I have a file a.txt and I want to find slashes(/ and \ ) in it, so I used this command in my a.ksh script slashcheck=cat a.txt | grep '/'.
but how do I check for '' backward slash at the same time?
you can try this: cat a.txt | grep "[\/]"
Please try with:
cat a.txt | egrep '(\/|\\)'
Try using a regular expression.
You could use something like grep -E '[/\\]'.
Note that you will need two backslashes because the first one masks the second one.
Using squared brackets allows any character in these brackets to be a match.
The -E stands for extended and allows regular expressions to be used as seach patterns.
Related
Is there a way to make grep output "words" from files that match the search expression?
If I want to find all the instances of, say, "th" in a number of files, I can do:
grep "th" *
but the output will be something like (bold is by me);
some-text-file : the cat sat on the mat
some-other-text-file : the quick brown fox
yet-another-text-file : i hope this explains it thoroughly
What I want it to output, using the same search, is:
the
the
the
this
thoroughly
Is this possible using grep? Or using another combination of tools?
Try grep -o:
grep -oh "\w*th\w*" *
Edit: matching from Phil's comment.
From the docs:
-h, --no-filename
Suppress the prefixing of file names on output. This is the default
when there is only one file (or only standard input) to search.
-o, --only-matching
Print only the matched (non-empty) parts of a matching line,
with each such part on a separate output line.
Cross distribution safe answer (including windows minGW?)
grep -h "[[:alpha:]]*th[[:alpha:]]*" 'filename' | tr ' ' '\n' | grep -h "[[:alpha:]]*th[[:alpha:]]*"
If you're using older versions of grep (like 2.4.2) which do not include the -o option, then use the above. Else use the simpler to maintain version below.
Linux cross distribution safe answer
grep -oh "[[:alpha:]]*th[[:alpha:]]*" 'filename'
To summarize: -oh outputs the regular expression matches to the file content (and not its filename), just like how you would expect a regular expression to work in vim/etc... What word or regular expression you would be searching for then, is up to you! As long as you remain with POSIX and not perl syntax (refer below)
More from the manual for grep
-o Print each match, but only the match, not the entire line.
-h Never print filename headers (i.e. filenames) with output lines.
-w The expression is searched for as a word (as if surrounded by
`[[:<:]]' and `[[:>:]]';
The reason why the original answer does not work for everyone
The usage of \w varies from platform to platform, as it's an extended "perl" syntax. As such, those grep installations that are limited to work with POSIX character classes use [[:alpha:]] and not its perl equivalent of \w. See the Wikipedia page on regular expression for more
Ultimately, the POSIX answer above will be a lot more reliable regardless of platform (being the original) for grep
As for support of grep without -o option, the first grep outputs the relevant lines, the tr splits the spaces to new lines, the final grep filters only for the respective lines.
(PS: I know most platforms by now would have been patched for \w.... but there are always those that lag behind)
Credit for the "-o" workaround from #AdamRosenfield answer
It's more simple than you think. Try this:
egrep -wo 'th.[a-z]*' filename.txt #### (Case Sensitive)
egrep -iwo 'th.[a-z]*' filename.txt ### (Case Insensitive)
Where,
egrep: Grep will work with extended regular expression.
w : Matches only word/words instead of substring.
o : Display only matched pattern instead of whole line.
i : If u want to ignore case sensitivity.
You could translate spaces to newlines and then grep, e.g.:
cat * | tr ' ' '\n' | grep th
Just awk, no need combination of tools.
# awk '{for(i=1;i<=NF;i++){if($i~/^th/){print $i}}}' file
the
the
the
this
thoroughly
grep command for only matching and perl
grep -o -P 'th.*? ' filename
I was unsatisfied with awk's hard to remember syntax but I liked the idea of using one utility to do this.
It seems like ack (or ack-grep if you use Ubuntu) can do this easily:
# ack-grep -ho "\bth.*?\b" *
the
the
the
this
thoroughly
If you omit the -h flag you get:
# ack-grep -o "\bth.*?\b" *
some-other-text-file
1:the
some-text-file
1:the
the
yet-another-text-file
1:this
thoroughly
As a bonus, you can use the --output flag to do this for more complex searches with just about the easiest syntax I've found:
# echo "bug: 1, id: 5, time: 12/27/2010" > test-file
# ack-grep -ho "bug: (\d*), id: (\d*), time: (.*)" --output '$1, $2, $3' test-file
1, 5, 12/27/2010
cat *-text-file | grep -Eio "th[a-z]+"
You can also try pcregrep. There is also a -w option in grep, but in some cases it doesn't work as expected.
