CODEWITHSUNDEEP
bashsed
user1841901
user1841901

Reputation: 35

BASH/SED with '$' substitution doesn't work

I have a bash script that lists subprograms/processes that could if the user chooses to, insert startupflags to a specific program. I want to match strings in the below formats and depending on which pgm the user chooses I want to insert/replace the string with the new flag infront of {PGMPATH}/pgm. The existing programs are listed in a startupfile according to something like this:

start -existingFlag ${PGMPATH}/pgm
start -existingFlag -anotherExistingFlag ${PGMPATH}/anotherPgm
start -existingFlag -anotherFlag ${PGMPATH}/yetAnotherPgm otherStuff

But to start with I try to match toward a hardcoded string (in the future toward the lines in the startup file):

start -existingFlag ${PGMPATH}\/pgm*

and replace it with a new line looking like this:

*start -existingFlag -newFlag ${PGMPATH}\/pgm*

From script:

existingString="start -existingFlag ${PGMPATH}\/pgm"
newString="start -existingFlag -newFlag ${PGMPATH}\/pgm"
sed 's/$replaceString/$newString/g' $STARTUPCONFFILE

This works (the string is replaced) as long as there is no '$' (just before {PGMPATH}) in the strings, but as soon as I add '$' as in ${PGMPATH} SED doesn't replace. I have tried a lot but I can't get it to work.

Suggestions?

Upvotes: 0

Views: 450

Answers (3)

anishsane
anishsane

Reputation: 20980

I think, there will be slashes in $PGMPATH. They will interfere with sed syntax.

You can use other character, like | or % as separator, instead of usual /.
e.g. try:

sed "s|$replaceString|$newString|g"

Alternately, you can use \%regexp% syntax, given in sed manual here. (I have not used it myself though..)

Another alternate option, is to escape all the slashes in $PGMPATH, using another line of sed; but that would be more difficult.

Also, as pointed to by sudo_O, I have changed the single quote to double quotes, since the variables won't expand, when quoted with single quotes.

Upvotes: 0

Chris Seymour
Chris Seymour

Reputation: 85805

You need double quotes for the shell to expand variables: 

$ set newString=1                                                         
$ set replaceString=one

# using single quotes: no expansion -> no replacement!              
$ echo one | sed 's/$replaceString/$newString/g'
one

# using double quotes: expansion -> replacement!
$ echo one | sed "s/$replaceString/$newString/g"
1

Upvotes: 1

Jens
Jens

Reputation: 72667

What does echo $PATH print? Are you aware that PATH is usually a colon separated list of directories? Is this really what you want? It would expand to, e.g.

start -existingFlag /usr/bin:/bin:/usr/local/bin/pgm

which is most certainly not what you expect. Maybe you have a variable name clash and should use another name than PATH.

Upvotes: 0

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