I have tried the below
history -d $(history | grep "echo.*" |awk '{print $1}')
But it is not deleting all the commands from the history with echo
I want to delete any commands start with echo
like
echo "mamam"
echoaaa
echo "hello"
echooooo
You can use this to remove echo entries :
for d in $(history | grep -Po '^\s*\K(\d+)(?= +echo)' | sort -nr); do history -d $d; done
I would do a
history -d $(history | grep -E "^ *[0-9]+ *echo" | awk '{print $1})
The history command produces one column of event number, followed by the command. We need to match an echo, which is following such a event number. The awk then prints just the event number.
An alternative without reverting to awk would be:
history -d $(history | grep -E "^ *[0-9]+ *echo" | grep -Eow '[0-9]+)
history -w
sed -i '/^echo.*/d' ~/.bash_history
history -c
history -r
Related
I was trying to filter all the files from the URLs and get only paths.
echo -e "http://sub.domain.tld/secured/database_connect.php\nhttp://sub.domain.tld/section/files/image.jpg\nhttp://sub.domain.tld/.git/audio-files/top-secret/audio.mp3" | grep -Ei "(http|https)://[^/\"]+" | sort -u
http://sub.domain.tld
But I want the result like this
http://sub.domain.tld/secured/
http://sub.domain.tld/section/files/
http://sub.domain.tld/.git/audio-files/top-secret/
Is there any way to do it with sed or grep
Using grep
$ echo ... | grep -o '.*/'
http://sub.domain.tld/secured/
http://sub.domain.tld/section/files/
http://sub.domain.tld/.git/audio-files/top-secret/
with grep
If your grep has the -o option:
... | grep -Eio 'https?://.*/'
If there could be multiple URLs per line:
... | grep -Eio 'https?://[^[:space:]]+/'
with sed
If the input is always precisely one URL per line and nothing else, you can just delete the filename part:
... | sed 's/[^/]*$//'
You could use match function of awk, will work in any version of awk. Simple explanation would be, passing echo command's output to awk program. Using match matching everything till last occurrence of / and then printing the sub-string to print just before /(with -1 to RLENGTH).
your_echo_command | awk 'match($0,/.*\//){print substr($0,RSTART,RLENGTH-1)}'
GNU Awk
$ echo ... | awk 'match($0,/.*\//,a){print a[0]}'
$ echo ... | awk '{print gensub(/(.*\/).*/,"\\1",1)}'
$ echo ... | awk 'sub(/[^/]*$/,"")'
http://sub.domain.tld/secured/
http://sub.domain.tld/section/files/
http://sub.domain.tld/.git/audio-files/top-secret/
xargs
$ echo ... | xargs -i sh -c 'echo $(dirname "{}")/'
http://sub.domain.tld/secured/
http://sub.domain.tld/section/files/
http://sub.domain.tld/.git/audio-files/top-secret/
I have a bash script as below:
curl -s "$url" | grep "https://cdn" | tail -n 1 | awk -F[\",] '{print $2}'
which is working fine, when i run run it, i able to get the cdn url as:
https://cdn.some-domain.com/some-result/
when i put it as variable :
myvariable=$(curl -s "$url" | grep "https://cdn" | tail -n 1 | awk -F[\",] '{print $2}')
and i echo it like this:
echo "CDN URL: '$myvariable'"
i get blank result. CDN URL:
any idea what could be wrong? thanks
If your curl command produces a trailing DOS carriage return, that will botch the output, though not exactly like you describe. Still, maybe try this.
myvariable=$(curl -s "$url" | awk -F[\",] '/https:\/\/cdn/{ sub(/\r/, ""); url=$2} END { print url }')
Notice also how I refactored the grep and the tail (and now also tr -d '\r') into the Awk command. Tangentially, see useless use of grep.
The result could be blank if there's only one item after awk's split.
You might try grep -o to only return the matched string:
myvariable=$(curl -s "$url" | grep -oP 'https://cdn.*?[",].*' | tail -n 1 | awk -F[\",] '{print $2}')
echo "$myvariable"
I have a string,
var=refs/heads/testing/branch
I want to get rid of refs/heads/ in the string using shell script, such that I have only:
var=testing/branch
Commands I tried (one per line):
echo $(var) | awk -F\\ {'print $2'}
echo $var | sed -e s,refs/heads/,,
echo "refs/heads/testing/branch" | grep -oP '(?<=refs/heads/\)\w+'
echo "refs/heads/testing/branch" | LC_ALL=C sed -e 's/.*\\//'
echo "refs/heads/testing/branch" | cut -d'\' -f2
echo refs/heads/testing/branch | sed -e s,refs/heads/,,
there are lots of options out there ,try easy ones:
echo $var | cut -d "/" -f 3,4
echo $var | awk -F"/" '{print $3"/"$4}'
Shell parameter expansion: remove the prefix "refs/heads/" from the variable contents
$ var=refs/heads/testing/branch
$ echo "${var#refs/heads/}"
testing/branch
I'm working on a shell script.
OUT=$1
here, the OUT variable is my filename.
I'm using grep search as follows:
l=`grep "$pattern " -A 15 $OUT | grep -w $i | awk '{print $8}'|tail -1 | tr '\n' ','`
The issue is that the filename parameter I must pass is test.log.However, I have the folder structure :
test.log
test.log.001
test.log.002
I would ideally like to pass the filename as test.log and would like it to search it in all log files.I know the usual way to do is by using test.log.* in command line, but I'm facing difficulty replicating the same in shell script.
My efforts:
var-$'.*'
l=`grep "$pattern " -A 15 $OUT$var | grep -w $i | awk '{print $8}'|tail -1 | tr '\n' ','`
However, I did not get the desired result.
Hopefully this will get you closer:
#!/bin/bash
for f in "${1}*"; do
grep "$pattern" -A15 "$f"
done | grep -w $i | awk 'END{print $8}'
My bash script is:
output=$(curl -s http://www.espncricinfo.com/england-v-south-africa-2012/engine/current/match/534225.html | sed -nr 's/.*<title>(.*?)<\/title>.*/\1/p')
score=echo"$output" | awk '{print $1}'
echo $score
The above script prints just a newline in my console whereas my required output is
$ curl -s http://www.espncricinfo.com/england-v-south-africa-2012/engine/current/match/534225.html | sed -nr 's/.*<title>(.*
?)<\/title>.*/\1/p' | awk '{print $1}'
SA
So, why am I not getting the output from my bash script whereas it works fine in terminal am I using echo"$output" in the wrong way.
#!/bin/bash
output=$(curl -s http://www.espncricinfo.com/england-v-south-africa-2012/engine/current/match/534225.html | sed -nr 's/.*<title>(.*?)<\/title>.*/\1/p')
score=$( echo "$output" | awk '{ print $1 }' )
echo "$score"
Score variable was probably empty, since your syntax was wrong.