From Wikipedia:
cat fruitlist.txt
apple
apples
pineapple
apple-
apple-fruit
fruit-apple
grep -w apple fruitlist.txt
apple
apple-
apple-fruit
fruit-apple
I had a similar problem, looking for grep/pattern regex and the "matched pattern found" as output.
At the end I used egrep (same regex on grep -e or -G didn't give me the same result of egrep) with the option -o
so, I think that could be something similar to (I'm NOT a regex Master) :
egrep -o "the*|this{1}|thoroughly{1}" filename
To search all the words with start with "icon-" the following command works perfect. I am using Ack here which is similar to grep but with better options and nice formatting.
ack -oh --type=html "\w*icon-\w*" | sort | uniq
You could pipe your grep output into Perl like this:
grep "th" * | perl -n -e'while(/(\w*th\w*)/g) {print "$1\n"}'
grep --color -o -E "Begin.{0,}?End" file.txt
? - Match as few as possible until the End
Tested on macos terminal
$ grep -w
Excerpt from grep man page:
-w: Select only those lines containing matches that form whole words. The test is that the matching substring must either be at the beginning of the line, or preceded by a non-word constituent character.
ripgrep
Here are the example using ripgrep:
rg -o "(\w+)?th(\w+)?"
It'll match all words matching th.
I am trying to find alphanumeric string including these two characters "/+" with at least 30 characters in length.
I have written this code,
grep "[a-zA-Z0-9\/\+]{30,}" tmp.txt
cat tmp.txt
> array('rWmyiJgKT8sFXCmMr639U4nWxcSvVFEur9hNOOvQwF/tpYRqTk9yWV2xPFBAZwAPRVs/s
ddd73ZEjfy+airfy8DtqIqKI9+dd 6hdd7soJ9iG0sGs/ld5f2GHzockoYHfh
+pAzx/t17Crf0T/2+8+reo+MU39lqCr02sAkcC1k/LzyBvSDEtu9N/9NHicr jA3SvDqg5s44DFlaNZ/8BW37fGEf2rk13S/q68OVVyzac7IT7yE7PIL9XZ/6LsmrY
KEsAmN4i/+ym8be3wwn KWGYaIB908+7W98pI6qao3iaZB
3mh7Y/nZm52hyLa37978f+PyOCqUh0Wfx2PL3vglofi0l
QVrOM1pg+mFLEIC88B706UzL4Pss7ouEo+EsrES+/qJq9Y1e/UGvwefOWSL2TJdt
this does not work, Mainly I wanted to have minimum length of the string to be 30
In the syntax of grep, the repetition braces need to be backslashed.
grep -o '[a-zA-Z0-9/+]\{30,\}' file
If you want to constrain the match to lines containing only matches to this pattern, add line-start and line-ending anchors:
grep '^[a-zA-Z0-9/+]\{30,\}$' file
The -o option in the first command line causes grep to only print the matching part, not the entire matching line.
The repetition operator is not directly supported in Basic Regular Expression syntax. Use grep -E to enable Extended Regular Expression syntax, or backslash the braces.
You can use
grep -e "^[a-zA-Z0-9/+]\{30,\}" tmp.txt
grep -e "^[a-zA-Z0-9/+]\{30,\}" tmp.txt
+pAzx/t17Crf0T/2+8+reo+MU39lqCr02sAkcC1k/LzyBvSDEtu9N/9NHicr jA3SvDqg5s44DFlaNZ/8BW37fGEf2rk13S/q68OVVyzac7IT7yE7PIL9XZ/6LsmrY
3mh7Y/nZm52hyLa37978f+PyOCqUh0Wfx2PL3vglofi0l
QVrOM1pg+mFLEIC88B706UzL4Pss7ouEo+EsrES+/qJq9Y1e/UGvwefOWSL2TJdt
man grep
Read up about the difference between between regular and extended patterns. You need the -E option.
