past_date=`date +"%Y-%m-%d" -d "-60 day"`
initial_date= sed -n "/$past_date/p" 'logfile.txt' | head -1 | sed -e 's/\([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9] [0-9][0-9]:[0-9][0-9]:[0-9][0-9]\).*/\1/'
echo $initial_date
/*I am trying to store the result of sed command to initial_date variable. But nothing is stored in initial_date*/
To store a command output into a variable $var, use the var=$(command) syntax:
initial_date=$(sed -n "/$past_date/p" 'logfile.txt' | head -1 | sed -e 's/\([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9] [0-9][0-9]:[0-9][0-9]:[0-9][0-9]\).*/\1/')
Then to print the result it is always recommended to quote the variable:
echo "$initial_date"
Update
If you are looking for the first date hour in logfile.txt, being date the $past_date, then you can use:
grep -o -m1 '2013-11-14 [0-9][0-9]:[0-9][0-9]:[0-9][0-9]' logfile.txt
Given this sample file:
$ cat logfile.txt
hello 2013-11-14 11:12:33
2013-11-14 11:12:33
2013-21-14 11:12:33
2013-r2:33
2013-19-14
2013-11-10 adf
$ grep -o -m1 '2013-11-14 [0-9][0-9]:[0-9][0-9]:[0-9][0-9]' logfile.txt
2013-11-14 11:12:33
$ data=$(grep -o -m1 '2013-11-14 [0-9][0-9]:[0-9][0-9]:[0-9][0-9]' logfile.txt)
$ echo $data
2013-11-14 11:12:33
initial_time=echo $line | sed -e 's/\([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9] [0-9][0-9]:[0-9][0-9]:[0-9][0-9]\).*/\1/'
Related
#!/bin/bash
MA=$(bt-device -l | cut -d " " -f 3)
MAC=${MA:1: -1}
bluetoothctl connect $MAC
Expected Result
98:9E:63:18:00:88
Actual result
(98:9E:63:18:00:88
A few alternatives:
$ echo 'Denny’s Tunez (98:9E:63:18:00:88)' | sed -En 's/^[^(]*\(([^)]*)\).*/\1/p'
98:9E:63:18:00:88
$ echo 'Denny’s Tunez (98:9E:63:18:00:88)' | cut -d'(' -f2 | cut -d')' -f1
98:9E:63:18:00:88
$ echo 'Denny’s Tunez (98:9E:63:18:00:88)' | awk -F'[)(]' '{print $2}'
98:9E:63:18:00:88
$ echo 'Denny’s Tunez (98:9E:63:18:00:88)' | grep -Eow '(..)(:..){5}'
98:9E:63:18:00:88
$ x='Denny’s Tunez (98:9E:63:18:00:88)'
$ y="${x//*\(/}"
$ y="${y//\)*}"
$ echo $y
98:9E:63:18:00:88
With GNU bash and its Parameter Expansion:
s="(98:9E:63:18:00:88)"
s="${s/#?/}" # remove first character
s="${s/%?/}" # remove last character
echo "$s"
Output:
98:9E:63:18:00:88
Using sed it can be done in a single step:
s='Denny’s Tunez (98:9E:63:18:00:88)'
echo "$s" | sed -E 's/.* \(|)//g'
98:9E:63:18:00:88
So for your example you can use:
mac=$(bt-device -l | sed -E 's/.* \(|)//g')
You can use parameter expansion:
offset and length
echo ${MA:1: -1}
prefix and suffix removal
tmp=${MA#(}
echo ${tmp%)}
parameter matching
tmp=${MA/#\(}
echo ${tmp/%\)}
Another approach is to:
whitelist what you do want
echo "$MA" | tr -dC '[0-9A-F:]'
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
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
So I have an expression that I want to extract some lines from a text and count them. I can grep them as follows:
$ cat medsCounts_totals.csv | grep -E 'NumMeds": 0' | wc -l
Which is fine. Now I want to loop over with the string ...
$ for i in {0..10}; do expr="NumMeds\": $i"; echo $expr; done
However, when I try to use $expr
for i in {0..10}; do expr="NumMeds:\" $i"; cat medsCounts_totals.csv | grep -E "$expr" | wc -l ; done
I get nothing. How do I solve this problem in an elegant manner?
there is a typo in
for i in {0..10}; do expr="NumMeds:\" $i"; cat medsCounts_totals.csv | grep -E "$expr" | wc -l ; done
it should be
"NumMeds\": $i"
I have problem with echo command I need export data to csv but its empty file
#!/bin/bash
while read domain
do
ownname= whois $domain | grep -A 1 -i "Administrative Contact:" |cut -d ":" -f 2 | sed 's/ //' | sed -e :a -e '$!N;s/ \n/,/;ta'
echo -e "$ownname" >> test.csv
done < dom.txt
You need to use command substitution to store command's output in a shell variable:
#!/bin/bash
while read domain; do
ownname=$(whois $domain | grep -A 1 -i "Administrative Contact:" |cut -d ":" -f 2 | sed 's/ //' | sed -e :a -e '$!N;s/ \n/,/;ta')
echo -e "$ownname" >> test.csv
done
PS: I haven't tested all the piped commands.