I need to replace a space ( ) with a dot (.) in a string in bash.
I think this would be pretty simple, but I'm new so I can't figure out how to modify a similar example for this use.
Use inline shell string replacement. Example:
foo=" "
# replace first blank only
bar=${foo/ /.}
# replace all blanks
bar=${foo// /.}
See http://tldp.org/LDP/abs/html/string-manipulation.html for more details.
You could use tr, like this:
tr " " .
Example:
# echo "hello world" | tr " " .
hello.world
From man tr:
DESCRIPTION
Translate, squeeze, and/or delete characters from standard input, writ‐
ing to standard output.
In bash, you can do pattern replacement in a string with the ${VARIABLE//PATTERN/REPLACEMENT} construct. Use just / and not // to replace only the first occurrence. The pattern is a wildcard pattern, like file globs.
string='foo bar qux'
one="${string/ /.}" # sets one to 'foo.bar qux'
all="${string// /.}" # sets all to 'foo.bar.qux'
Try this
echo "hello world" | sed 's/ /./g'
Use parameter substitution:
string=${string// /.}
Try this for paths:
echo \"hello world\"|sed 's/ /+/g'|sed 's/+/\/g'|sed 's/\"//g'
It replaces the space inside the double-quoted string with a + sing, then replaces the + sign with a backslash, then removes/replaces the double-quotes.
I had to use this to replace the spaces in one of my paths in Cygwin.
echo \"$(cygpath -u $JAVA_HOME)\"|sed 's/ /+/g'|sed 's/+/\\/g'|sed 's/\"//g'
The recommended solution by shellcheck would be the following:
string="Hello World" ; echo "${string// /.}"
output: Hello.World
Related
How would you go about removing everything after x number of characters? For example, cut everything after 15 characters and add ... to it.
This is an example sentence should turn into This is an exam...
GnuTools head can use chars rather than lines:
head -c 15 <<<'This is an example sentence'
Although consider that head -c only deals with bytes, so this is incompatible with multi-bytes characters like UTF-8 umlaut ü.
Bash built-in string indexing works:
str='This is an example sentence'
echo "${str:0:15}"
Output:
This is an exam
And finally something that works with ksh, dash, zsh…:
printf '%.15s\n' 'This is an example sentence'
Even programmatically:
n=15
printf '%.*s\n' $n 'This is an example sentence'
If you are using Bash, you can directly assign the output of printf to a variable and save a sub-shell call with:
trim_length=15
full_string='This is an example sentence'
printf -v trimmed_string '%.*s' $trim_length "$full_string"
Use sed:
echo 'some long string value' | sed 's/\(.\{15\}\).*/\1.../'
Output:
some long strin...
This solution has the advantage that short strings do not get the ... tail added:
echo 'short string' | sed 's/\(.\{15\}\).*/\1.../'
Output:
short string
So it's one solution for all sized outputs.
Use cut:
echo "This is an example sentence" | cut -c1-15
This is an exam
This includes characters (to handle multi-byte chars) 1-15, c.f. cut(1)
-b, --bytes=LIST
select only these bytes
-c, --characters=LIST
select only these characters
Awk can also accomplish this:
$ echo 'some long string value' | awk '{print substr($0, 1, 15) "..."}'
some long strin...
In awk, $0 is the current line. substr($0, 1, 15) extracts characters 1 through 15 from $0. The trailing "..." appends three dots.
Todd actually has a good answer however I chose to change it up a little to make the function better and remove unnecessary parts :p
trim() {
if (( "${#1}" > "$2" )); then
echo "${1:0:$2}$3"
else
echo "$1"
fi
}
In this version the appended text on longer string are chosen by the third argument, the max length is chosen by the second argument and the text itself is chosen by the first argument.
