I tried the command below on terminal
echo "aa" | sed -r 's/o*/_/g'
The result was like this
_a_a_
What I expected was
__
When the first a is read, it means that there's zero o so that a would be replaced by _
If I use
echo "aoooa" | sed -r 's/o+/_/g'
The result is as what I expected
a_a
But when I use
echo "aoooa" | sed -r /s/o*/_/g
The result is
_a_a_
I think it should be _a_a and without the last _
I totally had no idea why there were 3 underscores. Could anyone tell me the running process of this specific case?
[From my comment:] o* does not match the "a", it matches the zero-length string before the "a" (and the zero-length string between the "a"s, and the zero-length string after the second "a"). The "a" marks the end of the match, but it is not part of the match, and hence does not get replaced.
To make this clearer, consider what happens when there are runs of "o"s (and I used capital "A"s to make them more visible):
$ echo "ooooAooooAoooo" | sed 's/o*/_/g'
_A_A_
...each group of "o"s gets replaced by a single "_"; the "A"s end the first two groups, but they aren't part of the groups (they're between the groups), so they get left alone.
If there's just one "o" in a group, the same thing happens:
$ echo "oAoAo" | sed 's/o*/_/g'
_A_A_
...and finally, with no "o"s (i.e. with zero-length groups of "o"s):
$ echo "AA" | sed 's/o*/_/g'
_A_A_
...again, the "A"s aren't part of the matches, so they don't get replaced.
If the output desired is anything starting with o, with underscore on letters that are not o, you could use:
echo "abcdefhijklmnopqrstuvwxyzoo" | sed -r 's/[^o]/_/g'
Output: _____________o___________oo
Related
"$" should not be immediately followed by digits [0-9]. It should only show the
output- "$" which is immediately followed by aphabet/alphanumeric/alphacharacter.
Input: dirname $0/../bin/$12JAVA_INV/$FILE12NAME
Output: $FILE12NAME
grep -o '[$][a-zA-z_]*'
Using this I'm receiving an output as: $ $ $FILENAME
You're getting $ in the result because * means to match zero or more of the preceding pattern. $0 matches because it has a $ followed by 0 letters.
If you want at least 1 letter, use + instead, it means one or more.
But if you want to be able to match $FILE12NAME, you also need to allow digits after the first character. So use:
grep -i -o '\$[a-z_][a-z_0-9]*'
This matches $, followed by a letter or underscore, followed by zero or more letters, underscores, or numbers.
It looks like you want:
$ echo 'dirname $0/../bin/$12JAVA_INV/$FILE12NAME' | awk '{print $NF}' FS=/
$FILE12NAME
But if you really want to parse it the way you describe, you could do either of:
$ echo 'dirname $0/../bin/$12JAVA_INV/$FILE12NAME' | sed -e 's/.*\(\$[^0-9]\)/\1/'
$FILE12NAME
$ echo 'dirname $0/../bin/$12JAVA_INV/$FILE12NAME' | sed -E 's/.*(\$[^0-9])/\1/'
$FILE12NAME
I'm trying to convert a string that has either - (hyphen) or _ (underscore) to Capital_Case string.
#!/usr/bin/env sh
function cap_case() {
[ $# -eq 1 ] || return 1;
_str=$1;
_capitalize=${_str//[-_]/_} | sed -E 's/(^|_)([a-zA-Z])/\u\2/g'
echo "Capitalize:"
echo $_capitalize
return 0
}
read string
echo $(cap_case $string)
But I don't get anything out.
First I am replacing any occurrence of - and _ with _ ${_str//[-_]/_}, and then I pipe that string to sed which finds the first letter, or _ as the first group, and then the letter after the first group in the second group, and I want to uppercase the found letter with \u\2. I tried with \U\2 but that didn't work as well.
I want the string some_string to become
Some_String
And string some-string to become
Some_String
I'm on a mac, using zsh if that is helpful.
EDIT: More generic solution here to make each field's first letter Capital.
echo "some_string_other" | awk -F"_" '{for(i=1;i<=NF;i++){$i=toupper(substr($i,1,1)) substr($i,2)}} 1' OFS="_"
Following awk may help you.
echo "some_string" | awk -F"_" '{$1=toupper(substr($1,1,1)) substr($1,2);$2=toupper(substr($2,1,1)) substr($2,2)} 1' OFS="_"
Output will be as follows.
echo "some_string" | awk -F"_" '{$1=toupper(substr($1,1,1)) substr($1,2);$2=toupper(substr($2,1,1)) substr($2,2)} 1' OFS="_"
Some_String
This being zsh, you don't need sed (or even a function, really):
$ s=some-string-bar
$ print ${(C)s:gs/-/_}
Some_String_Bar
The (C) flag capitalizes words (where "words" are defined as sequences of alphanumeric characters separated by other characters); :gs/-/_ replaces hyphens with underscores.