This question already has answers here:
Colorized grep -- viewing the entire file with highlighted matches
(24 answers)
Closed 7 years ago.
When using grep, it will highlight any text in a line with a match to your regular expression.
What if I want this behaviour, but have grep print out all lines as well? I came up empty after a quick look through the grep man page.
Use ack. Checkout its --passthru option here: ack. It has the added benefit of allowing full perl regular expressions.
$ ack --passthru 'pattern1' file_name
$ command_here | ack --passthru 'pattern1'
You can also do it using grep like this:
$ grep --color -E '^|pattern1|pattern2' file_name
$ command_here | grep --color -E '^|pattern1|pattern2'
This will match all lines and highlight the patterns. The ^ matches every start of line, but won't get printed/highlighted since it's not a character.
(Note that most of the setups will use --color by default. You may not need that flag).
You can make sure that all lines match but there is nothing to highlight on irrelevant matches
egrep --color 'apple|' test.txt
Notes:
egrep may be spelled also grep -E
--color is usually default in most distributions
some variants of grep will "optimize" the empty match, so you might want to use "apple|$" instead (see: https://stackoverflow.com/a/13979036/939457)
EDIT:
This works with OS X Mountain Lion's grep:
grep --color -E 'pattern1|pattern2|$'
This is better than '^|pattern1|pattern2' because the ^ part of the alternation matches at the beginning of the line whereas the $ matches at the end of the line. Some regular expression engines won't highlight pattern1 or pattern2 because ^ already matched and the engine is eager.
Something similar happens for 'pattern1|pattern2|' because the regex engine notices the empty alternation at the end of the pattern string matches the beginning of the subject string.
[1]: http://www.regular-expressions.info/engine.html
FIRST EDIT:
I ended up using perl:
perl -pe 's:pattern:\033[31;1m$&\033[30;0m:g'
This assumes you have an ANSI-compatible terminal.
ORIGINAL ANSWER:
If you're stuck with a strange grep, this might work:
grep -E --color=always -A500 -B500 'pattern1|pattern2' | grep -v '^--'
Adjust the numbers to get all the lines you want.
The second grep just removes extraneous -- lines inserted by the BSD-style grep on Mac OS X Mountain Lion, even when the context of consecutive matches overlap.
I thought GNU grep omitted the -- lines when context overlaps, but it's been awhile so maybe I remember wrong.
You can use my highlight script from https://github.com/kepkin/dev-shell-essentials
It's better than grep cause you can highlight each match with it's own color.
$ command_here | highlight green "input" | highlight red "output"
Since you want matches highlighted, this is probably for human consumption (as opposed to piping to another program for instance), so a nice solution would be to use:
less -p <your-pattern> <your-file>
And if you don't care about case sensitivity:
less -i -p <your-pattern> <your-file>
This also has the advantage of having pages, which is nice when having to go through a long output
You can do it using only grep by:
reading the file line by line
matching a pattern in each line and highlighting pattern by grep
if there is no match, echo the line as is
which gives you the following:
while read line ; do (echo $line | grep PATTERN) || echo $line ; done < inputfile
If you want to print "all" lines, there is a simple working solution:
grep "test" -A 9999999 -B 9999999
A => After
B => Before
If you are doing this because you want more context in your search, you can do this:
cat BIG_FILE.txt | less
Doing a search in less should highlight your search terms.
Or pipe the output to your favorite editor. One example:
cat BIG_FILE.txt | vim -
Then search/highlight/replace.
If you are looking for a pattern in a directory recursively, you can either first save it to file.
ls -1R ./ | list-of-files.txt
And then grep that, or pipe it to the grep search
ls -1R | grep --color -rE '[A-Z]|'
This will look of listing all files, but colour the ones with uppercase letters. If you remove the last | you will only see the matches.
I use this to find images named badly with upper case for example, but normal grep does not show the path for each file just once per directory so this way I can see context.
Maybe this is an XY problem, and what you are really trying to do is to highlight occurrences of words as they appear in your shell. If so, you may be able to use your terminal emulator for this. For instance, in Konsole, start Find (ctrl+shift+F) and type your word. The word will then be highlighted whenever it occurs in new or existing output until you cancel the function.
