I have the following printf command that works correctly adding 5 spaces after the string ABC and then prints string "DEF".
printf '%s%*s' "ABC" 5 '' "DEF"
I'd like to add a newline character at the end, after the string DEF, but I don't know how to do it. I've tried without success in these ways:
user /d
$ printf '%s%*s' "ABC" 5 '' "DEF\n"
ABC DEF\n
user /d
$ printf '%s%*s\n' "ABC" 5 '' "DEF"
ABC
DEF
How should be done? thanks in advance
Add another String placeholder for DEF and a new line character:
printf '%s%*s%s\n' "ABC" 5 '' "DEF"
You may use $'\n' to get a newline:
printf '%s%*s' "ABC" 5 '' $'DEF\n'
It didn't work with
printf '%s%*s\n' "ABC" 5 '' "DEF"
because printf format is one short of total arguments. First %s is used for ABC and then %*s is used to output 5 spaces. But there is no format for DEF hence \n is appended to both the arguments.
Related
I have the following variable in Terminal
echo $VAR1
abc def ghi
<- Separated by tab
How can I convert this to
"abc" "def" "ghi"
with single space in-between?
In zsh, you can break your string up into an array of quoted words with
var1=$'abc\tdef\tghi'
words=( "${(qqq)=var1}" )
and then turn it back into a single string if wanted with
var2="${words[*]}"
printf "%s\n" "$var2" # prints "abc" "def" "ghi"
Or to skip the intermediate array if you don't need it:
var2=${(A)${(qqq)=var1}}
Assuming the variable contains actual tab characters (not backslashes followed by "t"), you can replace tabs with " " while expanding the variable (see here, a ways down in the list of modifiers), and also add quotes at the beginning and end, like this:
"\"${VAR1//$'\t'/\" \"}\""
There's a rather complex mix of quoting and escaping modes here. The double-quotes at the very beginning and end make the whole thing a double-quoted string, so the shell doesn't do anything weird to whitespace in it. The various escaped double-quotes in it will all be treated as literal characters (because they're escaped), and just be part of the output. And the pattern string, $'\t', is in ANSI-C quoting mode, so that the \t gets converted to an actual tab character.
Here's a couple of examples of using it:
% VAR1=$'abc\tdef\tghi' # Define a variable with actual tab characters
% echo "\"${VAR1//$'\t'/\" \"}\"" # Pass the converted version to a command
"abc" "def" "ghi"
% VAR2="\"${VAR1//$'\t'/\" \"}\"" # Store converted version in another variable
% echo "$VAR2"
"abc" "def" "ghi"
This could do what you want
echo -e "abc\tdef\tghi\tjhg\tmnb" | sed -ne 's/\t/" "/g; s/.*/"\0"/p'
Result:
"abc" "def" "ghi" "jhg" "mnb"
You may leverage awk. For example:
user$ $var='abc\tdef\tghi'
user$ $echo -e ${var}
(output)>>> abc def ghi
user$ $echo -e ${var} | awk -F '\t' '{ for (i=1; i <NF; i++) {printf "\"%s\" ", $i}; printf "\"%s\"\n", $NF}'
(output)>>> "abc" "def" "ghi"
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 $(cap_case "snake_case")
Note that the \u / \U replacement is a GNU extension to sed - if you're using a non-GNU implementation, check whether it supports this feature.
How can I just replace the last character (it's a }) from a string? I need everything before the last character but replace the last character with some new string.
I tried many things with awk and sed but didn't succeed.
