I am trying to match a line and use sed command to substitute it. Some thing like
aaa = 10
aaa =10
aaa=10
My sed regular expression should match all those patterns and should replace with something like bbb=5. I tried with
sed -i '/ *aaa *= */bbb=5'
But this is not properly working for all the patterns. Any help will be really appreciable.
sed -i 's/\s*aaa\s*=\s*[0-9]*/bbb=5/' input_file
cat a | sed -e '1s/aaa =10/bbb=10/' -e '2s/ aaa =10/bbb=10/' -e '3s/aaa=10/bbb=10/'
cat myfile | sed 's/\s*aaa\s*=\s*\(.*\)/bbb = \1/'
The \s character class matches both tab and space
Related
(Need in bash linux)I have a file with numbers like this
1.415949602
91.09582241
91.12042924
91.40270349
91.45625033
91.70150341
91.70174342
91.70660043
91.70966213
91.72597066
91.7287678315
91.7398645966
91.7542977976
91.7678146465
91.77196659
91.77299733
abcdefghij
91.7827827
91.78288651
91.7838959
91.7855
91.79080605
91.80103075
91.8050505
sed 's/^91\.//' file (working)
Any way possible I can do these 3 steps?
1st I try this
cat input | tr -d 91. > 1.txt (didnt work)
cat input | tr -d "91." > 1.txt (didnt work)
cat input | tr -d '91.' > 1.txt (didnt work)
then
grep -x '.\{10\}' (working)
then
grep "^[6-9]" (working)
Final 1 line solution
cat input.txt | sed 's/\91.//g' | grep -x '.\{10\}' | grep "^[6-9]" > output.txt
Your "final" solution:
cat input.txt |
sed 's/\91.//g' |
grep -x '.\{10\}' |
grep "^[6-9]" > output.txt
should avoid the useless cat, and also move the backslash in the sed script to the correct place (and I added a ^ anchor and removed the g flag since you don't expect more than one match on a line anyway);
sed 's/^91\.//' input.txt |
grep -x '.\{10\}' |
grep "^[6-9]" > output.txt
You might also be able to get rid of at least one useless grep but at this point, I would switch to Awk:
awk '{ sub(/^91\./, "") } /^[6-9].{9}$/' input.txt >output.txt
The sub() does what your sed replacement did; the final condition says to print lines which match the regex.
The same can conveniently, but less readably, be written in sed:
sed -n 's/^91\.([6-9][0-9]\{9\}\)$/\1/p' input.txt >output.txt
assuming your sed dialect supports BRE regex with repetitions like [0-9]\{9\}.
Want to convert the following pattern
ab
cd
de
fg as
'ab','cd','de','fg' using unix / linux command .
----Guys -------- The patern is as following
QRTC1065173134
QRTC3988977812
QRTC0889556882
QUTR1641276912
ABCD1763495154
QRTC3991601819
and this is the required pattern 'QRTC1065173134','QRTC3988977812','QRTC0889556882','QUTR1641276912','ABCD1763495154','QRTC3991601819'
I agree with the comments that it's a bit unclear but, for fun sake:
A="ab cd ef gh "
echo $A | sed -e "s/^\s*/'/; s/ \{1,\}/','/g; s/\s*$/'/g"
Since the question wasn't precise, I only worked on spaces & string boundaries. So it will work with any number of characters, separated space(s). The result is also trimmed at both ends. HTH.
From the clarification, it seems that this is what you want.
$ echo "QRTC1065173134 QRTC3988977812 QRTC0889556882 QUTR1641276912 ABCD1763495154 QRTC3991601819" | sed -E "s/ /', '/g" | sed -E "s/$/'/" | sed -E "s/^/'/"
'QRTC1065173134', 'QRTC3988977812', 'QRTC0889556882', 'QUTR1641276912', 'ABCD1763495154', 'QRTC3991601819'
Here "E" is for extended regex so that we do not need to escape the regex metacharacters.
COMMENT 1: Removing an extra whitespace is left as an exercise for you.
If I understand correctly, that you have groups (e.g. ab bc de ...) separated by spaces, where you want to include a ' at the beginning/end of everything, and replace the spaces with ',' then sed can handle this with relative ease. Below ab cd ... can be any string of characters, such as QRTC1065173134. There are several ways to piece together a matching regular expression, but the following is fairly simple:
sed -e "s/\s/','/g" -e "s/^/'/" -e "s/$/'/"
example
$ echo "ab cd de fg hi" | sed -e "s/\s/','/g" -e "s/^/'/" -e "s/$/'/"
'ab','cd','de','fg','hi'
or
$ echo "QRTC1065173134 QRTC3988977812 QRTC0889556882" | sed -e "s/\s/','/g" -e "s/^/'/" -e "s/$/'/"
'QRTC1065173134','QRTC3988977812','QRTC0889556882'
I have this data in file.txt:
1234-abca-dgdsf-kds-2;abc dfsfds 2
123-abcdegfs-sdsd;dsfdsf dfd f
12523-cvjbsvndv-dvd-dvdv;dsfdsfpage
I want to replace the string after "-" and up to ";" with just ";", so that I get:
1234;abc dfsfds 2
123;dsfdsf dfd f
12523;dsfdsfpage
I tried with the command:
sed -e "s/-.*;/;" file.txt
But it gives me the following error:
sed command garbled
Why is this happening?
sed replacement commands are defined as (source):
's/REGEXP/REPLACEMENT/[FLAGS]'
(substitute) Match the regular-expression against the content of the pattern space. If found, replace matched string with REPLACEMENT.
