I am trying to use the cut command in my bash script. I need to get the name of a rpm package containing 'sys' in its name, without its full path. So I tried this:
packageName=$(find temp/noarch/ -name '*sys*noarch.rpm' | cut -d "\/" -f 3)
I thought this way could return the package's name without the temp/noarch/, but it only tells me.
I know i could just remove the substring "temp\/noarch\/", but i'd like to make it with cut.
"\/" doesn't expand to /, so cut complains:
cut: the delimiter must be a single character
Use plain / instead:
packageName=$(find temp/noarch/ -name '*sys*noarch.rpm' | cut -d/ -f3)
Related
I am trying to find all instances of "type":"FollowEvent", and then within those instances, if the string "actor": is not followed by {, then capture the string enclosed in " that comes immediately after "actor":. Else, capture the string enclosed in " that comes immediately after "login:".
What I have so far:
zgrep -e '"type":"FollowEvent"' /path/to/dir/* | zgrep -o '"actor":(?!{)*' | cut -f2- -d: | cut -d',' -f1 > results_file.txt
What this does:
For all files in /path/to/dir, for all lines that contain "type":"FollowEvent", find "actor:" not followed by {. Then take everything after the :, and before the next ,. Put the results in results_file.txt.
A single line in the files that are being grep'd could look like this:
{"repo":{"url":"https://url","name":"/"},"type":"FollowEvent","public":true,"created_at":"2011-05-29","payload":{"target":{"gravatar_id":"73","id":64,"repos":35,"followers":58,"login":"username3"}},"actor":{"gravatar_id":"06","id":439,"url":"https://url","avatar_url":"https://.png","login":"username4"},"id":"14"}
or like this:
{"repo":{"url":"https://url/","name":"/"},"type":"FollowEvent","public":true,"created_at":"2011-04-01","payload":{"target":{"gravatar_id":"40","repos":2,"followers":1,"login":"username2"},"actor":"username1","actor_gravatar":"de4"},"actor":{"gravatar_id":"de4","id":716,"url":"https://url","avatar_url":"https://.png","login":"username2"},"id":"12"}
What I want:
a file containing only the usernames of actors. Here, I want, in results_file.txt:
username4
username1
Let's say:
JSON='{"repo":{"url":"https://url","name":"/"},"type":"FollowEvent","public":true,"created_at":"2011-05-29","payload":{"target":{"gravatar_id":"73","id":64,"repos":35,"followers":58,"login":"username3"}},"actor":{"gravatar_id":"06","id":439,"url":"https://url","avatar_url":"https://.png","login":"username4"},"id":"14"}'
For a simple answer, I do suggest you to use jq: https://stedolan.github.io/jq/
$ echo "$JSON" | jq -r '. | select(.type=="FollowEvent") | .actor.login'
username4
You can install it in most of distros with the default package manager.
Anyway if you need to do it with GNU tools.
$ echo "$JSON" | grep '"type":"FollowEvent"' | sed 's/.*"login":"\([^"]*\).*/\1/g'
username4
I have a CSV file with this lines:
----------+79975532211,----------+79975532212
4995876655,4995876658
I try to grep this lines in Bash script
#!/bin/bash
config='/test/config.conf'
sourcecsv=/test/sourse.csv
cat $sourcecsv | while read line
do
Oldnumber=$(echo $line | cut -d',' -f1)
cat $config | grep "\\$Oldnumber" -B 8
done
But when script grep value 4995876655 I get error:
grep: Invalid back reference
How I can grep all values in my file?
Instead of:
cat $config | grep "\\$Oldnumber" -B 8
You should do:
grep -B 8 -F -- "$Oldnumber" "$config"
If you really mean to grep for all strings between commas, you can do it all in one go.
tr ',' '\n' </test/sourse.csv |
grep -F -f - -B 8 /test/config.conf
If you need to obtain the matches in sequence (all matches for the first string followed by all matches for the second, etc) then maybe loop over them with a proper while loop:
tr ',' '\n' </test/sourse.csv |
while read -r Oldnumber; do
grep -F -B 8 -e "$Oldnumber" /test/config.conf
done
Keeping the file names in variables does not seem to offer any advantage here.
