sed replacing line confusion

freeindy · 08-06-2008, 10:49 AM

Hi,

I have the following lines in a bash script:

Code:

sudo sed -i '19s/.*/'"\t\t\thardware ethernet $MAC_ADDR;/" $DHCPD_CONF
sudo sed -i '26s/.*/'"\t\t\t\"$TFTP_ROOT_PATH\/targets\/cb6\";/" DHCPD_CONF

The first line works but not the second, I get the following error for the second:

Code:

/bin/sed: -e expression #1, char 16: unknown option to `s'

I can't figure out why. Anyone?

Thanks,
Indy

Kenhelm · 08-06-2008, 12:17 PM

You will get an error if the variable $TFTP_ROOT_PATH contains any unescaped '/'
If so, try using a different delimiter for the sed s command e.g.
s# # # instead of s/ / /
or escape all the '/' in $TFTP_ROOT_PATH before using it in sed:-

Code:

TFTP_ROOT_PATH=/path/to/dir
TFTP_ROOT_PATH_esc=${TFTP_ROOT_PATH//\//\\/}
echo $TFTP_ROOT_PATH_esc
\/path\/to\/dir

chakka.lokesh · 08-06-2008, 11:31 PM

Quote:

Originally Posted by Kenhelm

Code:

TFTP_ROOT_PATH_esc=${TFTP_ROOT_PATH//\//\\/}

hi,

I am almost new to linux. can u plz explain me that line.

thx

trey85stang · 08-06-2008, 11:36 PM

Quote:

Originally Posted by chakka.lokesh

hi,

I am almost new to linux. can u plz explain me that line.

thx

That is a variable, He is using a system variable with special charaters and makeing another variable without them so that sed can read it, since sed wants special charters to be proceeded with a "\".

I hope that didnt add more confusion to your question

freeindy · 08-07-2008, 03:14 AM

thanks lads. changing the variable helped.
I see the confusion on my part now. I was dealing with $ sign but actually forgot that it is replaced when sed is dealing with it.

However, this line:

Quote:

TFTP_ROOT_PATH_esc=${TFTP_ROOT_PATH//\//\\/}

What does it actually do? Say if my variable $VAR has following data:
echo $VAR
/usr/local/share/images

What are all these slashes doing? Is there a description? Just curious.

Indy

jschiwal · 08-07-2008, 03:22 AM

It escapes the forward slash: / -> \/

You could have tried it out yourself in the console. That's the best way to learn.

freeindy · 08-07-2008, 03:49 AM

jschiwal,

I get that, but why so many //\//\\/? I mean, it must have some logic somewhere.
If it would have been $VAR_esc=${VAR/\/} I could have understood say it is taking / -> \/ but this is not the case. so why '//\//\\/'?

indy

burschik · 08-07-2008, 03:55 AM

Look up parameter expansion in the bash manpage.

freeindy · 08-07-2008, 04:17 AM

ok, got it.
thanks

jschiwal · 08-07-2008, 04:25 AM

You might be interested in downloading the "Advanced Bash Scripting Guide" from the www.tldp.org website. It consists entirely of commented examples that you can try yourself, making it easier to learn even though it covers advanced subjects.

chakka.lokesh · 08-07-2008, 04:34 AM

Quote:

Originally Posted by trey85stang

I hope that didnt add more confusion to your question

sorry it didn't.

the thing is he added the slashes at the end of the variable. How this changed the contents inside the variable

chakka.lokesh · 08-07-2008, 04:36 AM

Quote:

Originally Posted by burschik

Look up parameter expansion in the bash manpage.

trying....

may be related to ${parameter/pattern/string}

will read it and will be back....

chakka.lokesh · 08-07-2008, 04:49 AM

seems to be got it.

I will explain it here......

after TFTP_ROOT_PATH

1. / indicates the beginning of the pattern
2. / indicates that all the patterns should be replaced
3. \/ will get evaluated to \
4. / indicates the end of the pattern
5. \\/ will become \/ which is the string.