However, you are saying:
sed "s/-.*;/;"
That is:
sed "s/REGEXP/REPLACEMENT"
And hence missing a "/" at the end of the expression. Just add it to have:
sed "s/-.*;/;/"
# ^
You are missing a slash at the end of the sed command:
Should be "s/-.*;/;/"
-.* here the * greedy, so this would fail if there are more than one ;
echo "12523-cvjbsvndv-dvd-dvdv;dsfdsfpage;test" | sed -e "s/-.*;/;/"
12523;test
Change to -[^;]*
echo "12523-cvjbsvndv-dvd-dvdv;dsfdsfpage;test" | sed -e "s/-[^;]*;/;/"
12523;dsfdsfpage;test
This should work :
sed 's/-.*;/;/g' file > newFile
I have a text file which contains some lines as the following:
ASDASD2W 3ASGDD12 SDADFDFDDFDD W11 ACC=PNO23 DFSAEFEA EAEDEWRESAD ASSDRE
AERREEW2 3122312 SDADDSADADAD W12 ACC=HH34 23SAEFEA EAEDEWRESAD ASEEWEE
A15ECCCW 3XCXXF12 SDSGTRERRECC W43 ACC=P11 XXFSAEFEA EAEDEWRESAD ASWWWW
ASDASD2W 3122312 SDAFFFDEEEEE SD3 ACC=PNI22 ABCEFEA EAEDEWRESAD ASWEDSSAD
...
I have to extract the substring between the '=' character and the following blank space for each line , i.e.
PNO23
HH34
P11
PNI22
I've been using the sed command but cannot figure out how to ignore all characters following the blank space.
Any help?
Use the right tool for the job.
$ awk -F '[= ]+' '{ print $6 }' input.txt
PNO23
HH34
P11
PNI22
Sorry, but have to add another one because I feel the existing answers are just to complicated
sed 's/.*=//; s/ .*//;' inputfile
This might work for you:
sed -n 's/.*=\([^ ]*\).*/\1/p' file
or, if you prefer:
sed 's/.*=\([^ ]*\).*/\1/p;d' file
Put the string you want to capture in a backreference:
sed 's/.*=\([^ =]*\) .*/\1/'
or do the substitution piecemeal;
sed -e 's/.*=//' -e 's/ .*//'
sed 's/[^=]*=\([^ ]*\) .*/\1/' inputfile
Match all the non-equal-sign characters and an equal sign. Capture a sequence of non-space characters. Match a space and the rest of the line. Substitute the captured string.
A chain of grep can do the trick.
grep -o '[=][a-zA-Z0-9]*' file | grep -o '[a-zA-Z0-9]*'
I want to delete a fixed number of some backspace characters ocurrences ( \b ) from stdin. So far I have tried this:
echo -e "1234\b\b\b56" | sed 's/\b{3}//'
But it doesn't work. How can I achieve this using sed or some other unix shell tool?
You can use the hexadecimal value for backspace:
echo -e "1234\b\b\b56" | sed 's/\x08\{3\}//'
You also need to escape the braces.
You can use tr:
echo -e "1234\b\b\b56" | tr -d '\b'
123456
If you want to delete three consecutive backspaces, you can use Perl:
echo -e "1234\b\b\b56" | perl -pe 's/(\010){3}//'
sed interprets \b as a word boundary. I got this to work in perl like so:
echo -e "1234\b\b\b56" | perl -pe '$b="\b";s/$b//g'
With sed:
echo "123\b\b\b5" | sed 's/[\b]\{3\}//g'
You have to escape the { and } in the {3}, and also treat the \b special by using a character class.
[birryree#lilun ~]$ echo "123\b\b\b5" | sed 's/[\b]\{3\}//g'
1235
Note if you want to remove the characters being deleted also, have a look at ansi2html.sh which contains processing like:
printf "12..\b\b34\n" | sed ':s; s#[^\x08]\x08##g; t s'
No need for Perl here!
# version 1
echo -e "1234\b\b\b56" | sed $'s/\b\{3\}//' | od -c
# version 2
bvar="$(printf '%b' '\b')"
echo -e "1234\b\b\b56" | sed 's/'${bvar}'\{3\}//' | od -c