How to Remove Leading Whitespace Chars from Ruby Heredoc

How do I remove leading whitespace chars from Ruby HEREDOC?

The <<- form of heredoc only ignores leading whitespace for the end delimiter.

With Ruby 2.3 and later you can use a squiggly heredoc (<<~) to suppress the leading whitespace of content lines:

def test
  <<~END
    First content line.
      Two spaces here.
    No space here.
  END
end

test
# => "First content line.\n  Two spaces here.\nNo space here.\n"

From the Ruby literals documentation:

The indentation of the least-indented line will be removed from each
line of the content. Note that empty lines and lines consisting solely
of literal tabs and spaces will be ignored for the purposes of
determining indentation, but escaped tabs and spaces are considered
non-indentation characters.

How to avoid last \n character in heredoc

As sagarpandya82 points out, you will need an additional action (i.e. chomp) to remove that extra \n.

You can use chomp along with heredoc like this:

a = <<~TEXT.chomp
  asd
  asd
TEXT
#=> "asd\nasd"

Ruby: Split, then remove leading/trailing whitespace in place?

s = "one thing, two things, three things, four things"
s.split(",").map(&:strip)
# => ["one thing", "two things", "three things", "four things"]

In my Ubuntu 13.04 OS,using Ruby 2.0.0p0

require 'benchmark'

s = "one thing, two things, three things, four things"
result = ""

Benchmark.bmbm do |b|
  b.report("strip/split: ") { 1_000_000.times {result = s.split(",").map(&:strip)} }
  b.report("regex: ") { 1_000_000.times {result = s.split(/\s*,\s*/)} }
end

Rehearsal -------------------------------------------------
strip/split:    6.260000   0.000000   6.260000 (  6.276583)
regex:          7.310000   0.000000   7.310000 (  7.320001)
--------------------------------------- total: 13.570000sec

                    user     system      total        real
strip/split:    6.350000   0.000000   6.350000 (  6.363127)
regex:          7.290000   0.000000   7.290000 (  7.302163)

Why does strip not remove the leading whitespace?

Where did the string " Bagsværd" come from?

It’s likely that the space character at the start of the string is not a “normal” space, but a non-breaking space (U+00A0):

2.0.0p353 :001 > " Bagsværd".strip
 => "Bagsværd" 
2.0.0p353 :002 > "\u00a0Bagsværd".strip
 => " Bagsværd" 

You could remove it with gsub rather than strip:

2.0.0p353 :003 > "\u00a0Bagsværd".gsub(/\A\p{Space}*/, '')
 => "Bagsværd" 

This uses the \A anchor, and the \p{Space} character property to emulate lstrip. To strip both leading and trailing whitespace, use:

2.0.0p353 :007 > "\u00a0Bagsværd\u00a0".gsub(/\A\p{Space}*|\p{Space}*\z/, '')
 => "Bagsværd" 

string formatting, remove leading chars

We call these "leading" characters, not trailing, since they're at the beginning, but the regex for this is very easy

x.sub(/^[0:]*/,"")

That works exactly as you phrased it: starting at the beginning of the string, remove all the 0s and :s.

How to separate a heredoc from other key-value pairs in a hash literal in Ruby?

This seems to work

{
  foo: 123,
  bar: <<-HEREDOC,
    a longer text
  HEREDOC
  baz: 456
}

How do I remove everything after first whitespace or certain number of characters?

I prefer to use as simple a solution as I can. Using gsub is needlessly complex. Either of these will do it:

str = '2008-10-09 20:30:40'
str[/(\S+)/, 1] #=> "2008-10-09"
str[0, 10] #=> "2008-10-09"

What is the Perl syntax to remove extraneous spaces from a Perl heredoc in 'use constant'?

Here's an example that will strip the leading whitespace from each line of a heredoc before assigning it to a constant:

use constant FOO => <<'STRING' =~ s/^\s+//mgr;
    abcde
    fghij
STRING

print FOO, "\n";

The /r modifier was new in Perl 5.14, and signifies that the target string isn't modified, but instead that it's copied internally, the copy is modified, and returned as the return value of the substitution operator. For example:

my $new_string = $string =~ s/foo/bar/r;

$new_string would contain the modified string, and $string would be left alone.

Applying this to your code, it might look like this:

use constant FIND_OBJECTS_SQL => <<'    END_SQL' =~ s/\s+/ /gr;
    ...
    END_SQL

This is very similar to the last example you demonstrated, but adds the /r modifier so that the substitution has a useful return value, and so that it doesn't attempt to modify the HERE-doc's string literal.

Remove multiple spaces and new lines inside of String

check out Rails squish method:

http://apidock.com/rails/String/squish

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