I want to grep a line that includes a quotation mark, more specifically I want to grep lines that include a " mark.
more specifically I want to grep lines like:
#include "something.h"
then pipe into sed to just return something.h
A single grep will do this job.
grep -oP '(?<=")[^"]*(?=")' file
Example:
$ echo '#include "something.h"' | grep -oP '(?<=")[^"]*(?=")'
something.h
sed '#n
/"/ s/.*"\([^"]*\)" *$/\1/p' YourFile
No need of grep (unless performance on huge file is wanted) with a sed. Sed could filter and adapt directly the content
In your case, /"/ is certainly modified by /#include *"/
in case of several string between quote
sed '#n
/"/ {s/"[^"]*$/"/;s/[^"]*"\([^"]*\)" */\1/gp;}' YourFile
You can use awk to get included filename:
awk -F'"' '{print $2}' file.c
something.h
Related
The file is like this
aaa&123
bbb&234
ccc&345
aaa&456
aaa$567
bbb&678
I want to output:(contain "aaa" and text after &)
123
456
I want to do in in shell script,
Follow code be consider
#!/bin/bash
raw=$(grep 'aaa' 1.txt)
var=$(cut -f2 -d"&" "$raw")
echo $var
It give me a error like
cut: aaa&123
aaa&456
aaa$567: No such file or directory
How to fix it? and how to cut (or grep or other) from exist variables?
Many thanks!
With GNU grep:
grep -oP 'aaa&\K.*' file
Output:
123
456
\K: ignore everything before pattern matching and ignore pattern itself
From man grep:
-o, --only-matching
Print only the matched (non-empty) parts of a matching line,
with each such part on a separate output line.
-P, --perl-regexp
Interpret PATTERN as a Perl compatible regular expression (PCRE)
Cyrus has my vote. An awk alternative if GNU grep is not available:
awk -F'&' 'NF==2 && $1 ~ /aaa/ {print $2}' file
Using & as the field separator, for lines with 2 fields (i.e. & must be present) and the first field contains "aaa", print the 2nd field.
The error with your answer is that you are treating the grep output like a filename in the cut command. What you want is this:
grep 'aaa.*&' file | cut -d'&' -f2
The pattern means "aaa appears before an &"
I get the following result in my script when I run it against my services. The result differs depending on the service but the text pattern showing below is similar. The result of my script is assigned to var1. I need to extract data from this variable
$var1=HOST1*prod*gem.dot*serviceList : svc1 HOST1*prod*kem.dot*serviceList : svc3, svc4 HOST1*prod*fen.dot*serviceList : svc5, svc6
I need to strip the name of the service list from $var1. So the end result should be printed on separate line as follow:
svc1
svc2
svc3
svc4
svc5
svc6
Can you please help with this?
Regards
Using sed and grep:
sed 's/[^ ]* :\|,\|//g' <<< "$var1" | grep -o '[^ ]*'
sed deletes every non-whitespace before a colon and commas. Grep just outputs the resulting services one per line.
Using gnu grep and gnu sed:
grep -oP ': *\K\w+(, \w+)?' <<< "$var1" | sed 's/, /\n/'
svc1
svc3
svc4
svc5
svc6
grep is the perfect tool for the job.
From man grep:
-o, --only-matching
Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
Sounds perfect!
As far as I'm aware this will work on any grep:
echo "$var1" | grep -o 'svc[0-9]\+'
Matches "svc" followed by one or more digits. You can also enable the "highly experimental" Perl regexp mode with -P, which means you can use the \d digit character class and don't have to escape the + any more:
grep -Po 'svc\d+' <<<"$var1"
In bash you can use <<< (a Here String) which supplies "$var1" to grep on the standard input.
