sed one-liner to convert all uppercase to lowercase? - regex

I have a textfile in which some words are printed in ALL CAPS. I want to be able to just convert everything in the textfile to lowercase, using sed. That means that the first sentence would then read, 'i have a textfile in which some words are printed in all caps.'

With tr:
# Converts upper to lower case
$ tr '[:upper:]' '[:lower:]' < input.txt > output.txt
# Converts lower to upper case
$ tr '[:lower:]' '[:upper:]' < input.txt > output.txt
Or, sed on GNU (but not BSD or Mac as they don't support \L or \U):
# Converts upper to lower case
$ sed -e 's/\(.*\)/\L\1/' input.txt > output.txt
# Converts lower to upper case
$ sed -e 's/\(.*\)/\U\1/' input.txt > output.txt

If you have GNU extensions, you can use sed's \L (lower entire match, or until \L [lower] or \E [end - toggle casing off] is reached), like so:
sed 's/.*/\L&/' <input >output
Note: '&' means the full match pattern.
As a side note, GNU extensions include \U (upper), \u (upper next character of match), \l (lower next character of match). For example, if you wanted to camelcase a sentence:
$ sed -E 's/\w+/\u&/g' <<< "Now is the time for all good men..." # Camel Case
Now Is The Time For All Good Men...
Note: Since the assumption is we have GNU extensions, we can use sequences such as \w (match a word character) and the -E (extended regex) option, which relieves you of having to escape the one-or-more quantifier (+) and certain other special regex characters.

You also can do this very easily with awk, if you're willing to consider a different tool:
echo "UPPER" | awk '{print tolower($0)}'

Here are many solutions :
To upercaser with perl, tr, sed and awk
perl -ne 'print uc'
perl -npe '$_=uc'
perl -npe 'tr/[a-z]/[A-Z]/'
perl -npe 'tr/a-z/A-Z/'
tr '[a-z]' '[A-Z]'
sed y/abcdefghijklmnopqrstuvwxyz/ABCDEFGHIJKLMNOPQRSTUVWXYZ/
sed 's/\([a-z]\)/\U\1/g'
sed 's/.*/\U&/'
awk '{print toupper($0)}'
To lowercase with perl, tr, sed and awk
perl -ne 'print lc'
perl -npe '$_=lc'
perl -npe 'tr/[A-Z]/[a-z]/'
perl -npe 'tr/A-Z/a-z/'
tr '[A-Z]' '[a-z]'
sed y/ABCDEFGHIJKLMNOPQRSTUVWXYZ/abcdefghijklmnopqrstuvwxyz/
sed 's/\([A-Z]\)/\L\1/g'
sed 's/.*/\L&/'
awk '{print tolower($0)}'
Complicated bash to lowercase :
while read v;do v=${v//A/a};v=${v//B/b};v=${v//C/c};v=${v//D/d};v=${v//E/e};v=${v//F/f};v=${v//G/g};v=${v//H/h};v=${v//I/i};v=${v//J/j};v=${v//K/k};v=${v//L/l};v=${v//M/m};v=${v//N/n};v=${v//O/o};v=${v//P/p};v=${v//Q/q};v=${v//R/r};v=${v//S/s};v=${v//T/t};v=${v//U/u};v=${v//V/v};v=${v//W/w};v=${v//X/x};v=${v//Y/y};v=${v//Z/z};echo "$v";done
Complicated bash to uppercase :
while read v;do v=${v//a/A};v=${v//b/B};v=${v//c/C};v=${v//d/D};v=${v//e/E};v=${v//f/F};v=${v//g/G};v=${v//h/H};v=${v//i/I};v=${v//j/J};v=${v//k/K};v=${v//l/L};v=${v//m/M};v=${v//n/N};v=${v//o/O};v=${v//p/P};v=${v//q/Q};v=${v//r/R};v=${v//s/S};v=${v//t/T};v=${v//u/U};v=${v//v/V};v=${v//w/W};v=${v//x/X};v=${v//y/Y};v=${v//z/Z};echo "$v";done
Simple bash to lowercase :
while read v;do echo "${v,,}"; done
Simple bash to uppercase :
while read v;do echo "${v^^}"; done
Note that ${v,} and ${v^} only change the first letter.
You should use it that way :
(while read v;do echo "${v,,}"; done) < input_file.txt > output_file.txt

I like some of the answers here, but there is a sed command that should do the trick on any platform:
sed 'y/ABCDEFGHIJKLMNOPQRSTUVWXYZ/abcdefghijklmnopqrstuvwxyz/'
Anyway, it's easy to understand. And knowing about the y command can come in handy sometimes.

