The goal is to use a static variable and combine it with the output of a pipe.
I am looking for an one line solution !
This is an complete overhaul of the written examples, to keep it more clean!
The old post I have removed cause of to mutch noise.
link="http://example.com/"
echo "index.html" > test.file
echo "search.php" >> test.file
echo "login.js" >> test.file
cat test.file | awk 'BEGIN {var=ARGV[1];ARGV[1]=""} {print var, $0}' "$link"
http://example.com/ index.html
http://example.com/ search.php
http://example.com/ login.js
cat test.file | awk -v var="$link" '{print var, $0}'
http://example.com/ index.html
http://example.com/ search.php
http://example.com/ login.js
This looks like what I need, but without the nasty white space in it.
I need that field separation gone!
Option -F'' result in error message.
Adding | tr or | sed to remove the while space seems like error correction to me.
The following solution works so far ...
for string in $(cat test.file)
do
printf "${link}%s\n" "$string"
done
Ouptut:
http://example.com/index.html
http://example.com/search.php
http://example.com/login.js
In a loop I can combine all the output of a pipe with my static variable.
But isn't there a better solution without using a loop ?
It seems approaching the pipe with printf doesn't work by default and I have to put xargs to it.
echo "World" | xargs printf 'Hello %s\n' "$1"
Hello
Hello World
Even then the output is doubled, but why ?
I am still looking for an one line solution !
You pipe something into printf, but printf does not process its standard input. You can easily verify this by doing a
echo x | printf foo%s bar
which just prints foobar, and ignores the x.
Related
How can I remove all text after a character, in this case a colon (":"), in bash? Can I remove the colon, too? I have no idea how to.
In Bash (and ksh, zsh, dash, etc.), you can use parameter expansion with % which will remove characters from the end of the string or # which will remove characters from the beginning of the string. If you use a single one of those characters, the smallest matching string will be removed. If you double the character, the longest will be removed.
$ a='hello:world'
$ b=${a%:*}
$ echo "$b"
hello
$ a='hello:world:of:tomorrow'
$ echo "${a%:*}"
hello:world:of
$ echo "${a%%:*}"
hello
$ echo "${a#*:}"
world:of:tomorrow
$ echo "${a##*:}"
tomorrow
An example might have been useful, but if I understood you correctly, this would work:
echo "Hello: world" | cut -f1 -d":"
This will convert Hello: world into Hello.
$ echo 'hello:world:again' |sed 's/:.*//'
hello
I know some solutions:
# Our mock data:
A=user:mail:password
With awk and pipe:
$ echo $A | awk -v FS=':' '{print $1}'
user
Via bash variables:
$ echo ${A%%:*}
user
With pipe and sed:
$ echo $A | sed 's#:.*##g'
user
With pipe and grep:
$ echo $A | egrep -o '^[^:]+'
user
With pipe and cut:
$ echo $A | cut -f1 -d\:
user
egrep -o '^[^:]*:'
trim off everything after the last instance of ":"
grep -o '^.*:' fileListingPathsAndFiles.txt
and if you wanted to drop that last ":"
grep -o '^.*:' file.txt | sed 's/:$//'
#kp123: you'd want to replace : with / (where the sed colon should be \/)
Let's say you have a path with a file in this format:
/dirA/dirB/dirC/filename.file
Now you only want the path which includes four "/". Type
$ echo "/dirA/dirB/dirC/filename.file" | cut -f1-4 -d"/"
and your output will be
/dirA/dirB/dirC
The advantage of using cut is that you can also cut out the uppest directory as well as the file (in this example), so if you type
$ echo "/dirA/dirB/dirC/filename.file" | cut -f1-3 -d"/"
your output would be
/dirA/dirB
Though you can do the same from the other side of the string, it would not make that much sense in this case as typing
$ echo "/dirA/dirB/dirC/filename.file" | cut -f2-4 -d"/"
results in
dirA/dirB/dirC
In some other cases the last case might also be helpful. Mind that there is no "/" at the beginning of the last output.
How can I remove all text after a character, in this case a colon (":"), in bash? Can I remove the colon, too? I have no idea how to.
In Bash (and ksh, zsh, dash, etc.), you can use parameter expansion with % which will remove characters from the end of the string or # which will remove characters from the beginning of the string. If you use a single one of those characters, the smallest matching string will be removed. If you double the character, the longest will be removed.
$ a='hello:world'
$ b=${a%:*}
$ echo "$b"
hello
$ a='hello:world:of:tomorrow'
$ echo "${a%:*}"
hello:world:of
$ echo "${a%%:*}"
hello
$ echo "${a#*:}"
world:of:tomorrow
$ echo "${a##*:}"
tomorrow
An example might have been useful, but if I understood you correctly, this would work:
echo "Hello: world" | cut -f1 -d":"
This will convert Hello: world into Hello.
