?
(15 answers)
Closed 11 months ago.
I have a filename with the format yyyymmdd.txt. How can I output only yyyymmdd without the .txt extension?
Example
20220414.txt (before output)
20220414 (after the output)
basename has an option to remove a suffix:
basename -s .txt 20220414.txt
gives:
20220414
Or, if your filename is stored in a variable, bash can help:
a=20220414.txt
echo ${a%.*}
gives:
20220414
You can user awk with flag -F to specify the separator . and then print the first part with $1
echo "20220414.txt" | awk -F "." ' {print $1}'
output
20220414
grep doesn't manipulate anything, it shows what you have in a file. So, you can't modify that file using grep, but you can modify what it shows, using the -o switch, as you can see here:
Prompt> echo "20220414.txt" | grep -o "[0-9]*"
20220414
The [0-9]* means "a list of integers, going from character '0' to character '9'.
This question already has answers here:
How to pass parameters to a Bash script?
(4 answers)
Closed 1 year ago.
At the beginning I have a file.txt, which contains several informations that I will take using the grep command as you see in the script.
What I want is to give the script the file I want instead of file.txt but without changing the file name each time in the script for example if the file is named Me.txt I don’t want to go into the script and write Me.txt in each grep command especially if I have dozens of orders.
Is there a way to do this?
#!/bin/bash
grep teste file.txt > testline.txt
awk '{print $2}' testline.txt > test.txt
echo '#'
echo '#'
grep remote file.txt > remoteline.txt
awk '{print $3}' remoteline.txt > remote.txt
echo '#'
echo '#'
grep adresse file.txt > adresseline.txt
awk '{print $2}' adresseline.txt > adresse.txt
Using a parameter, as many contributors here suggested, is of course the obvious approach, and the one which is usually taken in such case, so I want to extend this idea:
If you do it naively as
filename=$1
you have to supply the name on every invocation. You can improve on this by providing a default value for the case the parameter is missing:
filename=${1:-file.txt}
But sometimes you are in a situation, where for some time (working on a specific task), you always need the same filename over and over, and the default value happens to be not the one you need. Another possibility to pass information to a program is via the environment. If you set the filename by
filename=${MOOFOO:-file.txt}
it means that - assuming your script is called myscript.sh - if you invoke your script by
MOOFOO=myfile.txt myscript.sh
it uses myfile.txt, while if you call it by
myscript.sh
it uses the default file.txt. You can also set MOOFOO in your shell, as
export MOOFOO=myfile.txt
and then, even a lone execution of
myscript.sh
with use myfile.txt instead of the default file.txt
The most flexible approach is to combine both, and this is what I often do in such a situation. If you do in your script a
filename=${1:-${MOOFOO:-file.txt}}
it takes the name from the 1st parameter, but if there is no parameter, takes it from the variable MOOFOO, and if this variable is also undefined, uses file.txt as the last fallback.
You should pass the filename as a command line parameter so that you can call your script like so:
script <filename>
Inside the script, you can access the command line parameters in the variables $1, $2,.... The variable $# contains the number of command line parameters passed to the script, and the variable $0 contains the path of the script itself.
As with all variables, you can choose to put the variable name in curly brackets which has advantages sometimes: ${1}, ${2}, ...
#!/bin/bash
if [ $# = 1 ]; then
filename=${1}
else
echo "USAGE: $(basename ${0}) <filename>"
exit 1
fi
grep teste "${filename}" > testline.txt
awk '{print $2}' testline.txt > test.txt
echo '#'
echo '#'
grep remote "${filename}" > remoteline.txt
awk '{print $3}' remoteline.txt > remote.txt
echo '#'
echo '#'
grep adresse "${filename}" > adresseline.txt
awk '{print $2}' adresseline.txt > adresse.txt
By the way, you don't need two different files to achieve what you want, you can just pipe the output of grep straight into awk, e.g.:
grep teste "${filename}" | awk '{print $2}' > test.txt
but then again, awk can do the regex match itself, reducing it all to just one command:
awk '/teste/ {print $2}' "${filename}" > test.txt
This question already has answers here:
How do I use shell variables in an awk script?
(7 answers)
Closed 2 years ago.
I would like to use awk to replace any occurrence of a given word with another word in a .sh
The script I wrote is;
#!/bin/bash
read -p "Enter the word to change: " wrd1
read -p "Enter the new word: " wrd2
awk '{sub(/"$wrd1 /, $wrd2")}1' file.txt
but it's not working.
Use the -v option.
awk -v w1="$wrd1" -v w2="$wrd2" '{sub(w1, w2)}1' file.txt
A demonstration:
$ awk -v w1="$wrd1" -v w2="$wrd2" '{sub(w1, w2)}1' <<EOF
foo=3
hi=9
bye=2
EOF
bar=3
hi=9
bye=2
This question already has answers here:
Difference between sh and Bash
(11 answers)
Closed 3 years ago.
What is wrong with this script?
#!/bin/sh
for file in *.p; do awk -f get_p.awk "$file" > "$file"er; done
awk -f phase-comp.awk file1 file.per # výpočet fází
paste <(awk '{print $1}' output1) <(awk '{print $2}' file1) > out
The error is:
5: skript: Syntax error: "(" unexpected
When I write command separately to terminal. It works well. Thank you
The problem is your shebang:
#!/bin/sh
It means this script is meant to be interpreted by sh, but your script uses process substitution which is a bash-specific feature. So, you should change it to:
#!/bin/bash
to make your script work.
This question already has answers here:
Using awk with variables
(3 answers)
Closed 8 years ago.
As a part of my bash, I want to pass some constant from command line to awk. For example, I want to subtract constant1 from column 1 and constant2 from column 5
$ sh bash.sh infile 0.54 0.32
#!/bin/bash
#infile = $1
#constant1 = $2
#constant2 = $3
cat $1 | awk '{print $1"\t"$2"\t"$3"\t"$4"\t"$5"\t"$6}'
thank you very much for your help
As awk is it's own language, by default it does not share the same variables as Bash. To use Bash variables in an awk command, you should pass the variables to awk using the -v option.
#!/bin/bash
awk -v constant1=$2 -v constant2=$3 '{print($1-constant1),($5-constant2)}' $1
You'll notice I removed cat as there is no need to pipe cat into awk since awk can read from files.
you need to remove gaps when defining vaariables:
#!/bin/bash
infile=$1
constant1=$2
constant2=$3
cat $1 | awk '{print $1 $2 $3 $4 $5 $6}'