Replace one character with another in Bash
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 fi开发者_JAVA百科gure 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
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