regexp help for vi, grep, or sed

[TeMpT]

Anyone know how you go about quantifying the end of line or carriage return mark in regexp?
In vi, I am trying to do a global replace of characters (xxx, for example) with a line carriage and don't know what to put in the replacement part....so far, I have:

:g/xxx/s//LINE_CARRIAGE/g

anyone?

Want to use this in my grep too....so that I can get rid of all the blank lines

 

[Moohooya]

A basic regular expression doe not span lines. Sed should use \n as an extension to a basic regular epression to allow you to do this.

If you want to 'grep' over a new line, you may be able to do this with egrep, but I'm nt certain. grep will not do this.

 

[dszd0g]

This is not highly technical. You could get this from any regex manual.

But since no one has answered your question:

I do not know of a way for vi to match multiple lines. You can use sed, awk, or perl.

Sed:

The uppercase versions of the commands are for multi-line mode. D deletes the first line, P prints the first line, etc.

sed '/^foobar$/ {N; s/^foobar\nfoo/bar/;}'

Awk:

See the getline and next commands. You can also store the previous line in a variable.

Perl:

You can store the previous line in a variable, or you could do something like the following:

my @array = <IN>;
# Joins each line with a : separating them.
# You probably would want to use something other than a colon and actually check to make sure that (control?) character
# does not exist in your input, I'm just providing an example.
my $longline = join ':', @array;

Then do whatever matches you want on $longline. This is pretty bad code, but I don't know what in particular you are trying to do so I can't offer better.

 

[Cuular]

Well for a simple find characters and replace them with a carriage return in vi the following will work. For example if I wanted to replace the word "example" in the middle of that with a carriage return:

This is an example of replacing.

Then the following vi command would do that:
:g/example/s//Ctrl-VCtrl-M/g

Now that is exactly what it looks like, vi uses a Ctrl-V character to precede other Ctrl characters in it's regular expressions, so you type Ctrl-V, and then Ctrl-M. This will replace every occurrence of example with a carriage return.

In response to moohooya's:


<< If you want to 'grep' over a new line, you may be able to do this with egrep, but I'm nt certain. grep will not do this. >>



In most un*x regexp's '^' denotes the beginning of a line, and '$' denotes the end of a line, keeping that in mind then the following will do your remove empty lines in grep...

cat file | grep -v "^$"

and see how many carriage returns you see! That will strip out all blank lines, like the original poster wanted I believe.

 

[dszd0g]

<< 'XXX$^' >>



That is not going to work. For the general form in multi-line regex's ^ means the beginning of the regex and not the beginning of the line. $ is the end of the regex, and not the end of the line. If you read in two lines ^ is the beginning of the first line, and $ is the end of the second. Now when in single-line mode, the ends are the line so ^ is the beginning of the line and $ is the end of the line.

I have heard that you can do multi-line regex's in vi if you use the keystroke recorder, match the first line and play back the keystrokes to match the second and perform the action. It seems like a big hack to me and probably more work than its worth.

 

[Fallen Kell]

Hmmm....are you just replacing a single character with a newline? Cause if you are just use "tr".

tr '\015' '\012' <input filename> <output filename>

The above will replace all Windows and Mac style carrage returns (ASCI # 015) with Unix style carrage returns (ASCI # 012). Not sure if you need the ASCI number of the character if its a normal character that you can type, but since these two are not, you have to use the ASCI number.