replace variable from for loop into a sed command

Question

I tried the following snippet of code, but the sed command replaces with $var explicitly shown rather than as thr value of the var at that time. How can I solve this?s

for var in 01 02
do       
sed -e 's/%\\include{removeMe}/\\include{chapter_$var}/g'
done

Can you provide a sample of your input data (btw sed command has missing file name). — anubhava
– anubhava, Commented Jan 30, 2014 at 14:36

Beta · Accepted Answer · 2014-01-30 13:04:10Z

2

Try double quotes:

sed -e "s/%\\include{removeMe}/\\include{chapter_$var}/g"

(Better still, try single/double quotes with a simpler example first.)

answered Jan 30, 2014 at 13:04

Beta

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4 Comments

fedorqui Over a year ago

Yes, the solution is to use double quotes to get the variables expanded.

Geoff Over a year ago

@Beta Using double quotes didn't do any replace! The single quotes did the replace of the text but not the substitution on the $var.

Geoff Over a year ago

@Beta The problem is the % symbol. Without it it works. But with it, no replace takes place.

NeronLeVelu Over a year ago

add option --posix or escape % with \%

NeronLeVelu · Accepted Answer · 2014-01-30 16:42:53Z

0

for var in 01 02
 do       
   sed -e "/[%]\(\\include{\}removeMe}/ s//\1chapter_${var}}/g"
 done

use the back reference and encapsulation

but it could be replace by if var is not used with a filename association (so sed is a pipe) like in your sample (also because replacement by 01 will remove any possibility of finding a pattern for 02 replacement)

YourStream | sed -e 's/%\(\\include{\}removeMe}/\1chapter_01}/g'

answered Jan 30, 2014 at 16:42

NeronLeVelu

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