For example:
...\\tx4535\\tx5102\\tx5669\\tx6236\\tx6803\\pardirnatural
\\f0
}'
should become:
...\\tx4535\\tx5102\\tx5669\\tx6236\\tx6803\\pardirnatural
\\f0
\\cf2 Its red now
}'
This replaces the last occurrence of:
}
with
\\cf2 Its red now
}
sed would do this:
# replace '}' in the end
echo '\tx4535\tx5102\tx5669\tx6236\tx6803\pardirnatural \f0 }' | sed 's/}$/\\cf2 Its red now}/'
# replace any last character
echo '\tx4535\tx5102\tx5669\tx6236\tx6803\pardirnatural \f0 }' | sed 's/\(.\)$/\\cf2 Its red now\1/'
Replacing the trailing } could be done like this (with $ as the PS1 prompt and > as the PS2 prompt):
$ str="...\\tx4535\\tx5102\\tx5669\\tx6236\\tx6803\\pardirnatural
> \\f0
> }"
$ echo "$str"
...\tx4535\tx5102\tx5669\tx6236\tx6803\pardirnatural
\f0
}
$ echo "${str%\}}\cf2 It's red now
}"
...\tx4535\tx5102\tx5669\tx6236\tx6803\pardirnatural
\f0
\cf2 It's red now
}
$
The first 3 lines assign your string to my variable str. The next 4 lines show what's in the string. The 2 lines:
echo "${str%\}}\cf2 It's red now
}"
contain a (grammar-corrected) substitution of the material you asked for, and the last lines echo the substituted value.
Basically, ${str%tail} removes the string tail from the end of $str; I remember % ends in 't' for tail (and the analogous ${str#head} has hash starting with 'h' for head).
See shell parameter expansion in the Bash manual for the remaining details.
If you don't know the last character, you can use a ? metacharacter to match the end instead:
echo "${str%?}and the extra"
First make a string with newlines
str=$(printf "%s\n%s\n%s" '\\tx4535\\tx5102\\tx5669\\tx6236\\tx6803\\pardirnatural' '\\f0' "}'")
Now you look for the last } in your string and replace it including a newline.
The $ makes sure it will only replace it at the last line, & stands for the matches string.
echo "${str}" |sed '$ s/}[^}]$/\\\\cf2 Its red now\n&/'
The above solution only works when the } is at the last line. It becomes more difficult when you also want to support str2:
str2=$(printf "Extra } here.\n%s\nsome other text" "${str}")
You can not match the } on the last line. Removing the address $ for the last line will result in replacing all } characters (I added a } at the beginning of str2). You only want to replace the last one.
Replacing once is forced with ..../1. Replacing the last and not the first is done by reversing the order of lines with tac. Since you will tac again after the replacement, you need to use a different order in your sedreplacement string.
echo "${str2}" | tac |sed 's/}[^}]$/&\n\\\\cf2 Its red now/1' |tac
In awk:
$ awk ' BEGIN { RS=OFS=FS="" } $NF="\\\\cf2 Its red now\n}"' file
RS="" sets RS to an empty record (change it to suit your needs)
OFS=FS="" separates characters each to its own field
$NF="\\\\cf2 Its red now\n}" replaces the char in the last field ($NF=}) with the quoted text
awk '{sub(/\\f0/,"\\f0\n\\\\\cfs Its red now")}1' file
...\\tx4535\\tx5102\\tx5669\\tx6236\\tx6803\\pardirnatural
\\f0
\\cfs Its red now
}'
Looking for the ruby one liner substitute to print out a substitution only if the line matches the regular expression:
echo -e "Line 1\nLine 2\nLine 3" | perl -ne "print if s/Line 2/Line 2 replaced, others discarded/g"
Input:
Line 1
Line 2
Line 3
Output:
Line 2 replaced, others discarded
As I know, there is no equivalent to -ne shorthand in ruby. So it will be little longer:
echo -e "Line 1\nLine 2\nLine 3" | ruby -e 'puts $<.read.lines.map {|l| l =~ /Line 2/ ? l.gsub(/Line 2/, "Line 2 replaced, others discarded") : nil }.compact'
Where:
$< also ARGF (docs) is Stream for file argument or STDIO
$<.read will read it all to string
$<.read.lines split by new line character, returns array
map {|l| ... } will collect result of expression in a block to new array
l =~ /Line 2/ check if string match Regex
l.gsub(/Line 2/, "Line 2 replaced") will replace all "Line 2" to "Line 2 replaced"
.compact will remove nil values from array (return new array without nil's)
puts [] will print each element of array on new line
Probably ruby is not a best chose for this task, I would choose sed or do it in text editor. Most of text editors can find and replace by regex nowdays
I have a document containing many " marks, but I want to convert it for use in TeX.
