I am trying to use one or two lines of Bash (that can be run in a command line) to read a folder-name and return the version inside of the name.
So if I have myfolder_v1.0.13 I know that I can use echo "myfolder_v1.0.13" | awk -F"v" '{ print $2 }' and it will return with 1.0.13.
But how do I get the shell to read the folder name and pipe with the awk command to give me the same result without using echo? I suppose I could always navigate to the directory and translate the output of pwd into a variable somehow?
Thanks in advance.
Edit: As soon as I asked I figured it out. I can use
result=${PWD##*/}; echo $result | awk -F"v" '{ print $2 }'
and it gives me what I want. I will leave this question up for others to reference unless someone wants me to take it down.
But you don't need an Awk at all, here just use bash parameter expansion.
string="myfolder_v1.0.13"
printf "%s\n" "${string##*v}"
1.0.13
You can use
basename "$(cd "foldername" ; pwd )" | awk -Fv '{print $2}'
to get the shell to give you the directory name, but if you really want to use the shell, you could also avoid the use of awk completetly:
Assuming you have the path to the folder with the version number in the parameter "FOLDERNAME":
echo "${FOLDERNAME##*v}"
This removes the longest prefix matching the glob expression "*v" in the value of the parameter FOLDERNAME.
Related
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
I have a variable called $folder_name which contains the string
Release_2019_Config_V6_Standalone_PJ6678_Test
which is the name of a folder.
I'm trying to extract PJ6678 from the folder name.
I know the folder name will put the user id (the text I need) between the 5th and 6th underscore, I don't know what text/symbols will be present after the 6th underscore.
I'm using Bash script, i'd really appreciate the help if someone could help with this functionality as i'm completely lost trying to use sed (after reading for hours i'm assuming this is the correct tool for the job?
Here is a Bash only solution:
#!/bin/bash
INPUT="Release_2019_Config_V6_Standalone_PJ6678_Test"
IFS='_' read -ra IN <<< "$INPUT"
echo ${IN[5]}
Or use cut:
cut -d '_' -f 6 <<< "Release_2019_Config_V6_Standalone_PJ6678_Test"
Or use awk:
awk -F "_" '{ print $6 }' <<< "Release_2019_Config_V6_Standalone_PJ6678_Test"
If you want pure-bash solution, you can use tokenize the file name, and pick up the 5th element
IFS=_ read -a token <<< "$folder_name"
id=${token[5]}
Eliminating dependency and performance hit from launching additional programs per folder name.
Try this command:
echo $a | awk -F'_' '{print $6}'
Here, _ is the delimiter and $a is a variable that holds the value.
For completeness, here's a pure-shell solution that doesn't rely on bash extensions like arrays.
$ folder_name=Release_2019_Config_V6_Standalone_PJ6678_Test
$ tmp=${folder_name#*_*_*_*_*_} # Because we know how many _ to strip
$ echo ${tmp%_*}
PJ6678
Because the # operator strips the shortest prefix, this won't allow * to match any _ itself; if it did, we could shorten the prefix by making the underscore match one of the literal _ in the pattern instead.
I currently have the following line of code:
ls /some/dir/prefix* | sed -e 's/prefix.//' | tr '\n' ' '
Which does achieve what I want it to do:
Get list of files starting with prefix
Remove path and prefix from each string
Remove newlines and replace with spaces for later processing.
For example:
/some/dir/prefix.hello
/some/dir/prefix.world
Should become
hello world
But I feel like there's a nicer way of doing this. Is there a better way to do this in one line?
Here is a two-liner using just built-ins that does it:
fnames=(some/dir/prefix*)
echo "${fnames[#]##*.}"
And here's how this works:
fnames=(some/dir/prefix*) creates an array with all the files starting with prefix and avoids all the problems that come with parsing ls
echo "${fnames[#]##*.}" is a combination of two parameter expansions: ${fnames[#]} prints all array elements, and the ##*. part removes the longest match of anything that ends with . from each array element, leaving just the extension
If you're hell-bent on a one-liner, just join the two commands with &&.
passing ls output to external programs is not recommended, following bash solution may help you here.
for file in prefix*; do echo ${file##*.}; done
Adding a non-one liner form of solution too now.
for file in prefix*
do
echo ${file##*.}
done
Here is a very simple Awk one-liner to achieve this :
awk -F. '{$0=FILENAME; printf $NF" "; nextfile}' /some/dir/prefix*
It essentially does the following :
-F.: Set the field separator FS to a .. This way $NF represents the extension.
$0=FILENAME: Ignore the current record and set it to FILENAME, reparse everything this way.
print $NF; nextfile : print the extension and go to the next file.
The problem with this is that the file still reads a record of the current file. If that file is empty this will fail.
To make this work with empty files, you could use the gawk extension BEGINFILE
awk -F. 'BEGINFILE{$0=FILENAME; printf $NF" "; nextfile}' /some/dir/prefix*
Or you can loop over all the arguments :
awk -F. 'BEGIN{for(i in ARGV){$0=ARGV[i]; printf $NF" "};exit}' /some/dir/prefix*
One approach with awk:
ls /some/dir/prefix* | awk -F"." '{printf "%s ", $2} END {print ""}'
It might qualify as being "nicer" because there's only one command the output is piped through?!
