It works as expected here. I get the "Line is..." output. What are you seeing?

Of course, it's not working for exactly the reasons that you think it is. '|' is a special character in regexes and needs to be escaped (with a backslash) if you don't want it to mean "or".

In what way was the output different from your expected output? When I test the code, the output is exactly as you posted your expected output.

However, just because your snippet produced the expected output in the test case you provided doesn't mean the regexp is free of bugs. The '|' character has special meaning within regular expressions. It means "alternation" or "or". Only if you preceed it with a backslash, as in '\|' will it be interpreted as a plain old character within a regular expression:

$line =~ /^\|\s$str\s/
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...would probably be closer to your goal.

You'll be much more productive with regular expressions if you take an hour to read through perlretut and perlre. There's a mountain of information to learn, but those documents provide a great start.

Let me get this straight:
You're trying to match | br_summer and expect the output to have all the extra whitespace removed? Presuming this is a typo, not an expectation, your regex is incorrect. The pipe symbol is meaningful in a regexp,so it must be escaped, e.g.,

if($line =~ /^\|\s$str /){
    print "Line is $line (with $str)\n";
}
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Also, as a quibble, your regex won't match (even after the pipe symbol is escaped) if $str is followed by a non-blank whitespace character, e.g. a tab.

Although it isn't really what you are asking, have you considered using split instead of a regular expression? It looks to me as though you have a line with fields delimited by the pipe character ... often a good place to use split. You may still choose to apply a regular expression to a smaller part of the line, but you'll generally find less room for error when you have reached a more granular level.