How do I remove the first half of each line in a file? 2

[cpjust]

Hi,
I have a file like this:

blah blah blah myschema.table1 blah
blah blah myschema.table2 blah blah
...

and what I want to do is remove everything before the "myschema." on each line so that it becomes:

myschema.table1 blah
myschema.table2 blah blah

how would I do that?

I tried using sed like this, but it says "unknown command: `f'"

echo /Chris/myschema-find.txt | sed '/^.*myschema\..*$/myschema\..*$/'

Somebody said awk might be what I'm looking for, but I have no idea how to use awk and can barely use sed.

 

[p5wizard]

[tt]sed 's/^.*myschema/myschema/'[/tt]


HTH,

p5wizard

 

[cpjust]

Thanks.
Obviously the echo part didn't work either, but I got it working by passing the filename as the 2nd parameter to sed.

 

[cpjust]

Now what if I want to remove everything after the first space?
So change this:

myschema.table1 blah blah
myschema.table2 blah

to:

myschema.table1
myschema.table2

 

[cpjust]

Nevermind, I got it:

sed "s/ .*$//" /Chris/myschema-find1.txt > myschema-find2.txt

 

[PHV]

You don't need a temporary file:

sed 's/^.*myschema/myschema/;s/ .*$//' /Chris/myschema-find.txt > /Chris/myschema-find1.txt

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886