Replace one character with another in Bash

Question

I need to be able to do is replace a space () with a dot (.) in a string in bash.

I think this would be pretty simple, but I\'m new so I ca

Answer 1

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'

Answer 2

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.

Answer 3

Use parameter substitution:

string=${string// /.}

Answer 4

Try this

 echo "hello world" | sed 's/ /./g' 

Answer 5

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.

Answer 6

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'