Using escape characters inside grep

Question

I have the following regular expression for eliminating spaces, tabs, and new lines: [^ \n\t]

However, I want to expand this for certain additional characters, such as > and <.

I tried [^ \n\t<>], which works well for now, but I want the expression to not match if the < or > is preceded by a \.

I tried [^ \n\t[^\]<[^\]>], but this did not work.

Can anyone help?

Answer 1

Maybe you can use egrep and put your pattern string inside quotes. This should obliterate the need for escaping.

Answer 2

Can any one of the sequences below occur in your input?

\\>
\\\>
\\\\>
\blank
\tab
\newline
...

If so, how do you propose to treat them?

If not, then zero-width look-behind assertions will do the trick, provided that your regular expression engine supports it. This will be the case in any engine that supports Perl-style regular expressions (including Perl's, PHP, etc.):

 (?<!\\)[ \n\t<>]

The above will match any un-escaped space, newline, tab or angled braces. More generically (using \s to denote any space characters, including \r):

 (?<!\\)\s

Alternately, using complementary notation without the need for a zero-width look-behind assertion (but arguably less efficiently):

 (?:[^ \n\t<>]|\\[<>])

You may also use a variation of the latter to handle the \\>, \\\>, \\\\> etc. cases as well up to some finite number of preceding backslashes, such as:

 (?:[^ \n\t<>]|(?:^|[^<>])[\\]{1,3,5,7,9}[<>])

Cheers, V.

Answer 3

According to the grep man page:

A bracket expression is a list of characters enclosed by [ and ]. It matches any single character in that list; if the first character of the list is the caret ^ then it matches any character not in the list.

This means that you can't match a sequence of characters such as \< or \> only single characters.

Unless you have a version of grep built with Perl regex support then you can use lookarounds like one of the other posters mentioned. Not all versions of grep have this support though.