If you mean to search for the strings preceded by a literal backslash, you can add it back; the -F option I added turns all strings into literals. If you need metacharacters, and take out the -F option, you need to double the backslashes (inside double quotes, a single backslash needs to be represented as double; and to get a literal backslash in a regular expression, you need two of them).
I'm having a bit of an issue cutting the output up from egrep. I have output like:
From: First Last
From: First Last
From: First Last
I want to cut out the "From: " (essentially leaving the "First Last").
I tried
cut -d ":" -f 7
but the output is just a bunch of blank lines.
I would appreciate any help.
Here's the full code that I am trying to use if it helps:
egrep '^From:' $file | cut -d ":" -f 7
NOTE: I've already tested the egrep portion of the code and it works as expected.
The cut command lines in your question specify colon-separated fields and that you want the output to consist only of field 7; since there is no 7th field in your input, the result you're getting isn't what you intend.
Since the "From:" prefix appears to be identical across all lines, you can simply cut from the 7th character onward:
egrep '^From:' $file | cut -c7-
and get the result you intend.
you were really close.
I think you only need to replace ":" with " " as separator and add "-" after the "7": like this:
cut -d " " -f 2-
I tested and works pretty well.
The -f argument is for what fields. Since there is only one : in the line, there's only two fields. So changing -f 7 to -f 2- will give you want you want. Albeit with a leading space.
You can combine the egrep and cut parts into one command with sed:
sed -n 's/^From: //gp' $file
sed -n turns off printing by default, and then I am using p in the sed command explicitly to print the lines I want.
You can use sed:
sed 's/^From: *//'
OR awk:
awk -F ': *' '$1=="From"{print $2}'
OR grep -oP
grep -oP '^From: *\K.*'
Here is a Bash one-liner:
grep ^From file.txt | while read -a cols; do echo ${cols[#]:1}; done
See: Handling positional parameters at wiki.bash-hackers.org
cut itself is a very handy tool in bash
cut -d (delimiter character) -f (fields that you want as output)
a single field is given directly as -f 3 ,
range of fields can be selected as -f 5-9
so in your this particular case code would be
egrep '^From:' $file | cut -d\ -f 2-3
the delimiter is space here and can be escaped using a \
-f 1 corresponds to " From " and 2-3 corresponds to " First Last "
This approach is not finding the file I think I specified.
SHELL = /bin/bash
PKG_NAME = test
PKG_VERSION := $(shell grep -i '^version' $(PKG_NAME)/DESCRIPTION | cut -d ':' -f2 | cut -d ' ' -f2)
In the shell itself, grep -i '^version' test/DESCRIPTION | cut -d ':' -f2 | cut -d ' ' -f2 does return the version successfully, e.g. 0.4-7
But, running via the makefile returns:
grep: test: Is a directory
grep: /DESCRIPTION: No such file or directory
test is a directory, that's true, but test/DESCRIPTION does exist, so I'm guessing $(PKG_NAME)/DESCRIPTION wasn't the right way to assemble the file name.
Suggestions? Thanks.
That error indicates that grep is seeing test and /DESCRIPTION as two separate arguments. Do you have extra spaces on the PKG_NAME assignment line or an errant space between $(PKG_NAME) and /DESCRIPTION in the $(shell ...) line?
As a general rule you might want to start putting quotes around arguments to shell commands (i.e. '$(PKG_NAME)/DESCRIPTION') to prevent this sort of word splitting issue (though without spaces you generally don't have that sort of problem).
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2`
outputs something like this
"Abc Inc";
What I want to do is I want to remove the trailing ";" as well. How can i do that? I am a beginner to bash. Any thoughts or suggestions would be helpful.
