I'm not getting the result expect from the following code:
#!/bin/bash
cat /home/opmeitle/html/fiesta-one.html | grep -oiE '([$][0-9.]{1,7})'
this is output:
$90.850
$0
$389
$469
$670
$750
$684
$21.744
$604
Here is the result I desire, in console.
$90.850 $0 $389 $469 $670 $750 $684 $21.744 $604
I appreciate your answers. thanks
luis.
There are many other solutions:
$ echo $(grep -oiE '([$][0-9.]{1,7})' /home/opmeitle/html/fiesta-one.html)
$ grep -oiE '([$][0-9.]{1,7})' /home/opmeitle/html/fiesta-one.html | xargs echo
$ grep -oiE '([$][0-9.]{1,7})' /home/opmeitle/html/fiesta-one.html | tr '\n' ' '
$ grep -oiE '([$][0-9.]{1,7})' /home/opmeitle/html/fiesta-one.html | perl -pe 's/\n/ /;'
And without grep:
$ perl -ne 'print "$1 " if /([\$][0-9.]{1,7})/' /home/opmeitle/html/fiesta-one.html
A simple fix would be to translate the newlines into spaces (I also removed your unnecessary use of cat):
grep -oiE '([$][0-9.]{1,7})' /home/opmeitle/html/fiesta-one.html | tr '\n' ' '
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 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 $8}'
I'm trying to make a small function that removes all the chars that are not digits.
123a45a ---> will become ---> 12345
I've came up with :
temp=$word | grep -o [[:digit:]]
echo $temp
But instead of 12345 I get 1 2 3 4 5. How to I get rid of the spaces?
Pure bash:
word=123a45a
number=${word//[^0-9]}
Here's a pure bash solution
var='123a45a'
echo ${var//[^0-9]/}
12345
is this what you are looking for?
kent$ echo "123a45a"|sed 's/[^0-9]//g'
12345
grep & tr
echo "123a45a"|grep -o '[0-9]'|tr -d '\n'
12345
I would recommend using sed or perl instead:
temp="$(sed -e 's/[^0-9]//g' <<< "$word")"
temp="$(perl -pe 's/\D//g' <<< "$word")"
Edited to add: If you really need to use grep, then this is the only way I can think of:
temp="$( grep -o '[0-9]' <<< "$word" \
| while IFS= read -r ; do echo -n "$REPLY" ; done
)"
. . . but there's probably a better way. (It uses grep -o, like your solution, then runs over the lines that it outputs and re-outputs them without line-breaks.)
Edited again to add: Now that you've mentioned that you use can use tr instead, this is much easier:
temp="$(tr -cd 0-9 <<< "$word")"
What about using sed?
$ echo "123a45a" | sed -r 's/[^0-9]//g'
12345
As I read you are just allowed to use grep and tr, this can make the trick:
$ echo "123a45a" | grep -o [[:digit:]] | tr -d '\n'
12345
In your case,
temp=$(echo $word | grep -o [[:digit:]] | tr -d '\n')
tr will also work:
echo "123a45a" | tr -cd '[:digit:]'
# output: 12345
Grep returns the result on different lines:
$ echo -e "$temp"
1
2
3
4
5
So you cannot remove those spaces during the filtering, but you can afterwards, since $temp can transform itself like this:
temp=`echo $temp | tr -d ' '`
$ echo "$temp"
12345
I run this bash command to display contents of somefile.cf in a Weblogic domain directory.
find $(/usr/ucb/ps auwwx | grep weblogic | tr ' ' '\n' | grep security.policy | grep domain | awk -F'=' '{print $2}' | sed -e 's/weblogic.policy//' -e 's/security\///' -e 's/dep\///' | awk -F'/' '{print "/"$2"/"$3"/"$4"/somefile.cf"}' | sort | uniq) 2> /dev/null -exec ls {} \; -exec cat {} \;
I tried incorporating this in an expect script and also escaped some special characters and double quotes too but it throws an error "extra characters after close-quote"
send "echo ; echo 'Weblogic somefile.cf:' ; find \$(/usr/ucb/ps auwwx | grep weblogic | tr ' ' '\n' | grep security.policy | grep domain | awk -F'=' '{print \$2}' | sed -e 's/weblogic.policy//' -e 's/security\\///' -e 's/dep\\///' | awk -F'/' '{print \"/\"\$2\"/\"\$3\"/\"\$4\"/somefile.cf\"}' | sort | uniq) 2> /dev/null -exec ls {} \\; -exec cat {} \\;
I guess it needs some more escaping of special characters or probably I dint escape the existing ones correctly.
Any help would be appreciated.
give us the syntax error find or bash threw on the other side.
and try adding an extra \ or 2 before the semicolons at the end.
The problem with expect is the number of layers of escapes you need when it get's ugly.
In the awk statement, go escape all the doublequotes ( " -> \" )
and get me an error message :)
If you have a command line with complex quoting that you know works in bash then it's often easier to just go ahead and use bash. Like this:
set cmd {find $(/usr/ucb/ps auwwx | grep weblogic | tr ' ' '\n' | grep security.policy | grep domain | awk -F'=' '{print $2}' | sed -e 's/weblogic.policy//' -e 's/security\///' -e 's/dep\///' | awk -F'/' '{print "/"$2"/"$3"/"$4"/somefile.cf"}' | sort | uniq) 2> /dev/null -exec ls {} \; -exec cat {} \;}
spawn /bin/bash -c $cmd
expect ... whatever is appropriate ...
Notice that I used the Tcl {} operator instead of "" around the command string. This operator is like single quote in bash, it means "literal string, do not interpret the contents in any way" and is appropriate here because I want to pass this string verbatim to the spawned bash subprocess.
There is a " missing at the end of your send line.