If you really want a function, it's cap_case () { print ${(C)1:gs/-/_} }.
pure bash:
#!/bin/bash
camel_case(){
local d display string
declare -a strings # = scope local
[ "$2" ] && d="$2" || d=" " # optional output delimiter
ifs_ini="$IFS"
IFS+='_-' # we keep initial IFS
strings=( "$1" ) # array
for string in ${strings[#]} ; do
display+="${string^}$d"
done
echo "${display%$d}"
IFS="$ifs_ini"
}
camel_case "some-string_here" "_"
camel_case "some-string_here some strings here" "+"
camel_case "some-string_here some strings here"
echo "$BASH_VERSION"
exit
output:
Some_String_Here
Some+String+Here+Some+Strings+Here
Some String Here Some Strings Here
4.4.18(1) release
You can try this gnu sed
echo 'some_other-string' | sed -E 's/(^.)/\u&/;s/[_-](.)/_\u\1/g'
Explains :
s/(^.)/\u&/
(^.) match the first char and \u& put the match in capital letter.
s/[_-](.)/_\u\1/g
[_-](.) capture a char preceded by _ or - and replace it by _ and the matched char in capital letter.
The g at the end tell sed to make the replacement for each char which meet the criteria
You didn't assign to _capitalize - you set a _capitalize environment variable for the empty command that you piped into sed.
You probably meant
_capitalize=$(<<<"${_str//[-_]/_}" sed -E 's/(^|_)([a-zA-Z])/\1\u\2/g')
Note also that ${//} isn't standard shell, so you really ought to specify an interpreter other than sh.
A simpler approach would be simply:
#!/bin/sh
cap_case() {
printf "Capitalize: "
echo "$*" | sed -e 'y/-/_/' -e 's/\(^\|_\)[[:alpha:]]/\U&/g'
}
echo $(cap_case "snake_case")
Note that the \u / \U replacement is a GNU extension to sed - if you're using a non-GNU implementation, check whether it supports this feature.
How can I reverse a four length of letters with sed?
For example:
the year was 1815.
Reverse to:
the raey was 5181.
This is my attempt:
cat filename | sed's/\([a-z]*\) *\([a-z]*\)/\2, \1/'
But it does not work as I intended.
not sure it is possible to do it with GNU sed for all cases. If _ doesn't occur immediately before/after four letter words, you can use
sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
\b is word boundary, word definition being any alphabet or digit or underscore character. So \b will ensure to match only whole words not part of words
$ echo 'the year was 1815.' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
the raey was 5181.
$ echo 'two time five three six good' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
two emit evif three six doog
$ # but won't work if there are underscores around the words
$ echo '_good food' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
_good doof
tool with lookaround support would work for all cases
$ echo '_good food' | perl -pe 's/(?<![a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])(?!=[a-z0-9])/$4$3$2$1/gi'
_doog doof
(?<![a-z0-9]) and (?!=[a-z0-9]) are negative lookbehind and negative lookahead respectively
Can be shortened to
perl -pe 's/(?<![a-z0-9])[a-z0-9]{4}(?!=[a-z0-9])/reverse $&/gie'
which uses the e modifier to place Perl code in substitution section. This form is suitable to easily change length of words to be reversed
Possible shortest sed solution even if a four length of letters contains _s.
sed -r 's/\<(.)(.)(.)(.)\>/\4\3\2\1/g'
Following awk may help you in same. Tested this in GNU awk and only with provided sample Input_file
echo "the year was 1815." |
awk '
function reverse(val){
num=split(val, array,"");
i=array[num]=="."?num-1:num;
for(;i>q;i--){
var=var?var array[i]:array[i]
};
printf (array[num]=="."?var".":var);
var=""
}
{
for(j=1;j<=NF;j++){
printf("%s%s",j==NF||j==2?reverse($j):$j,j==NF?RS:FS)
}}'
This might work for you (GNU sed):
sed -r '/\<\w{4}\>/!b;s//\n&\n/g;s/^[^\n]/\n&/;:a;/\n\n/!s/(.*\n)([^\n])(.*\n)/\2\1\3/;ta;s/^([^\n]*)(.*)\n\n/\2\1/;ta;s/\n//' file
If there are no strings of the length required to reverse, bail out.
Prepend and append newlines to all required strings.
Insert a newline at the start of the pattern space (PS). The PS is divided into two parts, the first line will contain the current word being reversed. The remainder will contain the original line.
Each character of the word to be reversed is inserted at the front of the first line and removed from the original line. When all the characters in the word have been processed, the original word will have gone and only the bordering newlines will exist. These double newlines are then replaced by the word in the first line and the process is repeated until all words have been processed. Finally the newline introduced to separate the working line and the original is removed and the PS is printed.
N.B. This method may be used to reverse strings of varying string length i.e. by changing the first regexp strings of any number can be reversed. Also strings between two lengths may also be reversed e.g. /\<w{2,4}\>/ will change all words between 2 and 4 character length.
It's a recurrent problem so somebody created a bash command called "rev".
echo "$(echo the | rev) $(echo year | rev) $(echo was | rev) $(echo 1815 | rev)".