By the way, if your data was originally on separate lines, like:
HOST1*prod*gem.dot*serviceList : svc1
HOST1*prod*kem.dot*serviceList : svc3, svc4
HOST1*prod*fen.dot*serviceList : svc5, svc6
This would be a good job for awk:
awk -F': ' '{split($2,a,", "); for (i in a) print a[i]}'
Given the following file
$ cat a.txt
FOO='hhh';BAR='eee';BAZ='ooo'
I can easily parse out one item with tr and grep
$ tr ';' '\n' < a.txt | grep BAR
BAR='eee'
However if I try this using sed it just prints everything
$ sed 's/;/\n/g; /BAR/!d' a.txt
FOO='hhh'
BAR='eee'
BAZ='ooo'
With awk you could do this:
awk '/BAR/' RS=\; file
But if in the case of BAZ this would produce an extra newline, because the is no ; after the last word. If you want to remove that newline as well you would need to do something like:
awk '/BAZ/{sub(/\n/,x); print}' RS=\; file
or with GNU awk or mawk you could use:
awk '/BAZ/' RS='[;\n]'
If your grep has the -o option then you could also try this:
grep -o '[^;]*BAZ[^;]*' file
sed can do it just as you want:
sed -n 's/.*\(BAR[^;]*\).*/\1/gp' <<< "FOO='hhh';BAR='eee';BAZ='ooo'"
The point here is that you must suppress sed's default output -- the whole line --, and print only the substitutions you want to performed.
Noteworthy points:
sed -n suppresses the default output;
s/.../.../g operates in the entire line, even if already matched -- greedy;
s/.1./.2./p prints out the substituted part (.2.);
the tr part is given as the delimiter in the expression \(BAR[^;]*\);
the grep job is represented by the matching of the line itself.
awk 'BEGIN {RS=";"} /BAR/' a.txt
The following grep solution might work for you:
grep -o 'BAR=[^;]*' a.txt
$ sed 's/;/\n/g;/^BAR/!D;P;d' a.txt
BAR='eee'
replace all ; with \n
delete until BAR line is at the top
print BAR line
delete pattern space
I have a file as show below
1.2.3.4.ask
sanma.nam.sam
c.d.b.test
I want to remove the last field from each line, the delimiter is . and the number of fields are not constant.
Can anybody help me with an awk or sed to find out the solution. I can't use perl here.
Both these sed and awk solutions work independent of the number of fields.
Using sed:
$ sed -r 's/(.*)\..*/\1/' file
1.2.3.4
sanma.nam
c.d.b
Note: -r is the flag for extended regexp, it could be -E so check with man sed. If your version of sed doesn't have a flag for this then just escape the brackets:
sed 's/\(.*\)\..*/\1/' file
1.2.3.4
sanma.nam
c.d.b
The sed solution is doing a greedy match up to the last . and capturing everything before it, it replaces the whole line with only the matched part (n-1 fields). Use the -i option if you want the changes to be stored back to the files.
Using awk:
$ awk 'BEGIN{FS=OFS="."}{NF--; print}' file
1.2.3.4
sanma.nam
c.d.b
The awk solution just simply prints n-1 fields, to store the changes back to the file use redirection:
$ awk 'BEGIN{FS=OFS="."}{NF--; print}' file > tmp && mv tmp file
Reverse, cut, reverse back.
rev file | cut -d. -f2- | rev >newfile
Or, replace from last dot to end with nothing:
sed 's/\.[^.]*$//' file >newfile
The regex [^.] matches one character which is not dot (or newline). You need to exclude the dot because the repetition operator * is "greedy"; it will select the leftmost, longest possible match.
With cut on the reversed string
cat youFile | rev |cut -d "." -f 2- | rev
If you want to keep the "." use below:
awk '{gsub(/[^\.]*$/,"");print}' your_file
I want to extract the string from a line which starts with a specific pattern from a file in shell script.
For example: I want the strings from lines that start with hello:
hi to_RAm
hello to_Hari
hello to_kumar
bye to_lilly
output should be
to_Hari
to_kumar
Can anyone help me?
sed is the most appropriate tool:
sed -n 's/^hello //p'
Use grep:
grep ^hello file | awk '{print $2}'
^ is to match lines that starts with "hello". This is assuming you want to print the second word.
If you want to print all words except the first then:
grep ^hello file | awk '{$1=""; print $0}'
You could use GNU grep's perl-compatible regexes and use a lookbehind:
grep -oP '(?<=hello ).*'