No need for variables :)
Using Bash Shell Expansions (No External Commands)
If you don't care about shell portability, you can do this entirely within Bash using a number of different shell expansions in the printf builtin. This avoids shelling out to external commands. For example:
trim () {
local str ellipsis_utf8
local -i maxlen
# use explaining variables; avoid magic numbers
str="$*"
maxlen="15"
ellipsis_utf8=$'\u2026'
# only truncate $str when longer than $maxlen
if (( "${#str}" > "$maxlen" )); then
printf "%s%s\n" "${str:0:$maxlen}" "${ellipsis_utf8}"
else
printf "%s\n" "$str"
fi
}
trim "This is an example sentence." # This is an exam…
trim "Short sentence." # Short sentence.
trim "-n Flag-like strings." # Flag-like strin…
trim "With interstitial -E flag." # With interstiti…
You can also loop through an entire file this way. Given a file containing the same sentences above (one per line), you can use the read builtin's default REPLY variable as follows:
while read; do
trim "$REPLY"
done < example.txt
Whether or not this approach is faster or easier to read is debatable, but it's 100% Bash and executes without forks or subshells.
PATH="/dir1/subdir1"
Want to replace "/" with "-" from the above string
and the result should be
-dir1-subdir1
You can check the more detailed version.
$ var="001244abc"
$ replace="polo"
$ echo ${var//001244/$replace}
poloabc
Hope it helps..
You can use tr for that since you want to replace a single character with another one:
echo $PATH | tr "/" "-"
Result is: -dir1-subdir1
In bash you can use a pattern substitution of the form ${parameter//pattern/string} to replace all matches of pattern with string:
echo "${PATH//\//-}"
Note that your pattern string / must be escaped with \/.
hi trying to replace the following string with a long one :
#x#
with string that I got from the command line:
read test
sed -i --backup 's/#x#/'${test}'/g' file.json README.md
but it is working only for 1 word, it is not working if there is space between word . even between quotes
sed: 1: "s/#x#/string test string: unterminated substitute in regular expression
if case you run it on MacOS and struggling with "unterminated substitute in regular expression", there is an easier explanation for this:
MacOS has slightly other version of sed than usually is on linux. -i requires a parameter. If you have none, just add "" after -i
sed -i "" --backup 's/#x#/'${test}'/g' file.json README.md
or for example if you just have to delete dome line, this works on linux, but brings “invalid command code” on MacOS
sed -i 39d filenamehere.log
and this works on MacOS
sed -i "" 39d filenamehere.log
The problem originates from the way you are using the single-quotes. Currently you are terminating your input behind the 2. single-quote. See the Error message, it makes you aware of the fact that it is missing something.
If you have a file with the following content:
foo #x# foo
Than you can replace the content e.g. with the following command:
sed 's/#x#/bar foo bar/' foo.txt > foo2.txt
And get:
foo bar foo bar foo
If you need to pass in a variable the comment from Gordon Davisson shows you the right way.
By the way, if you want to use the inplace option, on my linux you would need to use the command like this:
sed -i.old "s/#x#/${test}/" foo.txt
But I think this might depends on your enviroment (mac?).
sed doesn't understand strings where a string is a series of literal characters. It replaces a regexp (not a string) with a backreference-enabled "string" (also not a string) all within a set of delimiters (which ALSO require careful handling in both the regexp and the replacement). See Is it possible to escape regex metacharacters reliably with sed for more info.
To replace a string with another string the simplest approach is to just use a tool that understands strings such as awk:
$ cat file
before stuff
foo #x# bar
after stuff
$ cat tst.awk
BEGIN {
old = ARGV[1]
new = ARGV[2]
ARGV[1] = ARGV[2] = ""
}
s = index($0,old) { $0 = substr($0,1,s-1) new substr($0,s+length(old)) }
{ print }
$ test='a/\t/&"b'
$ awk -f tst.awk '#x#' "$test" file
before stuff
foo a/\t/&"b bar
after stuff
The above will work no matter what characters test contains, even newlines:
$ test='contains a
newline'
$ awk -f tst.awk '#x#' "$test" file
before stuff
foo contains a
newline bar
after stuff
I thought my bash-fu was strong enough but apparently it isn't. I can't seem to figure this out. I would like to do something like this:
var="XXXX This is a line"
word_to_replace="XXXX"
# ...do something
echo "Done:${var}"
Done: This is a line
Basically I want to quickly replace all characters in a word with spaces, preferably in one step. Note, if it makes things easier var currently will be at the start of the string although it may have leading spaces (which would need to be retained).
