I need to detect if array elements contain line breaks and substitute them for nothing.  I guess there is a special reference for these but can not find them.  The end aim is clean a MySQL db of EOL chars.

if(@row =~ m/(\\n)/){ print "Found EOL char in @row"; }

What if the EOL originated in an html page.

I would use an s operator in a for loop on the list (as an alternative you could use a map which would be even shorter). Here's a sample program that reads a file into a list then takes all training white space off the end of each line and also the line separators, even when the line separators are embedded withing a multiline string within the @rows list.  The program replaces the new line with a single space - put what you like in there of course but I guess you don't want the lines to run toghether!

Code:

#!/usr/bin/perl open (FH,"callcosts"); @rows = <FH>; print "before\n",@rows[12..14],"\n"; foreach (@rows) {        s/\s*[\n\r]+/ /g;        } print "after\n",@rows[12..14],"\n";

By the way - I know the test data isn't very good as it only has one new line on each list element and the whole ting could have been done with a chomp in this case.

You also asked about HTML line ends ....
For a paragraph tag:
     s/\s*<p\s.*?>/ /igs;
will get rid of the "p tags" and replace them with a space

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