I want grep a word in my file, but only the part I gave to grep.
Example : in my file, i've "hell\nhell:o", i want grep hell but not hello.
How can I do that?
Give this a try:
grep -E "\bhell(\s|$)" file
add o if you just want the matched word:
kent$ echo "hell\nhell:o"|grep -oE "\bhell(\s|$)"
hell
or:
grep -oP '\bhell(?=\s|$)'
Using the -o option of grep :
grep -o hell your_file
From grep manual page :
-o, --only-matching
Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
Your question (and comments) aren't overly clear but you could try either:
"negative lookahead"
using the "end of line anchor"
Negative Lookahead
The following regex will match any hell that isn't followed by a : (adapt as suitable):
hell(?!:)
grep doesn't support this, so you'll need to use perl or something else:
echo -e "hell\nhell:o" \
| perl -ne 'print if /hell(?!:)/'
End of Line Anchor
The following will work with grep, matching only where the hell touches the end of the line ($):
hell$
echo -e "hell\nhell:o" \
| grep 'hell$'
Try to use -oP
echo "hell hello" | grep -oP '\bhell\b'
Related
How can I extract a word that comes after a specific word in bash ? More precisely, I have a file which has a line which looks like this:
Demo.txt
IN=../files/d
out=../files/d
dataload
name
i want to read "d" from above line.
sed -n '/\/files\// s~.*/files/\([^.]*\)\..*~\1~p' file
this code helping if line having "."
IN=../files/d.txt
so its printing "d"
here we have "d" without "." as end delimeter. So i want to read till end of line.
i/p :
Demo.txt
IN=../files/d
out=../files/d
dataload
name
output looking for:
d
d
code: in bash
You could use GNU grep with PCRE :
grep -oP '/files/\K[^.]+' file
The -P flag makes grep use PCRE, the -o makes it display only the matched part rather than the full line, and the \K in the regex omits what precedes from the displayed matched part.
Alternatively if you don't have access to GNU grep, the following perl command will have the same effect :
perl -nle 'print $& if m{/files/\K[^.]+}' file
Sample run.
This sed variant should work for you:
sed -n '/\/files\// s~.*/files/\([^.]*\).*~\1~p' file
d
d
Minor change from earlier sed is that it doesn't match \. right after first capture group.
When you don't want to think about a single command solution, you can use
grep -Eo "/files/." Demo.txt | cut -d/ -f3
Given the following content:
title="Bar=1; Fizz=2; Foo_Bar=3;"
I'd like to match the first occurrence of Bar value which is 1. Also I don't want to rely on soundings of the word (like double quote in the front), because the pattern could be in the middle of the line.
Here is my attempt:
$ grep -o -m1 'Bar=[ ./0-9a-zA-Z_-]\+' input.txt
Bar=1
Bar=3
I've used -m/--max-count which suppose to stop reading the file after num matches, but it didn't work. Why this option doesn't work as expected?
I could mix with head -n1, but I wondering if it is possible to achieve that with grep?
grep is line-oriented, so it apparently counts matches in terms of lines when using -m[1]
- even if multiple matches are found on the line (and are output individually with -o).
While I wouldn't know to solve the problem with grep alone (except with GNU grep's -P option - see anubhava's helpful answer), awk can do it (in a portable manner):
$ awk -F'Bar=|;' '{ print $2 }' <<<"Bar=1; Fizz=2; Foo_Bar=3;"
1
Use print "Bar=" $2, if the field name should be included.
Also note that the <<< method of providing input via stdin (a so-called here-string) is specific to Bash, Ksh, Zsh; if POSIX compliance is a must, use echo "..." | grep ... instead.
[1] Options -m and -o are not part of the grep POSIX spec., but both GNU and BSD/OSX grep support them and have chosen to implement the line-based logic.
This is consistent with the standard -c option, which counts "selected lines", i.e., the number of matching lines:
grep -o -c 'Bar=[ ./0-9a-zA-Z_-]\+' <<<"Bar=1; Fizz=2; Foo_Bar=3;" yields 1.
