I have a file, part of which I am greping into a variable in shell script. In this variable I am trying to replace part of the string through sed. But it is somehow replacing 'newline' as well.
Here is the script
#!/bin/bash
str=$(grep "my_stat" $1)
str=$(echo $str | sed -e "s/x/\./g")
echo "$str"
My str after the grep looks like this:
mystatx1x2x3 10
mystatx3x4x5 20
mystatx4x4x1 100
but after the replacement of "x" to ".". It is changing to like this:
mystats.1.2.3 10 mystat.3.4.5 20 mystat.4.4.1 100
Any idea why is it happening like this?
Thanks.
replace
str=$(echo $str | sed -e "s/x/\./g")
with
str=$(echo "$str" | sed -e "s/x/\./g")
but instead of grep then sed, you could have said
sed -n '/mystat/s/x/\./gp' $1
do your replacement in one shot
sed -e '/my_stat/s/x/\./g' file
Related
i have text like 1|2|3||| , and try to replace each || with |0|, my command is following
echo '1|2|3|||' | sed -e 's/||/|0|/g'
but get result 1|2|3|0||, the pattern is only replaced once.
could someone help me improve the command, thx
Just do it 2 times
l_replace='s#||#|0|#g'
echo '1|2|3||||||||4||5|||' | sed -e "$l_replace;$l_replace"
Using any sed or any awk in any shell on every Unix box:
$ echo '1|2|3|||' | sed -e 's/||/|0|/g; s/||/|0|/g'
1|2|3|0|0|
$ echo '1|2|3|||' | awk '{while(gsub(/\|\|/,"|0|"));}1'
1|2|3|0|0|
This might work for you (GNU sed):
sed 's/||/|0|/g;s//[0]/g' file
or:
sed ':a;s/||/|0|/g;ta' file
The replacement needs to actioned twice because part of the match is in the replacement.
I have a string something like this
xsd:import schemaLocation="AppointmentManagementService.xsd6.xsd" namespace=
I want to extract the following from it :
AppointmentManagementService.xsd6.xsd
I have tried using regex, bash and sed with no success. Can someone please help me out with this?
The regex that I used was this :
/AppointmentManagementService.xsd\d{1,2}.xsd/g
Your string is:
nampt#nampt-desktop:$ cat 1
xsd:import schemaLocation="AppointmentManagementService.xsd6.xsd" namespace=
Try with awk:
cat 1 | awk -F "\"" '{print $2}'
Output:
AppointmentManagementService.xsd6.xsd
sed doesn't recognize \d, use [0-9] or [[:digit:]] instead:
sed 's/^.*schemaLocation="\([^"]\+[[:digit:]]\{1,2\}\.xsd\)".*$/\1/g'
## or
sed 's/^.*schemaLocation="\([^"]\+[0-9]\{1,2\}\.xsd\)".*$/\1/g'
You can use bash native regex matching:
$ in='xsd:import schemaLocation="AppointmentManagementService.xsd6.xsd" namespace='
$ if [[ $in =~ \"(.+)\" ]]; then echo "${BASH_REMATCH[1]}"; fi
Output:
AppointmentManagementService.xsd6.xsd
Based on your example, if you want to grant, at least, 1 or, at most, 2 digits in the .xsd... component, you can fine tune the regex with:
$ if [[ $in =~ \"(AppointmentManagementService.xsd[0-9]{1,2}.xsd)\" ]]; then echo "${BASH_REMATCH[1]}"; fi
using PCRE in GNU grep
grep -oP 'schemaLocation="\K.*?(?=")'
this will output pattern matched between schemaLocation=" and very next occurrence of "
Reference:
https://unix.stackexchange.com/a/13472/109046
Also we can use 'cut' command for this purpose,
[root#code]# echo "xsd:import schemaLocation=\"AppointmentManagementService.xsd6.xsd\" namespace=" | cut -d\" -f 2
AppointmentManagementService.xsd6.xsd
s='xsd:import schemaLocation="AppointmentManagementService.xsd6.xsd" namespace='
echo $s | sed 's/.*schemaLocation="\(.*\)" namespace=.*/\1/'
I've read a lot of questions about how to replace spaces from a file but I have the following problem:
I have a file like so:
<foo>"crazy foo"</foo> <bar>dull-bar</bar>
and I'm trying to remove spaces between > < and only those ones so the file would be like:
`<foo>"crazy foo"</foo><bar>dull-bar</bar>`
So far I've tried to remove then by using sed and tr. Sed is not working by any chance and using tr '> <' '><' outputs:
<foo>"crazy foo"</foo><<bar>dull-bar</bar>
sed -i -e "s/> *</></g" YourFile
-i means YourFile is modified. Remove this option to test your command and display the result in shell output.
* matches n spaces.
The g at the end of sed expression means "Replace all the occurrences".