$ echo 'hello:world:again' |sed 's/:.*//'
hello
I know some solutions:
# Our mock data:
A=user:mail:password
With awk and pipe:
$ echo $A | awk -v FS=':' '{print $1}'
user
Via bash variables:
$ echo ${A%%:*}
user
With pipe and sed:
$ echo $A | sed 's#:.*##g'
user
With pipe and grep:
$ echo $A | egrep -o '^[^:]+'
user
With pipe and cut:
$ echo $A | cut -f1 -d\:
user
egrep -o '^[^:]*:'
trim off everything after the last instance of ":"
grep -o '^.*:' fileListingPathsAndFiles.txt
and if you wanted to drop that last ":"
grep -o '^.*:' file.txt | sed 's/:$//'
#kp123: you'd want to replace : with / (where the sed colon should be \/)
Let's say you have a path with a file in this format:
/dirA/dirB/dirC/filename.file
Now you only want the path which includes four "/". Type
$ echo "/dirA/dirB/dirC/filename.file" | cut -f1-4 -d"/"
and your output will be
/dirA/dirB/dirC
The advantage of using cut is that you can also cut out the uppest directory as well as the file (in this example), so if you type
$ echo "/dirA/dirB/dirC/filename.file" | cut -f1-3 -d"/"
your output would be
/dirA/dirB
Though you can do the same from the other side of the string, it would not make that much sense in this case as typing
$ echo "/dirA/dirB/dirC/filename.file" | cut -f2-4 -d"/"
results in
dirA/dirB/dirC
In some other cases the last case might also be helpful. Mind that there is no "/" at the beginning of the last output.
I have this string in a variable:
strVar="Hello World [randomSubstring].zip"
I would like to extract [randomSubstring], where that substring inside the brackets could be anything.
The expected result must be something like this:
echo "$strVar"
Hello World .zip
I tried several combinations with grep and awk but without success, I am using CentOS 7.
echo "Hello World [RNVE5Z].zip" | grep -oP '(?<=[).*(?=])'
echo "Hello World [RNVE5Z].zip" | awk -F"["" '{print $1}' | awk -F"]" '{print $2}'
Bash only:
echo ${strVar/\[*\]/}
Just use bash's substring manipulation:
echo ${strVar/\[*\]/}
I would prefer it over an external call to sed, except there is more to be done, why I use sed anyway:
echo $strVar | sed 's/\[.*\]//'
I don't think there is an elegant solution for grep but might be wrong. In awk I'm not that fluent.
I am banging my head against the keyboard on this simple piece of code.
#!/bin/bash
connstate="Connected"
vpnstatus=$(/opt/cisco/anyconnect/bin/vpn state | (grep -m 1 'state:'))
echo $vpnstatus
vpnconn=$(echo $vpnstatus | sed -e 's/>>\ state: //g' | sed "s/ //g")
echo "$vpnconn" "$connstate"
if [ "$vpnconn" = "$connstate" ];then
echo $vpnconn
else echo "this script still fails"
fi
echo done
This is the output from the above code:
>> state: Connected
Connected Connected
this script still fails
done
I believe the issue revolves around the vpnconn=$ if I comment that section of code out and fill the variable vpnconn="Connected" this code works fine. Something with how the sed is working on the input from vpnstatus and outputting the results to vpnconn is making what looks like a correct result incorrect when doing the compare in the if then.
I have tried splitting up the vpnconn line into two separate lines and that did not change anything, I took out the sed "s/ //g" and replaced it with a trim -d ' ' and that did not change the results. I know this is something small in this tiny piece of code that I am missing.
Did you try?
vpnconn=$(echo "$vpnstatus" | awk '{print $3}')
Something like:
vpnstatus=$(/opt/cisco/anyconnect/bin/vpn state|grep -m 1 'state:'|awk '{print 3}')
should do the work.
I'm trying to do some manipulation with Wordpress and I'm trying to write a script for it...
# cat /usr/local/uftwf/_wr.sh
#!/bin/sh
# $Id$
#
table_prefix=`grep ^\$table_prefix wp-config.php | awk -F\' '{print $2}'`
echo $table_prefix
#
Yet I'm getting following output
# /usr/local/uftwf/_wr.sh
ABSPATH ABSPATH wp-settings.php_KEY LOGGED_IN_KEY NONCE_KEY AUTH_SALT SECURE_AUTH_SALT LOGGED_IN_SALT NONCE_SALT wp_0zw2h5_ de_DE WPLANG WP_DEBUG s all, stop editing! Happy blogging. */
#
Running from command line, I get the correct output that I'm looking for:
# grep ^\$table_prefix wp-config.php | awk -F\' '{print $2}'
wp_0zw2h5_
#
What is going wrong in the script?
The problem is the grep command:
table_prefix=`grep ^\$table_prefix wp-config.php | awk -F\' '{print $2}'`
It either needs three backslashes - not one - or you need to use single quotes (which is much simpler):
table_prefix=$(grep '^$table_prefix' wp-config.php | awk -F"'" '{print $2}')
It's also worth using the $( ... ) notation in general.
The trouble is that the backquotes removes the backslash, so the shell variable is evaluated, and what's passed to grep is, most likely, just ^, and each line starts with a beginning of line.
This has all the appearance as though the grep is not omitting all the lines that are not matching, when you issue the echo $table_prefix without quotes it collapses all the white space into a single output line, if you issue an: echo "$table_prefix", you would see the match with all the other white-space that was output.
I'd recommend the following sed expression instead:
table_prefix=$(sed -n "s/^\$table_prefix.*'\([^']*\)'.*/\1/p" wp-config.php)
You should try
#!/bin/sh
table_prefix=$(awk -F"'" '/^\$table_prefix/{print $2}' wp-config.php)
echo $table_prefix
Does this one work for you?
awk -F\' '/^\$table_prefix/ {print $2}' wp-config.php
Update
If you are using shell scripting, there is no need to call up awk, grep:
#!/bin/sh
while read varName op varValue theRest
do
if [ "_$varName" = "_\$table_prefix" ]
then
table_prefix=${varValue//\'/} # Remove the single quotes
table_prefix=${table_prefix/;/} # Remove the semicolon
break
fi
done < wp-config.php
echo "Found: $table_prefix"