Replace special characters with Escape characters?

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# 1  
Old 01-05-2012
Replace special characters with Escape characters?

i need to replace the any special characters with escape characters like below.

test!=123-> test\!\=123

!@#$%^&*()-= to be replaced by


Last edited by Scott; 01-05-2012 at 12:01 PM.. Reason: Code tags
# 2  
Old 01-05-2012
echo '!@#$%^&*()-' | sed "s/[!@#$%^&*()-]/\\\&/g"

# 3  
Old 01-05-2012
You listed = to be enclosed by / / too, but in your example it is not. Remove it from the pattern list maybe. Anyway maybe this is what you are looking for:
$> echo 'test!=123'| sed 's_[!@#$%^&*()-=]_\/&\/_g'

# 4  
Old 01-05-2012
echo 'test!=123' | sed 's:[!@#$%^&*()=-]:\\&:g'

# 5  
Old 01-05-2012
And just one more.... Smilie

export var1='One Way To !@#$%^&*()-= Do It'

echo $var1 | sed 's#\([]\!\(\)\#\%\@\*\$\/&\-\=[]\)#\\\1#g'

One Way To \!\@\#\$\%^\&\*\(\)-\= Do It

# 6  
Old 01-05-2012

Just for education purpose :

The trick is that in between a list [.....] some character have a special meaning :

The hyphen - is usually used to define and interval example : [0-9]

That is the reason why, in order to make it interpreted as litteral , it should be placed at a edge of the list (beginning or end) consider :
$ echo '!@#$%^&*()-' | sed 's/[-!@#$%^&*()]/\\&/g'
$ echo '!@#$%^&*()-' | sed 's/[!@#$%^&*()-]/\\&/g'

There is another special case within a list [......] :
The closing square bracket ]
This special closing square bracket - to be taken as litteral (so that the substitution will apply)- needs to be setup at the very first position in the list otherwise it is take as end of the list or unexpected behaviour may occure.

Consider the following example :
$ echo '!@#[]$%^&*()-' | sed 's/[]!@#[$%^&*()-]/\\&/g'
$ echo '!@#[]$%^&*()-' | sed 's/[-!@#[$%^&*()]]/\\&/g'
$ echo '!@#[]$%^&*()-' | sed 's/[-!@#[]$%^&*()]/\\&/g'

Last edited by ctsgnb; 01-05-2012 at 01:57 PM..
# 7  
Old 01-05-2012
@laknar & off topic
Just wondered why you would want to do this?
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