TeX uses 2 ` marks for the beginning quote mark, and 2 ' mark for the closing quote mark.
I only want to make changes to these when " appears on a single line in an even number (e.g. there are 2, 4, or 6 "'s on the line). For e.g.
"This line has 2 quotation marks."
--> ``This line has 2 quotation marks.''
"This line," said the spider, "Has 4 quotation marks."
--> ``This line,'' said the spider, ``Has 4 quotation marks.''
"This line," said the spider, must have a problem, because there are 3 quotation marks."
--> (unchanged)
My sentences never break across lines, so there is no need to check on multiple lines.
There are few quotes with single quotes, so I can manually change those.
How can I convert these?
This is my one-liner which is works for me:
awk -F\" '{if((NF-1)%2==0){res=$0;for(i=1;i<NF;i++){to="``";if(i%2==0){to="'\'\''"}res=gensub("\"", to, 1, res)};print res}else{print}}' input.txt >output.txt
And there is long version of this one-liner with comments:
{
FS="\"" # set field separator to double quote
if ((NF-1) % 2 == 0) { # if count of double quotes in line are even number
res = $0 # save original line to res variable
for (i = 1; i < NF; i++) { # for each double quote
to = "``" # replace current occurency of double quote by ``
if (i % 2 == 0) { # if its closes quote replace by ''
to = "''"
}
# replace " by to in res and save result to res
res = gensub("\"", to, 1, res)
}
print res # print resulted line
} else {
print # print original line when nothing to change
}
}
You may run this script by:
awk -f replace-quotes.awk input.txt >output.txt
Here's my one-liner using repeated sed's:
cat file.txt | sed -e 's/"\([^"]*\)"/`\1`/g' | sed '/"/s/`/\"/g' | sed -e 's/`\([^`]*\)`/``\1'\'''\''/g'
(note: it won't work correctly if there are already back-ticks (`) in the file but otherwise should do the trick)
EDIT:
Removed back-tick bug by simplifying, now works for all cases:
cat file.txt | sed -e 's/"\([^"]*\)"/``\1'\'\''/g' | sed '/"/s/``/"/g' | sed '/"/s/'\'\''/"/g'
With comments:
cat file.txt # read file
| sed -e 's/"\([^"]*\)"/``\1'\'\''/g' # initial replace
| sed '/"/s/``/"/g' # revert `` to " on lines with extra "
| sed '/"/s/'\'\''/"/g' # revert '' to " on lines with extra "
Using awk
awk '{n=gsub("\"","&")}!(n%2){while(n--){n%2?Q=q:Q="`";sub("\"",Q Q)}}1' q=\' in
Explanation
awk '{
n=gsub("\"","&") # set n to the number of quotes in the current line
}
!(n%2){ # if there are even number of quotes
while(n--){ # as long as we have double-quotes
n%2?Q=q:Q="`" # alternate Q between a backtick and single quote
sub("\"",Q Q) # replace the next double quote with two of whatever Q is
}
}1 # print out all other lines untouched'
q=\' in # set the q variable to a single quote and pass the file 'in' as input
Using sed
sed '/^\([^"]*"[^"]*"[^"]*\)*$/s/"\([^"]*\)"/``\1'\'\''/g' in
This might work for you:
sed 'h;s/"\([^"]*\)"/``\1''\'\''/g;/"/g' file
Explanation:
Make a copy of the original line h
Replace pairs of "'s s/"\([^"]*\)"/``\1''\'\''/g
Check for odd " and if found revert to original line /"/g