I'm trying to figure out how to grep for lines that are made up of A-Z and a-z exclusively, that is, the "American" alphabet of letters. I would expect this to work, but it does not:
$ echo -e "Jutland\nJastrząb" | grep -x '[A-Za-z]*'
Jutland
Jastrząb
I want this to only print "Jutland", because ą is not a letter in the American alphabet. How can I achieve this?
You need to add LC_ALL=C before grep:
printf '%b\n' "Jutland\nJastrząb" | LC_ALL=C grep -x '[A-Za-z]*'
Jutland
You may also use -i switch to ignore case and reduce regex:
printf '%b\n' "Jutland\nJastrząb" | LC_ALL=C grep -ix '[a-z]*'
LC_ALL=C avoids locale-dependent effects otherwise your current LOCALE treats ą as [a-zA-Z].
You can use perl regex:
$ echo -e "Jutland\nJastrząb" | grep -P '^[[:ascii:]]+$'
Jutland
It's experimental though:
-P, --perl-regexp
Interpret the pattern as a Perl-compatible regular expression (PCRE). This is experimental and
grep -P may warn of unimplemented features.
EDIT
For letters only, use [A-Za-z]:
$ echo -e "L'Egyptienne\nJutland\nJastrząb" | grep -P '^[A-Za-z]+$'
Jutland
Related
I have a file that has a line stating
version = "12.0.08-SNAPSHOT"
The word version and quoted strings can occur on multiple lines in that file.
I am looking for a single line bash statement that can output the following string:
12.0.08-SNAPSHOT
The version can have RELEASE tag too instead of SNAPSHOT.
So to summarize, given
version = "12.0.08-SNAPSHOT"
expected output: 12.0.08-SNAPSHOT
And given
version = "12.0.08-RELEASE"
expected output: 12.0.08-RELEASE
The following command prints strings enquoted in version = "...":
grep -Po '\bversion\s*=\s*"\K.*?(?=")' yourFile
-P enables perl regexes, which allow us to use features like \K and so on.
-o only prints matched parts instead of the whole lines.
\b ensures that version starts at a word boundary and we do not match things like abcversion.
\s stands for any kind of whitespace.
\K lets grep forget, that it matched the part before \K. The forgotten part will not be printed.
.*? matches as few chararacters as possible (the matching part will be printed) ...
(?=") ... until we see a ", which won't be included in the match either (this is called a lookahead).
Not all grep implementations support the -P option. Alternatively, you can use perl, as described in this answer:
perl -nle 'print $& if m{\bversion\s*=\s*"\K.*?(?=")}' yourFile
Seems like a job for cut:
$ echo 'version = "12.0.08-SNAPSHOT"' | cut -d'"' -f2
12.0.08-SNAPSHOT
$ echo 'version = "12.0.08-RELEASE"' | cut -d'"' -f2
12.0.08-RELEASE
Portable solution:
$ echo 'version = "12.0.08-RELEASE"' |sed -E 's/.*"(.*)"/\1/g'
12.0.08-RELEASE
or even:
$ perl -pe 's/.*"(.*)"/\1/g'.
$ awk -F"\"" '{print $2}'
I have an output like Johny-Smith, Juarez-Hugo, etc. and I need instead S, H, etc. Basically, I need the last uppercase letter in a string and that's it. If this is possible in any built in Linux tools (ex awk, sed, grep, etc.) it would be greatly appreciated.
Do you need like this ?
echo "Johny-Smith" | sed 's/^.*\([A-Z]\)[^A-Z]*$/\1/g'
Test:
$ echo "Johny-Smith-Hello Johny-Smith" | sed 's/.*\([A-Z]\)[^A-Z]*/\1/g'
S
With GNU grep and if PCRE option is available
$ echo 'Johny-Smith' | grep -oP '.*\K[A-Z]'
S
$ echo 'Juarez-Hugo' | grep -oP '.*\K[A-Z]'
H
-o prints only matched portion
-P Perl regular expression
.*\K positive lookbehind, not part of output
[A-Z] any uppercase character
with perl, see perldoc for command line options explanation
$ # prints the string within captured group
$ echo 'Johny-Smith' | perl -lne 'print /.*([A-Z])/'
S
$ echo 'Juarez-Hugo' | perl -lne 'print /.*([A-Z])/'
H
In Bash:
$ var="Johny-Smith-Hello Johny-Smith"; var="${var//[^[:upper:]]/}";echo "${var: -1}"
S
${var//[^[:upper:]]/} remove all non-upper case letter chars
echo ${var: -1} output the last one
I'm trying to get a pattern over multiple lines. I would like to ensure the line I'm looking for ends in \r\n and that there is specific text that comes after it at some point. The two problems I've had are I often get unmatched parenthesis in groupings or I get a positive match when there is none. Here are two simple examples.
echo -e -n "ab\r\ncd" | grep -U -c -z -E $'(\r\n)+.*TEST'
grep: Unmatched ( or \(
What exactly is unmatched there? I don't get it.
echo -e -n "ab\r\ncd" | grep -U -c -z -E $'\r\n.*TEST'
1
There is no TEST in the string, so why does this return a count of 1 for matches?