I'm on a Mac, and I want to find a field in a CSV file adjacent to a search string
This is going to be a single file with a hard path; here's a sample of it:
84:a5:7e:6c:a6:b0, AP-ATC-151g84
84:a5:7e:6c:a6:b1, AP-A88-131g84
84:a5:7e:73:10:32, AP-AG7-133g56
84:a5:7e:73:10:30, AP-ADC-152g81
84:a5:7e:73:10:31, AP-D78-152e80
so if my search string is "84:a5:7e:73:10:32"
I want to get returned "AP-AG7-133g56"
I had been working within an Applescript, but maybe a shell script will do.
I just need the proper syntax for opening the file and having awk search it. Again, I'm weak conceptually on how shell commands run, how they must be executed, etc
This errors, gives me ("command not found"):
set the_file to "/Users/Paw/Desktop/AP-Decoder 3.app/Contents/Resources/BSSIDtable.csv"
set the_val to "70:56:81:cb:a2:dc"
do shell script "'awk $1 ~ the_val {print $2} the_file'"
Thank you for coddling me...
This is a relatively simple:
awk '$1 == "70:56:81:cb:a2:dc," {print "The answer is "$2}' 'BSSIDtable.csv'
(the "The answer is " text can be omitted if you only wish to see only the data, but this shows you how to get more user-friendly output if desired).
The comma is included since awk uses white space for separators so the comma becomes part of column 1.
If the thing you're looking for is in a shell variable, you can use -v to provide that to awk as an awk variable:
lookfor="70:56:81:cb:a2:dc,"
awk -v mac=$lookfor '$1 == mac {print "The answer is "$2}' 'BSSIDtable.csv'
As an aside, your AppleScript solution is probably not working because the $1/$2 are being interpreted as shell variable rather than awk variables. If you insist on using AppleScript, you will have to figure out how to construct a shell command that quotes the awk commands correctly.
My advice is to just use the shell directly, the number of people proficient in that almost certainly far outnumber those proficient in AppleScript :-)
if sed is available (normaly on mac, event if not tagged in OP)
simple but read all the file
sed -n 's/84:a5:7e:73:10:32,[[:blank:]]*//p' YourFile
quit after first occurence (so average of 50% faster on huge file)
sed -n -e '/84:a5:7e:73:10:32,[[:blank:]]*/!b' -e 's///p;q' YourFile
awk
awk '/^84:a5:7e:73:10:32/ {print $2}'
# OR using a variable for batch interaction
awk -v Src='84:a5:7e:73:10:32' '$1 == Src {print $2}'
# OR assuming that case is unknow
awk -v Src='84:a5:7e:73:10:32' 'BEGIN{IGNORECASE=1} $1 == Src {print $2}'
by default it take $0 as compare test if a regex is present, just add the ^ to take first field content
So, I have a file called 'dummy' which contains the string:
"There is 100% packet loss at node 1".
I also have a small script that I want to use to grab the percentage from this file. The script is below.
result=`grep 'packet loss' dummy` |
awk '{ first=match($0,"[0-9]+%")
last=match($0," packet loss")
s=substr($0,first,last-first)
print s}'
echo $result
I want the value of $result to basically be 100% in this case. But for some reason, it just prints out a blank string. Can anyone help me?
You would need to put the closing backtick after the end of the awk command, but it's preferable to use $() instead:
result=$( grep 'packet loss' dummy |
awk '{ first=match($0,"[0-9]+%")
last=match($0," packet loss")
s=substr($0,first,last-first)
print s}' )
echo $result
but you could just do:
result=$( grep 'packet loss' | grep -o "[0-9]\+%" )
Try
awk '{print $3}'
instead.
the solution below can be used when you don't know where the percentage numbers are( and there's no need to use awk with greps)
$ results=$(awk '/packet loss/{for(i=1;i<=NF;i++)if($i~/[0-9]+%$/)print $i}' file)
$ echo $results
100%
You could do this with bash alone using expr.
i=`expr "There is 98.76% packet loss at node 1" : '[^0-9.]*\([0-9.]*%\)[^0-9.]*'`; echo $i;
This extracts the substring matching the regex within \( \).
Here I'm assuming that the output lines you're interested in adhere strictly to your example, with the percentage value being the only variation.
With that assumption, you really don't need anything more complicated than:
awk '/packet loss/ { print $3 }' dummy
This quite literally means "print the 3rd field of any lines containing 'packet loss' in them". By default awk treats whitespace as field delimiters, which is perfect for you.
If you are doing more than simply printing the percentage, you could save the results to a shell variable using backticks, or redirect the output to a file. But your sample code simply echoes the percentages to stdout, and then exits. The one-liner does the exact same thing. No need for backticks or $() or any other shell machinations whatsoever.
NB: In my experience, piping the output of grep to awk is usually doing something that awk can do all by itself.