OR
echo "the year was 1815." | rev | tr ' ' '\n' | tac | tr '\n' ' '
I have a string that is sometimes
xxx.11_222_33_44_555.yyy
and sometimes
xxx.11_222_33_44.yyy
I would like to:
Check if has 4 occourances of _ (figured out how to do it).
If so - remove string's _33 (the 33 string changes, can be any number), so I am left with xxx.11_222_44.yyy.
Using sed :
sed 's/\(_[0-9]*\)_[0-9]*\(_[0-9]*_[0-9]*\)/\1\2/'
It matches the four underscores and replace the whole by the needed parts.
Test run :
$ echo "xxx.11_222_33_44_555.yyy" | sed 's/\(_[0-9]*\)_[0-9]*\(_[0-9]*_[0-9]*\)/\1\2/'
xxx.11_222_44_555.yyy
$ echo "xxx.11_222_33_44.yyy" | sed 's/\(_[0-9]*\)_[0-9]*\(_[0-9]*_[0-9]*\)/\1\2/'
xxx.11_222_33_44.yyy
perhaps something like this
echo "xxx.11_222_33_44.yyy" | sed -e's/\.\([0-9]\+\)_\([0-9]\+\)_\([0-9]\+\)_\([0-9]\+\)\./.\1_\2_\4./'
which checks if there are 4 groups of numbers separated by _ between the two dots and if yes, it leaves out the third group
try this;
echo "xxx.11_222_33_44_555.yyy" | awk -F'_' 'NF>4{print $1"_"$2"_"$4"_"$5};'
Solution using perl and Lookahead and Lookbehind
$ a="xxx.11_222_33_44_555.yyy"
$ perl -pe 's/\.\d+_\d+_\K\d+_(?=\d+_\d+\.)//' <<< "$a"
xxx.11_222_44_555.yyy
I having problems in extracting the word from a line. What i want is that it picks the first word before the symbol # but after the /. Which is the only delimiter that stand out.
A line looks like this:
,["https://picasaweb.google.com/111560558537332305125/Programming#5743548966953176786",1,["https://lh6.googleusercontent.com/-Is8rb8G1sb8/T7UvWtVOTtI/AAAAAAAAG68/Cht3FzfHXNc/s0-d/Geek.jpg",1920,1200]
I want the word Programming.
To get that line i am using this which narrows it down.
sed -n '/.*picasa.*.jpg/p' 5743548866439293105
So i want it to pretty much find # and then go backward until it hit the first /. Then print it out. In this case the word should be Programming but could be anything.
I want it to be as short as possible and have experimented with
sed -n '/.*picasa.*.jpg/p' 5743548866439293105 | awk '$0=$2' FS="/" RS="[$#]"
You can do that with sed (slightly shortened for formatting but works on your original string as well):
pax> echo ',["https://p.g.com/111/Prog#574' | sed 's/^[^#]*\/\([^#]*\)#.*$/\1/'
Prog
pax>
Explaining in more detail:
/---+------------------> greedy capture up to '/'.
/ |
| | /------+---------> capture the stuff between '/' and '#'.
| |/ |
| || | /-+-----> everything from '#' to end of line.
| || |/ |
| || || |
's/^[^#]*\/\([^#]*\)#.*$/\1/'
||
\+---> replace with captured group.
It basically searches for an entire line that has the pattern you want (first # following a /), whilst capturing (with the \( and \) brackets) just the stuff between / and #.
The substitution then replaces the entire line with just that captured text you're interested in (via \1).
Using grep with some Perl regex extensions:
echo $string | grep -P -o "(?<=/)[^/]+(?=#)"
-P tells grep to use Perl extensions. -o tells grep to display only the matched text. To understand what gets matched, break the regex into three parts: (?<=/), [^/]+?, and (?=#). The first part says that the matched text must follow a '/', without including the '/' in the match. The second parts matches a string of non-'/' characters. The last part says that the matched text must be immediately followed by a '#', without including the '#' in the match.
Another grep, using the "\K" feature to "throw away" the match up to the last '/' before the '#':
# Match as much as possible up to a '/', but throw it away, then match as much as you can
# up to the first #
echo $string | grep -oP ".*/\K.+(?=#)"
Using cut and awk to get the first field (splitting on #) followed by the last field (splitting on /):
echo $string | cut -d# -f1 | awk -F/ '{print $NF}'
Using some temporary variables and bash's parameter expansion facilities:
$ FOO=["https://picasaweb.google.com/111560558537332305125/Programming#5743548966953176786",1,["https://lh6.googleusercontent.com/-Is8rb8G1sb8/T7UvWtVOTtI/AAAAAAAAG68/Cht3FzfHXNc/s0-d/Geek.jpg",1920,1200]
$ BAR=${FOO%#*} # Strip the last # and everything after
$ echo $BAR
[https://picasaweb.google.com/111560558537332305125/Programming
$ BAZ=${BAR##*/} # Strip everything up to and including the last /
$ echo $BAZ
Programming
This might work for you:
sed '/.*\/\([^#]*\)#.*/{s//\1/;q};d' file