If you have GNU sed (likely on Linux, but not on *BSD or macOS):
echo "Hello MY name is SUJIT " | sed 's/./\L&/g'
Output:
hello my name is sujit

If you are using posix sed
Selection for any case for a pattern (converting the searched pattern with this sed than use the converted pattern in you wanted command using regex:
echo "${MyOrgPattern} | sed "s/[aA]/[aA]/g;s/[bB]/[bB]/g;s/[cC]/[cC]/g;s/[dD]/[dD]/g;s/[eE]/[eE]/g;s/[fF]/[fF]/g;s/[gG]/[gG]/g;s/[hH]/[hH]/g;s/[iI]/[iI]/g;s/[jJ]/[jJ]/g;s/[kK]/[kK]/g;s/[lL]/[lL]/g;s/[mM]/[mM]/g;s/[nN]/[nN]/g;s/[oO]/[oO]/g;s/[pP]/[pP]/g;s/[qQ]/[qQ]/g;s/[rR]/[rR]/g;s/[sS]/[sS]/g;s/[tT]/[tT]/g;s/[uU]/[uU]/g;s/[vV]/[vV]/g;s/[wW]/[wW]/g;s/[xX]/[xX]/g;s/[yY]/[yY]/g;s/[zZ]/[zZ]/g" | read -c MyNewPattern
YourInputStreamCommand | egrep "${MyNewPattern}"
convert in lower case
sed "s/[aA]/a/g;s/[bB]/b/g;s/[cC]/c/g;s/[dD]/d/g;s/[eE]/e/g;s/[fF]/f/g;s/[gG]/g/g;s/[hH]/h/g;s/[iI]/i/g;s/j/[jJ]/g;s/[kK]/k/g;s/[lL]/l/g;s/[mM]/m/g;s/[nN]/n/g;s/[oO]/o/g;s/[pP]/p/g;s/[qQ]/q/g;s/[rR]/r/g;s/[sS]/s/g;s/[tT]/t/g;s/[uU]/u/g;s/[vV]/v/g;s/[wW]/w/g;s/[xX]/x/g;s/[yY]/y/g;s/[zZ]/z/g"
same for uppercase replace lower letter between // by upper equivalent in the sed
Have fun

short, sweet and you don't even need redirection :-)
perl -p -i -e 'tr/A-Z/a-z/' file

Instead of typing this long expression:
sed 'y/ABCDEFGHIJKLMNOPQRSTUVWXYZ/abcdefghijklmnopqrstuvwxyz/' input
One could use this:
sed 'y/'$(printf "%s" {A..Z} "/" {a..z} )'/' input

Related

Get substring using either perl or sed

I can't seem to get a substring correctly.
declare BRANCH_NAME="bugfix/US3280841-something-duh";
# Trim it down to "US3280841"
TRIMMED=$(echo $BRANCH_NAME | sed -e 's/\(^.*\)\/[a-z0-9]\|[A-Z0-9]\+/\1/g')
That still returns bugfix/US3280841-something-duh.
If I try an use perl instead:
declare BRANCH_NAME="bugfix/US3280841-something-duh";
# Trim it down to "US3280841"
TRIMMED=$(echo $BRANCH_NAME | perl -nle 'm/^.*\/([a-z0-9]|[A-Z0-9])+/; print $1');
That outputs nothing.
What am I doing wrong?
Using bash parameter expansion only:
$: # don't use caps; see below.
$: declare branch="bugfix/US3280841-something-duh"
$: tmp="${branch##*/}"
$: echo "$tmp"
US3280841-something-duh
$: trimmed="${tmp%%-*}"
$: echo "$trimmed"
US3280841
Which means:
$: tmp="${branch_name##*/}"
$: trimmed="${tmp%%-*}"
does the job in two steps without spawning extra processes.
In sed,
$: sed -E 's#^.*/([^/-]+)-.*$#\1#' <<< "$branch"
This says "after any or no characters followed by a slash, remember one or more that are not slashes or dashes, followed by a not-remembered dash and then any or no characters, then replace the whole input with the remembered part."
Your original pattern was
's/\(^.*\)\/[a-z0-9]\|[A-Z0-9]\+/\1/g'
This says "remember any number of anything followed by a slash, then a lowercase letter or a digit, then a pipe character (because those only work with -E), then a capital letter or digit, then a literal plus sign, and then replace it all with what you remembered."
GNU's manual is your friend. I look stuff up all the time to make sure I'm doing it right. Sometimes it still takes me a few tries, lol.
An aside - try not to use all-capital variable names. That is a convention that indicates it's special to the OS, like RANDOM or IFS.
You may use this sed:
sed -E 's~^.*/|-.*$~~g' <<< "$BRANCH_NAME"
US3280841
Ot this awk:
awk -F '[/-]' '{print $2}' <<< "$BRANCH_NAME"
US3280841
sed 's:[^/]*/\([^-]*\)-.*:\1:'<<<"bugfix/US3280841-something-duh"
Perl version just has + in wrong place. It should be inside the capture brackets:
TRIMMED=$(echo $BRANCH_NAME | perl -nle 'm/^.*\/([a-z0-9A-Z]+)/; print $1');
Just use a ^ before A-Z0-9
TRIMMED=$(echo $BRANCH_NAME | sed -e 's/\(^.*\)\/[a-z0-9]\|[^A-Z0-9]\+/\1/g')
in your sed case.
Alternatively and briefly, you can use
TRIMMED=$(echo $BRANCH_NAME | sed "s/[a-z\/\-]//g" )
too.
type on shell terminal
$ BRANCH_NAME="bugfix/US3280841-something-duh"
$ echo $BRANCH_NAME| perl -pe 's/.*\/(\w\w[0-9]+).+/\1/'
use s (substitute) command instead of m (match)
perl is a superset of sed so it'd be identical 'sed -E' instead of 'perl -pe'
Another variant using Perl Regular Expression Character Classes (see perldoc perlrecharclass).
echo $BRANCH_NAME | perl -nE 'say m/^.*\/([[:alnum:]]+)/;'

How to toggle cases with sed command?