HI
I am not very good with linux shell scripting.I am trying following shell script to replace
revision number token $rev -<rev number> in all html files under specified directory
cd /home/myapp/test
set repUpRev = "`svnversion`"
echo $repUpRev
grep -lr -e '\$rev -'.$repUpRev.'\$' *.html | xargs sed -i 's/'\$rev -'.$repUpRev.'\$'/'\$rev -.*$'/g'
This seems not working, what is wrong with the above code ?
rev=$(svnversion)
sed -i.bak "s/$rev/some other string/g" *.html
What is $rev in the regexp string? Is it another variable? Or you're looking for a string '$rev'. If latter - I would suggest adding '\' before $ otherwise it's treated as a special regexp character...
This is how you show the last line:
grep -lr -e '\$rev -'.$repUpRev.'\$' *.html | xargs sed -i 's/'\$rev -'.$repUpRev.'\$'/'\$rev -.*$'/g'
It would help if you showed some input data.
The -r option makes the grep recursive. That means it will operate on files in the directory and its subdirectories. Is that what you intend?
The dots in your grep and sed stand for any character. If you want literal dots, you'll need to escape them.
The final escaped dollar sign in the grep and sed commands will be seen as a literal dollar sign. If you want to anchor to the end of the line you should remove the escape.
The .* works only as a literal string on the right hand side of a sed s command. If you want to include what was matched on the left side, you need to use capture groups. The g modifier on the s command is only needed if the pattern appears more than once in a line.
Using quote, unquote, quote, unquote is hard to read. Use double quotes to permit variable expansion.
Try your grep command by itself without the xargs and sed to see if it's producing a list of files.
This may be closer to what you want:
grep -lr -e "\$rev -.$repUpRev.$" *.html | xargs sed -i "s/\$rev -.$repUpRev.$/\$rev -REPLACEMENT_TEXT/g"
but you'll still need to determine if the g modifier, the dots, the final dollar signs, etc., are what you intend.
I have long file with the following list:
/drivers/isdn/hardware/eicon/message.c//add_b1()
/drivers/media/video/saa7134/saa7134-dvb.c//dvb_init()
/sound/pci/ac97/ac97_codec.c//snd_ac97_mixer_build()
/drivers/s390/char/tape_34xx.c//tape_34xx_unit_check()
(PROBLEM)/drivers/video/sis/init301.c//SiS_GetCRT2Data301()
/drivers/scsi/sg.c//sg_ioctl()
/fs/ntfs/file.c//ntfs_prepare_pages_for_non_resident_write()
/drivers/net/tg3.c//tg3_reset_hw()
/arch/cris/arch-v32/drivers/cryptocop.c//cryptocop_setup_dma_list()
/drivers/media/video/pvrusb2/pvrusb2-v4l2.c//pvr2_v4l2_do_ioctl()
/drivers/video/aty/atyfb_base.c//aty_init()
/block/compat_ioctl.c//compat_blkdev_driver_ioctl()
....
It contains all the functions in the kernel code. The notation is file//function.
I want to copy some 100 files from the kernel directory to another directory, so I want to strip every line from the function name, leaving just the filename.
It's super-easy in python, any idea how to write a 1-liner in the bash prompt that does the trick?
Thanks,
Udi
cat "func_list" | sed "s#//.*##" > "file_list"
Didn't run it :)
You can use pure Bash:
while read -r line; do echo "${line%//*}"; done < funclist.txt
Edit:
The syntax of the echo command is doing the same thing as the sed command in Eugene's answer: deleting the "//" and everything that comes after.
Broken down:
"echo ${line}" is the same as "echo $line"
the "%" deletes the pattern that follows it if it matches the trailing portion of the parameter
"%" makes the shortest possible match, "%%" makes the longest possible
"//*" is the pattern to match, "*" is similar to sed's ".*"
See the Parameter Expansion section of the Bash man page for more information, including:
using ${parameter#word} for matching the beginning of a parameter
${parameter/pattern/string} to do sed-style replacements
${parameter:offset:length} to retrieve substrings
etc.
here's a one liner in (g)awk
awk -F"//" '{print $1}' file
Here's one using cut and rev
cat file | rev | cut -d'/' -f2-| rev