I'm using grep (GNU grep) 2.16 on Ubuntu 14. Thanks
Instead of -E you can use -P for PCRE support in gnu grep to use advanced regex like this:
echo -ne "ab\r\ncd" | ggrep -UczP '\r\n.*TEST'
0
echo -ne "ab\r\ncd" | ggrep -UczP '\r\n.*cd'
1
grep -E matches only in single line input.
I'm writing a bash script which analyses a html file and
I want to get the content of each single <tr>...</tr>. So my command looks like:
$ tr -d \\012 < price.html | grep -oE '<tr>.*?</tr>'
But it seems that grep gives me the result of:
$ tr -d \\012 < price.html | grep -oE '<tr>.*</tr>'
How can I make .* non-greedy?
If you have GNU Grep you can use -P to make the match non-greedy:
$ tr -d \\012 < price.html | grep -Po '<tr>.*?</tr>'
The -P option enables Perl Compliant Regular Expression (PCRE) which is needed for non-greedy matching with ? as Basic Regular Expression (BRE) and Extended Regular Expression (ERE) do not support it.
If you are using -P you could also use look arounds to avoid printing the tags in the match like so:
$ tr -d \\012 < price.html | grep -Po '(?<=<tr>).*?(?=</tr>)'
If you don't have GNU grep and the HTML is well formed you could just do:
$ tr -d \\012 < price.html | grep -o '<tr>[^<]*</tr>'
Note: The above example won't work with nested tags within <tr>.
Non-greedy matching is not part of the Extended Regular Expression syntax supported by grep -E. Use grep -P instead if you have that, or switch to Perl / Python / Ruby / what have you. (Oh, and pcregrep.)
Of course, if you really mean
<tr>[^<>]*</tr>
you should say that instead; then plain old grep will work fine.
You could (tediously) extend the regex to accept nested tags which are not <tr> but of course, it's better to use a proper HTML parser than spend a lot of time rediscovering why regular expressions are not the right tool for this.
.*? is a Perl regular expression. Change your grep to
grep -oP '<tr>.*?</tr>'
Try perl-style-regexp
$ grep -Po '<tr>.*?</tr>' input
<tr>stuff</tr>
<tr>more stuff</tr>
I have following string:
{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}
and I need to get value of "scheme version", which is 1234 in this example.
I have tried
grep -Eo "\"scheme_version\":(\w*)"
however it returns
"scheme_version":1234
How can I make it? I know I can add sed call, but I would prefer to do it with single grep.
You'll need to use a look behind assertion so that it isn't included in the match:
grep -Po '(?<=scheme_version":)[0-9]+'
This might work for you:
echo '{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}' |
sed -n 's/.*"scheme_version":\([^}]*\)}/\1/p'
1234
Sorry it's not grep, so disregard this solution if you like.
Or stick with grep and add:
grep -Eo "\"scheme_version\":(\w*)"| cut -d: -f2
I would recommend that you use jq for the job. jq is a command-line JSON processor.
$ cat tmp
{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}
$ cat tmp | jq .scheme_version
1234
As an alternative to the positive lookbehind method suggested by SiegeX, you can reset the match starting point to directly after scheme_version": with the \K escape sequence. E.g.,
$ grep -Po 'scheme_version":\K[0-9]+'
This restarts the matching process after having matched scheme_version":, and tends to have far better performance than the positive lookbehind. Comparing the two on regexp101 demonstrates that the reset match start method takes 37 steps and 1ms, while the positive lookbehind method takes 194 steps and 21ms.
You can compare the performance yourself on regex101 and you can read more about resetting the match starting point in the PCRE documentation.
To avoid using greps PCRE feature which is available in GNU grep, but not in BSD version, another method is to use ripgrep, e.g.
$ rg -o 'scheme_version.?:(\d+)' -r '$1' <file.json
1234
-r Capture group indices (e.g., $5) and names (e.g., $foo).
Another example with Python and json.tool module which can validate and pretty-print:
$ python -mjson.tool file.json | rg -o 'scheme_version[^\d]+(\d+)' -r '$1'
1234
Related: Can grep output only specified groupings that match?
You can do this:
$ echo '{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}' | awk -F ':' '{print $4}' | tr -d '}'
Improving #potong's answer that works only to get "scheme_version", you can use this expression :
$ echo '{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}' | sed -n 's/.*"_id":["]*\([^(",})]*\)[",}].*/\1/p'
scheme_version
$ echo '{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}' | sed -n 's/.*"_rev":["]*\([^(",})]*\)[",}].*/\1/p'
4-cad1842a7646b4497066e09c3788e724
$ echo '{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}' | sed -n 's/.*"scheme_version":["]*\([^(",})]*\)[",}].*/\1/p'
1234