Using perl based regex flavor in gnu grep you can use:
grep -oP '^(.(?!Bar=\d+))*Bar=\d+' <<< "Bar=1; Fizz=2; Foo_Bar=3;"
Bar=1
(.(?!Bar=\d+))* will match 0 or more of any characters that don't have Bar=\d+ pattern thus making sure we match first Bar=\d+
If intent is to just print the value after = then use:
grep -oP '^(.(?!Bar=\d+))*Bar=\K\d+' <<< "Bar=1; Fizz=2; Foo_Bar=3;"
1
You can use grep -P (assuming you are on gnu grep) and positive look ahead ((?=.*Bar)) to achieve that in grep:
echo "Bar=1; Fizz=2; Foo_Bar=3;" | grep -oP -m 1 'Bar=[ ./0-9a-zA-Z_-]+(?=.*Bar)'
First use a grep to make the line start with Bar, and then get the Bar at the start of the line:
grep -o "Bar=.*" input.txt | grep -o -m1 "^Bar=[ ./0-9a-zA-Z_-]\+"
When you have a large file, you can optimize with
grep -o -m1 "Bar=.*" input.txt | grep -o -m1 "^Bar=[ ./0-9a-zA-Z_-]\+"
I'm trying to read a version number from between a set of parentheses, from this output of some command:
Test Application version 1.3.5
card 0: A version 0x1010000 (1.0.0), 20 ch
Total known cards: 1
What I'm looking to get is 1.0.0.
I've tried variations of sed and grep:
command.sh | grep -o -P '(?<="(").*(?=")")'
command.sh | sed -e 's/(\(.*\))/\1/'
and plenty of variations. No luck :-(
Help?
You were almost there! In pgrep, use backslashes to keep literal meaning of parentheses, not double quotes:
grep -o -P '(?<=\().*(?=\))'
Having GNU grep you can also use the \K escape sequence available in perl mode:
grep -oP '\(\K[^)]+'
\K removes what has been matched so far. In this case the starting ( gets removed from match.
Alternatively you could use awk:
awk -F'[()]' 'NF>1{print $2}'
The command splits input lines using parentheses as delimiters. Once a line has been splitted into multiple fields (meaning the parentheses were found) the version number is the second field and gets printed.
Btw, the sed command you've shown should be:
sed -ne 's/.*(\(.*\)).*/\1/p'
There are a couple of variations that will work. First with grep and sed:
grep '(' filename | sed 's/^.*[(]\(.*\)[)].*$/\1/'
or with a short shell script:
#!/bin/sh
while read -r line; do
value=$(expr "$line" : ".*(\(.*\)).*")
if [ "x$value" != "x" ]; then
printf "%s\n" "$value"
fi
done <"$1"
Both return 1.0.0 for your given input file.
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($2,a,", "); for (i in a) print a[i]}'
I have a command like this :
cat error | grep -o [0-9]
which is printing only numbers like 2,30 and so on. Now I wish to pass this number to sed.
Something like :
cat error | grep -o [0-9] | sed -n '$OutPutFromGrep,$OutPutFromGrepp'
Is it possible to do so?
I'm new to shell scripting. Thanks in advance
If the intention is to print the lines that grep returns, generating a sed script might be the way to go:
grep -E -o '[0-9]+' error | sed 's/$/p/' | sed -f - error
You are probably looking for xargs, particularly the -I option:
themel#eristoteles:~$ xargs -I FOO echo once FOO, twice FOO
hi
once hi, twice hi
there
once there, twice there
Your example:
themel#eristoteles:~$ cat error
error in line 123
error in line 234
errors in line 345 and 346
themel#eristoteles:~$ grep -o '[0-9]*' < error | xargs -I OutPutFromGrep echo sed -n 'OutPutFromGrep,OutPutFromGrepp'
sed -n 123,123p
sed -n 234,234p
sed -n 345,345p
sed -n 346,346p
For real-world use, you'll probably want to pass sed an input file and remove the echo.
(Fixed your UUOC, by the way. )
Yes you can pass output from grep to sed.
Please note that in order to match whole numbers you need to use [0-9]* not only [0-9] which would match only a single digit.
Also note you should use double quotes to get variables expanded(in the sed argument) and it seems you have a typo in the second variable name.
Hope this helps.