sed whole word search and replace

Question

How do I search and replace whole words using sed?

Doing

sed -i 's/[oldtext]/[newtext]/g' <file> 

will also replace partial matches of [oldtext] which I don't want it to do.

Answer 1

In one of my machine, delimiting the word with "\b" (without the quotes) did not work. The solution was to use "\<" for starting delimiter and "\>" for ending delimiter.

To explain with Joakim Lundberg's example:

$ echo "bar embarassment" | sed "s/\<bar\>/no bar/g"
no bar embarassment

Answer 2

\b in regular expressions match word boundaries (i.e. the location between the first word character and non-word character):

$ echo "bar embarassment" | sed "s/\bbar\b/no bar/g"
no bar embarassment

Answer 3

in shell command:

echo "bar embarassment" | sed "s/\bbar\b/no bar/g" 

or:

echo "bar embarassment" | sed "s/\<bar\>/no bar/g"

but if you are in vim, you can only use the later:

:% s/\<old\>/new/g

Answer 4

On Mac OS X, neither of these regex syntaxes work inside sed for matching whole words

• \bmyWord\b
• \<myWord\>

Hear me now and believe me later, this ugly syntax is what you need to use:

• /[[:<:]]myWord[[:>:]]/

So, for example, to replace mint with minty for whole words only:

• sed "s/[[:<:]]mint[[:>:]]/minty/g"

Source: re_format man page

Answer 5

Use \b for word boundaries:

sed -i 's/\boldtext\b/newtext/g' <file>

Answer 6

$ echo "bar embarassment"|awk '{for(o=1;o<=NF;o++)if($o=="bar")$o="no bar"}1'
no bar embarassment