I want to replace kAbcdedf with abcdef in a file. I try to use the sed command. Some suggestions said, if I input this command
echo 'abc'|sed 's/^../\u&/'
the result will be
Abc
but on macOS (using zsh), the result is
uabc
Does anyone know the correct way to toggle cases with the sed command?
How can I search kAbcd and then replace that string with abcd, i.e., remove k and uppercase the next letter after k?
$ # with GNU sed
$ echo 'kAbcd' | sed -E 's/k(.)/\l\1/'
abcd
$ # with perl if GNU sed is not available
$ echo 'kAbcd' | perl -pe 's/k(.)/\l$1/'
abcd
\l will lowercase the first character of given argument
use g modifier to replace all such occurrences in a line
As mentioned by Wiktor Stribiżew in comments, see also: How to use GNU sed on Mac OS X

Extract version using grep/regex in bash

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}'

sed - exchange words with delimiter

I'm trying swap words around with sed, not replace because that's what I keep finding on Google search.
I don't know if it's the regex that I'm getting wrong. I did a search for everything before a char and everything after a char, so that's how I got the regex.
echo xxx,aaa | sed -r 's/[^,]*/[^,]*$/'
or
echo xxx/aaa | sed -r 's/[^\/]*/[^\/]*$/'
I am getting this output:
[^,]*$,aaa
or this:
[^,/]*$/aaa
What am I doing wrong?
For the first sample, you should use:
echo xxx,aaa | sed 's/\([^,]*\),\([^,]*\)/\2,\1/'
For the second sample, simply use a character other than slash as the delimiter:
echo xxx/aaa | sed 's%\([^/]*\)/\([^/]*\)%\2/\1%'
You can also use \{1,\} to formally require one or more:
echo xxx,aaa | sed 's/\([^,]\{1,\}\),\([^,]\{1,\}\)/\2,\1/'
echo xxx/aaa | sed 's%\([^/]\{1,\}\)/\([^/]\{1,\}\)%\2/\1%'
This uses the most portable sed notation; it should work anywhere. With modern versions that support extended regular expressions (-r with GNU sed, -E with Mac OS X or BSD sed), you can lose some of the backslashes and use + in place of * which is more precisely what you're after (and parallels \{1,\} much more succinctly):
echo xxx,aaa | sed -E 's/([^,]+),([^,]+)/\2,\1/'
echo xxx/aaa | sed -E 's%([^/]+)/([^/]+)%\2/\1%'
With sed it would be:
sed 's#\([[:alpha:]]\+\)/\([[:alpha:]]\+\)#\2,\1#' <<< 'xxx/aaa'
which is simpler to read if you use extended posix regexes with -r:
sed -r 's#([[:alpha:]]+)/([[:alpha:]]+)#\2/\1#' <<< 'xxx/aaa'
I'm using two sub patterns ([[:alpha:]]+) which can contain one or more letters and are separated by a /. In the replacement part I reassemble them in reverse order \2/\1. Please also note that I'm using # instead of / as the delimiter for the s command since / is already the field delimiter in the input data. This saves us to escape the / in the regex.
Btw, you can also use awk for that, which is pretty easy to read:
awk -F'/' '{print $2,$1}' OFS='/' <<< 'xxx/aaa'

Getting defined substring with help of sed or egrep

Everyone!!
I want to get specific substring from stdout of command.
stdout:
{"response":
{"id":"110200dev1","success":"true","token":"09ad7cc7da1db13334281b84f2a8fa54"},"success":"true"}
I need to get a hex string after token without quotation marks, the length of hex string is 32 letters.I suppose it can be done by sed or egrep. I don't want to use awk here. Because the stdout is being changed very often.
This is an alternate gnu-awk solution when grep -P isn't available:
awk -F: '{gsub(/"/, "")} NF==2&&$1=="token"{print $2}' RS='[{},]' <<< "$string"
09ad7cc7da1db13334281b84f2a8fa54
grep's nature is extracting things:
grep -Po '"token":"\K[^"]+'
-P option interprets the pattern as a Perl regular expression.
-o option shows only the matching part that matches the pattern.
\K throws away everything that it has matched up to that point.
Or an option using sed...
sed 's/.*"token":"\([^"]*\)".*/\1/'
With sed:
your-command | sed 's/.*"token":"\([^"]*\)".*/\1/'
YourStreamOrFile | sed -n 's/.*"token":"\([a-f0-9]\{32\}\)".*/\1/p'
doesn not return a full string if not corresponding