I have a file "Input_file" with content like this
%name=ABC
%value=123
sample text in file
sample text in file
%name=XYZ
%value=789
sample text in file
I need to extract the lines of this file matching this pattern.
str="%name=*\n%value=*"
I was working this way
gawk -v st=$str '/"$st"/ {print}' $Input_file
I'm getting the error
gawk: ^ backslash not last character on line
Even with grep as in
grep -e "$str" $Input_file
it says there is no such matching pattern. Where am I going wrong.
Try this:
grep -A1 "^%name=" $Input_file | grep -B1 "^%value=" | grep -v "^--"
you cannot directly use your "pattern (str)" in awk. because awk default doesn't work in multi-line mode. However you could do this with awk:
awk '/^%name=/{n=$0;next}/^%value=/&&n{print n"\n"$0}{n=""}' file
with your example, the above one-liner outputs:
%name=ABC
%value=123
%name=XYZ
%value=789
You can use a different syntax in your $str variable, the '*' is useless in because you are searching a pattern not a literal value, for gawk I can't help sorry
try this:
str="\%name=|\%value="
egrep $str $input_file
So you can match the two criteria of you search
Related
so I have a file with the following:
username=jsmith
api=3434kjklj23j4l3kj4l34j3l4j
I would like to return using regular expression "jsmith" and "3434kjklj23j4l3kj4l34j3l4j"
I know the regular expression for it is:
(username=)(.*) > \2
(api=)(.*) > \2
however using grep or sed or awk. I can't seem to figure out the way to use them without return the entire line.
How would you go about doing that with a commandline command?
awk is made for this task:
awk -F= '{print$2}' file
If the file has other entries, you can limit the output with a condition:
awk -F= '$1=="username"||$1=="api"{print$2}' file
Here is one using bash, PCRE and positive lookbehind (where supported):
$ grep -Po "((?<=^username=)|(?<=^api=)).*" file
jsmith
3434kjklj23j4l3kj4l34j3l4j
ie. output everything that is preceeded by username= or api= that start the lines.
And one in awk:
$ awk 'sub(/^(username|api)=/,""){print}' file
jsmith
3434kjklj23j4l3kj4l34j3l4j
ie. print lines where preceeding ^username= or ^api= are removed first.
Since you want to see chess with the input game=chess, here some solutions without matching username= or api=
cut -d"=" -f2- file
# or
sed -n 's/[^=]*=//p' file
here's the answer that worked on the macos and RHEl7.
awk -F= '$1=="username"{print$2}' testfile.txt
awk -F= '$1=="api"{print$2}' testfile.txt
testfile.txt
username=user1
api=pass1
username=user2
api =pass2
Looking for suggestion to cat file | grep REGEX to get the lines with <version>anything</version>.
grep -F '<version>1.1.9-beta</version>' file
-F will match your pattern as literal text
you don't need that useless cat
if you really mean anything: try grep '<version>.*</version>' file or grep -P '<version>.*?</version>' file , however searching xml with regex is bad idea.
Use the -E option to match a regular expression:
grep -E "<version>.*</version>" file
Refer to these rules for the regular expression: https://www.gnu.org/savannah-checkouts/gnu/grep/manual/grep.html#Regular-Expressions
For example, to match the typical version format (3.14, or 13.14, or 0.1458) you can type:
grep -E "<version>[0-9]?\.[0-9]?</version>" file
You can do:
grep '<version>[^<]*</version>' file.xml
[^<]* will match zero or more characters upto next <.
Everyone!!
I want to get specific substring from stdout of command.
stdout:
{"response":
{"id":"110200dev1","success":"true","token":"09ad7cc7da1db13334281b84f2a8fa54"},"success":"true"}
I need to get a hex string after token without quotation marks, the length of hex string is 32 letters.I suppose it can be done by sed or egrep. I don't want to use awk here. Because the stdout is being changed very often.
This is an alternate gnu-awk solution when grep -P isn't available:
awk -F: '{gsub(/"/, "")} NF==2&&$1=="token"{print $2}' RS='[{},]' <<< "$string"
09ad7cc7da1db13334281b84f2a8fa54
grep's nature is extracting things:
grep -Po '"token":"\K[^"]+'
-P option interprets the pattern as a Perl regular expression.
-o option shows only the matching part that matches the pattern.
\K throws away everything that it has matched up to that point.
Or an option using sed...
sed 's/.*"token":"\([^"]*\)".*/\1/'
With sed:
your-command | sed 's/.*"token":"\([^"]*\)".*/\1/'
YourStreamOrFile | sed -n 's/.*"token":"\([a-f0-9]\{32\}\)".*/\1/p'
doesn not return a full string if not corresponding
please refer the file contents below.
#HD VN:1.0 SO:unsorted
#SQ SN:Chr1 LN:30427680
#PG ID:bowtie2 PN:bowtie2 VN:2.1.0
how can i extract just the number 30427680 using awk or any other unix command.
Using sed
sed -n 's/.*LN://p' < input.txt
This will erase everything up until LN:, and print what's left, and only if a substitution did take place.
Using awk
awk -v FS=: '/LN:/ { print $3; }' < input.txt
This will match lines that contain LN:, use : as field separator, and print the 3rd column.
Using grep
grep -o '[0-9]\{3,\}' < input.txt
This will match sequences of 3 or more digits, and print only the matched pattern thanks to the -o.
Depending on other cases not included in your question, you might have to make the patterns more strict.
Using grep:
grep -oP 'LN:\K.*' filename
Just use grep:
grep -o 30427680 file
-o, --only-matching
Prints only the matching part of the lines.
Using perl :
perl -ne 'print $& if /LN:\K.*/' filename
or
perl -ne 'print $1 if /LN:(.*)/' filename
Another awk
awk -F"LN:" 'NF>1 {print $2}' file
My input:
1:FAILED + *1 0 (8328832,AR,UNDECLARED)
This is what I expect:
8328832,AR,UNDECLARED
I am trying to find a general regular expression that allows to take any content between two brackets out.
My attempt is
grep -o '\[(.*?)\]' test.txt > output.txt
but it doesn't match anything.
Still using grep and regex
grep -oP '\(\K[^\)]+' file
\K means that use look around regex advanced feature. More precisely, it's a positive look-behind assertion, you can do it like this too :
grep -oP '(?<=\()[^\)]+' file
if you lack the -P option, you can do this with perl :
perl -lne '/\(\K[^\)]+/ and print $&' file
Another simpler approach using awk
awk -F'[()]' '{print $2}' file