find and replace each line - bash script

Question

I want to delete in a file each line beginning by '#'. I ran that (I am using osx)

 sed -i '' -e 's/#.*/d' file

but I get this error message :

sed: 1: "s/#.*/d
": unescaped newline inside substitute pattern

@gniourf_gniourf: sed -i '/^#/d' file → sed: 1: "file": invalid command code f — Gareth Rees
– Gareth Rees, Commented Oct 29, 2013 at 10:28
@GarethRees you probably don't have GNU sed that handles the -i option... — gniourf_gniourf
– gniourf_gniourf, Commented Oct 29, 2013 at 10:30
@GarethRees Funny that my GNU sed is happy with the command I gave... are you sure you didn't mistype it? Besides, that command is exactly the one you gave in your answer :D. — gniourf_gniourf
– gniourf_gniourf, Commented Oct 29, 2013 at 10:36
The OP is on OS X, where the sed is BSD-ish, and that means that the -i option requires an argument. — Gareth Rees
– Gareth Rees, Commented Oct 29, 2013 at 10:43

Gareth Rees · Accepted Answer · 2013-10-29 10:08:24Z

1

The s command in sed means "substitute" and it takes two arguments:

s/pattern/replacement/

What you want to do is just to match lines starting with # and delete them, so you need the sed program:

/^#/d

Note that the pattern needs to start with ^ (meaning "start of line") otherwise it will match a # anywhere in the line.

answered Oct 29, 2013 at 10:08

Gareth Rees

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Comments

gstandaert · Accepted Answer · 2013-10-29 10:16:38Z

1

As stated by Gareth Rees above, the correct command is:

sed '/^#/ d' file

This good sed tutorial contains your question as an example:

answered Oct 29, 2013 at 10:16

gstandaert

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