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I have a string "config"
and want to replace with
"server"
using sed in Linux. I tried the below one. But It did not work.
sed -i "s#$"config"#$"server"#g" setup.xml-->
How can I do that? If not sed other options are fine too.
before "config"
after "server"
One example:
sed 's/"config"/"\s\e\r\v\e\r"/' setup.xml
The replacement string has characters with special meaning in sed such as ; # and &. These will all need to be escaped and so:
sed -n 's/"config"/"\&\#115\;\&\#101\;\&\#114\;\&\#118\;\&\#101\;\&\#114\;"/p' <<< '"config"'
Related
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I am having a text file containing a list of date and time just like the sample below -
posted_at"
2012-06-09 11:48:31"
2012-08-09 12:40:02"
2012-04-09 13:10:00"
2012-03-09 13:40:00"
2012-10-09 14:30:01"
2012-12-09 15:30:00"
2012-11-09 16:20:00"
I want to extract the month from each line.
P.S - grep should not be used at any point of the code
Thanks in advance!
First select date pattern :
egrep '[0-9]{4}-[0-9]{2}-[0-9]{2} ' content_file
Second, extract the month :
awk -F '-' '{print $2}'
Third redirect to desired file :
>> desired_file
So mix of all this with | to final solution :
egrep '[0-9]{4}-[0-9]{2}-[0-9]{2} ' content_file| awk -F '-' '{print $2}'>> desired_file
Voilà
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Here's my config file
#comment 1
--longoption1
#comment 2
--longoption2
#comment 3
-s
#comment 4
--longoption4
I want to write a bash script that will read this .conf file, skip comments and serialize the commandline options like so.
./binary --longoption1 --longoption2 -s --longoption4
Working off of this post on sed, you just need to pipe the output from sed to xargs:
sed -e 's/#.*$//' -e '/^$/d' inputFile | xargs ./binary
As Wiimm points out, xargs can be finicky with a lot of arguments and it might split it up across multiple calls to binary. It may be better off to use sed directly:
./binary $(sed -e 's/#.*$//' -e '/^$/d' inputFile)
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I know how to print all letters
{a..z} and {A..Z} and {0..9}
But is there a way to print all possible ASCII Characters via a bash loop?
You don't need a loop
echo -e \\x{0..7}{{0..9},{A..F}}
It prints all chars from 0 to 127.
If it is okay to use awk:
awk 'BEGIN{for (i=32;i<127;i++) printf("%c", i)}'
Or using printf:
for((i=32;i<127;i++)) do printf "\x$(printf %x $i)"; done
use this:
for ((i=32;i<127;i++)) do printf "\\$(printf %03o "$i")"; done;printf "\n"
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How can I remove all the 'www.' with awk in my output file.
e.g.: my output file has multiple sites like
abc.com
www.def.com
blabla.org
www.zxc.net
I would like to remove all the www. in my output file:
abc.com
def.com
blabla.org
zxc.net
Probably better done in sed:
sed -i 's/^www\.//g' outputFile
In awk:
awk '{gsub(/^www\./,"",$0)}1' outputFile
This is probably what you're looking for:
$ cat file
abc.com
www.def.com
blabla.org
www.zxc.net
www.org
www.acl.lanl.gov
$ sed -E 's/^www\.(([^.]+(\.|$)){2,})/\1/' file
abc.com
def.com
blabla.org
zxc.net
www.org
acl.lanl.gov
The above uses a sed that has -E for ERE support, e.g. GNU or OSX sed. Note the need for a more comprehensive input file to test if a proposed solution really works or not.
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I have two text file like below:
File1.txt
A|234-211
B|234-244
C|234-351
D|999-876
E|456-411
F|567-211
File2.txt
234-244
999-876
567-211
And I want to compare both files and get containing values like below:
Dequired output
B|234-244
D|999-876
F|567-211
$ grep -F -f file2.txt file1.txt
B|234-244
D|999-876
F|567-211
The -F makes grep search for fixed strings (not patterns). Both -F and -f are POSIX options to grep.
Note that this assumes your file2.txt does not contain short strings like 11 which could lead to false positives.
Try:
grep -f File2.txt File1.txt