This will remove the last character contained in your COMPANY_NAME var regardless if it is or not a semicolon:
echo "$COMPANY_NAME" | rev | cut -c 2- | rev
I'd use sed 's/;$//'. eg:
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2 | sed 's/;$//'`
foo="hello world"
echo ${foo%?}
hello worl
I'd use head --bytes -1, or head -c-1 for short.
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2 | head --bytes -1`
head outputs only the beginning of a stream or file. Typically it counts lines, but it can be made to count characters/bytes instead. head --bytes 10 will output the first ten characters, but head --bytes -10 will output everything except the last ten.
NB: you may have issues if the final character is multi-byte, but a semi-colon isn't
I'd recommend this solution over sed or cut because
It's exactly what head was designed to do, thus less command-line options and an easier-to-read command
It saves you having to think about regular expressions, which are cool/powerful but often overkill
It saves your machine having to think about regular expressions, so will be imperceptibly faster
I believe the cleanest way to strip a single character from a string with bash is:
echo ${COMPANY_NAME:: -1}
but I haven't been able to embed the grep piece within the curly braces, so your particular task becomes a two-liner:
COMPANY_NAME=$(grep "company_name" file.txt); COMPANY_NAME=${COMPANY_NAME:: -1}
This will strip any character, semicolon or not, but can get rid of the semicolon specifically, too.
To remove ALL semicolons, wherever they may fall:
echo ${COMPANY_NAME/;/}
To remove only a semicolon at the end:
echo ${COMPANY_NAME%;}
Or, to remove multiple semicolons from the end:
echo ${COMPANY_NAME%%;}
For great detail and more on this approach, The Linux Documentation Project covers a lot of ground at http://tldp.org/LDP/abs/html/string-manipulation.html
Using sed, if you don't know what the last character actually is:
$ grep company_name file.txt | cut -d '=' -f2 | sed 's/.$//'
"Abc Inc"
Don't abuse cats. Did you know that grep can read files, too?
The canonical approach would be this:
grep "company_name" file.txt | cut -d '=' -f 2 | sed -e 's/;$//'
the smarter approach would use a single perl or awk statement, which can do filter and different transformations at once. For example something like this:
COMPANY_NAME=$( perl -ne '/company_name=(.*);/ && print $1' file.txt )
don't have to chain so many tools. Just one awk command does the job
COMPANY_NAME=$(awk -F"=" '/company_name/{gsub(/;$/,"",$2) ;print $2}' file.txt)
In Bash using only one external utility:
IFS='= ' read -r discard COMPANY_NAME <<< $(grep "company_name" file.txt)
COMPANY_NAME=${COMPANY_NAME/%?}
Assuming the quotation marks are actually part of the output, couldn't you just use the -o switch to return everything between the quote marks?
COMPANY_NAME="\"ABC Inc\";" | echo $COMPANY_NAME | grep -o "\"*.*\""
you can strip the beginnings and ends of a string by N characters using this bash construct, as someone said already
$ fred=abcdefg.rpm
$ echo ${fred:1:-4}
bcdefg
HOWEVER, this is not supported in older versions of bash.. as I discovered just now writing a script for a Red hat EL6 install process. This is the sole reason for posting here.
A hacky way to achieve this is to use sed with extended regex like this:
$ fred=abcdefg.rpm
$ echo $fred | sed -re 's/^.(.*)....$/\1/g'
bcdefg
Some refinements to answer above. To remove more than one char you add multiple question marks. For example, to remove last two chars from variable $SRC_IP_MSG, you can use:
SRC_IP_MSG=${SRC_IP_MSG%??}
cat file.txt | grep "company_name" | cut -d '=' -f 2 | cut -d ';' -f 1
I am not finding that sed 's/;$//' works. It doesn't trim anything, though I'm wondering whether it's because the character I'm trying to trim off happens to be a "$". What does work for me is sed 's/.\{1\}$//'.