How can we find two substrings within a line in particular order using grep?
For example:
grep -c "word1" | grep -r "word2" logs
gives if string has both word1 and word2. I am looking for string which has "... word1.... word2..."
Try a regex in grep like grep -E "word1.*word2"
$ echo -e 'both word1 and word2. \nI hich\n has "... word1.... word2..."' | grep -E "word1.*word2"
both word1 and word2.
has "... word1.... word2..."
You may need a better regex to match exactly the words, but that is not your question.
Related
Both of the regexes below work In my case.
grep \s
grep ^[[:space:]]
However all those below fail. I tried both in git bash and putty.
grep ^\s
grep ^\s*
grep -E ^\s
grep -P ^\s
grep ^[\s]
grep ^(\s)
The last one even produces a syntax error.
If I try ^\s in debuggex it works.
Debuggex Demo
How do I find lines starting with whitespace characters with grep ? Do I have to use [[:space:]] ?
grep \s works for you because your input contains s. Here, you escape s and it matches the s, since it is not parsed as a whitespace matching regex escape. If you use grep ^\\s, you will match a string starting with whitespace since the \\ will be parsed as a literal \ char.
A better idea is to enable POSIX ERE syntax with -E and quote the pattern:
grep -E '^\s' <<< "$s"
See the online demo:
s=' word'
grep ^\\s <<< "$s"
# => word
grep -E '^\s' <<< "$s"
# => word
Team,
I want to grep for a substring container - and then only output that string and not whole line. how can i? I know i can awk on space and pull using $ but want to know how to do in grep?
echo $test_pods_info | grep -F 'test-'
output
test-78ac951e-89a6-4199-87a4-db8a1b8b054f export-9b55f0d5-071d-431-1d2ux0-avexport-xavierisp-sjc4--a4dd85-102 1/1 Running 0 19h
expected output
test-78ac951e-89a6-4199-87a4-db8a1b8b054f
awk is more suitable for this as you want to get first field in a matching line:
awk '/test-/{print $1}' <<< "$taxIncluded"
test-78ac951e-89a6-4199-87a4-db8a1b8b054f
If you really want to use grep then this might be what you're looking for:
grep -o 'test-\S*' <<< "$taxIncluded"
or:
grep -o 'test-[^[:space:]]*' <<< "$taxIncluded"
Try
echo $test_pods_info | grep -o 'test-'
the -o option is:
show[ing] only the part of a line matching PATTERN
according to grep --help. Of course, this will only print test-, so you'll need to rework your regex:
grep -oE '(test).*[[:space:]]\b'
Figured it out..
echo $test_pods_info | grep -o "\test-\w*-\w*\-\w*\-\w*\-\w*"
outoput
test-78ac951e-89a6-4199-87a4-db8a1b8b054f
but i wish there is simple way. like \test-*\
I want to match all cyrillic characters, but print the ID to file. For example:
Author: Doe, John
Title: Оптимизация ресурсного потенциала промышленности города с учетом его конкурентных преимуществ
ID: 1234567
My current approach is to grep for cyrillic characters:
grep -i -r --include=*{rdf,redif,rdf~} --color="auto" -P -n '[\x{0400}-\x{04FF}]' > cyrillic.txt
How can I just print the ID line to a file and not the matching line?
Use -A1 option if the ID: line is right after the matching pattern. Then pipe it to another grep to get the line with ID:.
grep -A1 -i -r --include=*{rdf,redif,rdf~} --color="auto" -P -n '[\x{0400}-\x{04FF}]' \
| grep 'ID: ' > cryllic.txt
Use grep flag h - to suppress output file names - you'll have output like:
4:string with matching pattern
5:string with matching pattern
7:string with matching pattern
Now you can pipe this output into awk and print only fist column, which is matching string number:
{your_grep} | awk -F ':' '{print $1}' > cyrillic.txt
I will need to grep one variable for a variable pattern.
Like so
foo="--test2"
bar="--test"
echo "${foo}" | grep "'${bar}'"
Unfortunately it is not working.
Any ideas about how to achieve this?
If you use , single quotes it will take it as literal string. Remove the single quotes. Then it will throw the error for -- in your string. For that use -e option for mention that is a pattern to match.
echo "${foo}" | grep -e "${bar}"
your pattern is leading with -, -e option is needed.
this line should work for your example:
echo "${foo}" | grep -e "${bar}"
I am trying to use grep to print only lines that start with a specific pattern. Here is an example
$SERVER_IP = 2.2.2.2
$SERVER_IP_PORT = 1111
$SERVER_IP_XXX = blablabla
I want grep to print only SERVER_IP = 2.2.2.2 and not the other three lines.
I tried the command below but it did not work
grep -e "^\s*\$SERVER_IP$"
If I try:
grep -e "^\s*\$SERVER_IP"
grep will print all three lines
How can I accomplish this using grep -e or egrep? Thank you
grep -e "^\s*\$SERVER_IP\>"
The \> means "word-boundary", or "place where word characters meet non-word characters."
Use grep -e '^\$SERVER_IP =' to match any line that starts with $SERVER_IP =
If you have awk, you can do:
awk '$1=="$SERVER_IP"' file
$SERVER_IP = 2.2.2.2
The == makes it match only while field 1 is exact $SERVER_IP