In python I would possibly do this:
>>> var="XXXX This is a line"
>>> word_to_replace="XXXX"
>>> var=var.replace(word_to_replace, ' '*len(word_to_replace))
>>> print("Done:%s" % var)
Done: This is a line
Here's one way you could do it, using a combination of shell parameter expansion and the sed command.
$ var="XXXX This is a line"
$ word_to_replace="XXXX"
$ replacement=${word_to_replace//?/ }
$ sed "s/$word_to_replace/$replacement/" <<<"$var"
This is a line
? matches any character and ${var//find/replace} does a global substitution, so the variable $replacement has the same length as $word_to_replace, but is composed solely of spaces.
You can save the result to a variable in the usual way:
new_var=$(sed "s/$word_to_replace/$replacement/" <<<"$var")
In plain Bash:
If we know the word to be replaced:
$ line=" foo and some"
$ word=foo
$ spaces=$(printf "%*s" ${#word} "")
$ echo "${line/$word/$spaces}"
and some
If we don't, we could pick the string apart to find the leading word, but this gets a bit ugly:
xxx() {
shopt -s extglob # for *( )
local line=$1
local indent=${line%%[^ ]*} # the leading spaces
line=${line##*( )} # remove the leading spaces
local tail=${line#* } # part after first space
local head=${line%% *} # part before first space...
echo "$indent${head//?/ } $tail" # replace and put back together
}
$ xxx " word on a line"
on a line
That also fails if there is only one word on the line, head and tail both get set to that word, we'd need to check for if there is a space and handle the two cases separately.
Using sed:
#!/usr/bin/env sh
word_to_replace="XXXX"
var="$word_to_replace This is a line"
echo "Done: $var"
word_to_replace=$(echo "$word_to_replace" | sed 's,., ,g')
var="$word_to_replace This is a line"
echo "Done: $var"
I use GNU Awk:
echo "$title" | gawk '{gsub(/./, "*"); print}'
This replaces each character with an asterisk.
EDIT. Consolidated answer:
$ export text="FOO hello"
$ export sub="FOO"
$ export space=${sub//?/ }
$ echo "${text//$sub/$space}"
hello
Given a list of paths separated by a single space:
/home/me/src/test /home/me/src/vendor/a /home/me/src/vendor/b
I want to remove the prefix /home/me/src/ so that the result is:
test vendor/a vendor/b
For a single path I would do: ${PATH#/home/me/src/} but how do I apply it to this series?
You can use // to replace all occurrences of substring. Replace it with null string to remove them.
$ path="/home/me/src/test /home/me/src/vendor/a /home/me/src/vendor/b"
$ echo ${path//\/home\/me\/src\/}
test vendor/a vendor/b
Reference: ${parameter/pattern/string} in Bash reference manual
Using shell parameter expansion doesn't seem to be the solution for this, since it would remove everything up to / from a given point is useful, as nu11p01n73R's answer reveals.
For clarity, I would use sed with the syntax sed 's#pattern#replacement#g':
$ str="/home/me/src/test /home/me/src/vendor/a /home/me/src/vendor/b"
$ sed 's#/home/me/src/##g' <<< "$str"
test vendor/a vendor/b
Like always a grep solution from my side :
echo 'your string' | grep -Po '^/([^ /]*/)+\K.+'
Please note that the above regex do this for any string like /x/y/z/test ... But if you are interested only in replacing /home/me/src/, try the following :
echo 'your string' | grep -Po '^/home/me/src/\K.+' --color