You could try something like this
echo "<foo>"crazy foo"</foo> <bar>dull-bar</bar>" | sed 's/>[[:space:]]*</></g '
awk -F"\"" '{print $3}' file.txt | sed 's/ //g'
I need to replace characters from 10th to 20th in the string which looks like that:
123456789012345678901234567890
So far I've tried:
a)
Works for the 10th character ONLY:
echo "123456789012345678901234567890" | sed 's/./X/10'
b)
Doesn't work on the range:
echo "123456789012345678901234567890" | sed 's/./X/10,20'
echo "123456789012345678901234567890" | sed 's/./X/10\,20'
echo "123456789012345678901234567890" | sed 's/./X/\{10,20\}'
echo "123456789012345678901234567890" | sed 's/./X/\{10\,20\}'
Does not work and I get error
unknown option to `s'
So - the question is - how do I make this to work:
echo "123456789012345678901234567890" | sed 's/./X/10,20'
Try:
$ sed -r "s/^(.{9})(.{11})/\1XXXXXXXXXX/" <<< 123456789012345678901234567890
123456789XXXXXXXXXX1234567890
It is a complex sed problem, I could just find this solution:
$ sed 's/^\(.\{10\}\)\(.\{10\}\)/\1XXXXXXXXXX/' <<< 123456789012345678901234567890
1234567890XXXXXXXXXX1234567890
With awk it looks nicer:
$ awk 'BEGIN{FS=OFS=""} {for (i=10;i<=20;i++) $i="X"} {print}' <<< 123456789012345678901234567890
123456789XXXXXXXXXXX1234567890
You can do it with bash parameter substitution like this:
#!/bin/bash
s="123456789012345678901234567890"
l=${s:0:9} # Extract left part
m=${s:10:11} # Extract middle part
r=${s:20} # Extract right part
# Diddle with middle part to your heart's content and re-assemble "$l$m$r" when done
m=$(sed 's/./X/g' <<<$m)
See here for more explanation and examples.
Or, you can do this:
transform the row of letters into a column so each is on its own line
apply your edits to LINES 10 through 20 (as opposed to characters 10 through 20)
transform column of letters back into a row (by deleting linefeeds)
as shown in the one-liner below:
$ echo "123456789012345678901234567890" | sed "s/\(.\)/\1\n/g" | sed "10,20s/./X/" | tr -d "\n"
I know, that it looks ugly, but:
echo "123456789012345678901234567890" | \
sed 's/^\(.\{10\}\).\{10\}\(.*\)/\1XXXXXXXXXX\2/'
Without placing multiple X in sed command:
sed -r 's/^(.{9})(.{10,20})(.*)$/\1\n\2\n\3/' | sed -e '2s/./X/g' -e 'N;N;s/\n//g'
To replace the 10th to 20th characters, inclusive, try:
echo 123456789012345678901234567890 | sed 's/\(.\{9\}\).\{11\}/\1XXXXXXXXXX/'
123456789XXXXXXXXXX1234567890
With the GNU sed, you can use the -r switch to remove most of the backslashes:
echo 123456789012345678901234567890 | sed -r 's/(.{9}).{11}/\1XXXXXXXXXX/'
Or the naive approach also works here:
echo 123456789012345678901234567890 | sed 's/\(.........\).........../\1XXXXXXXXXX/'
This might work for you (GNU sed):
sed ':a;/.\{9\}X\{11\}/!s/\(.\{9\}X*\)./\1X/;ta' file
or with a bit of syntactic sugar:
sed -r ':a;/.{9}X{11}/!s/(.{9}X*)./\1X/;ta' file
I make a cat of a file and apply on it a grep with a regular expression like this
cat /tmp/tmp_file | grep "toto.titi\[[0-9]\+\].tata=55"
the command display the following output
toto.titi[12].tata=55
is it possible to modify my grep command in order to extract the number 12 as displayed output of the command?
You can grab this in pure BASH using its regex capabilities:
s='toto.titi[12].tata=55'
[[ "$s" =~ ^toto.titi\[([0-9]+)\]\.tata=[0-9]+$ ]] && echo "${BASH_REMATCH[1]}"
12
You can also use sed:
sed 's/toto.titi\[\([0-9]*\)\].tata=55/\1/' <<< "$s"
12
OR using awk:
awk -F '[\\[\\]]' '{print $2}' <<<"$s"
12
use lookahead
echo toto.titi[12].tata=55|grep -oP '(?<=\[)\d+'
12
without perl regex,use sed to replace "["
echo toto.titi[12].tata=55|grep -o "\[[0-9]\+"|sed 's/\[//g'
12
Pipe it to sed and use a back reference:
cat /tmp/tmp_file | grep "toto.titi\[[0-9]\+\].tata=55" | sed 's/.*\[(\d*)\].*/\1/'