regex to replace "foo-some white space-bar" with "fubar"

Question

I'm a newcomer to regular expressions, and am having a hard time with what appears to be a simple case.

I need to replace "foo bar" with "fubar", where there is any amount and variety of white space between foo and bar.

For what it's worth, I'm using php's eregi_replace() to accomplish this.

Thanks in advance for the help.

Answer 1

... = preg_replace('/foo\s+bar/', 'fubar', ...);

Answer 2

I'm not sure about the eregi_replace syntax, but you'd want something like this:

Pattern: foo\s*bar
Replace with: fubar

Answer 3

Try this:

find = '(foo)\s*(bar)'
replace = '\\1\\2'

\s is the metachar for any whitespace character.

Answer 4

I also prefer preg_replace, but for completeness, here it is with ereg_replace:

$pattern     = "foo[[:space:]]+bar";
$replacement = "fubar";
$string = "foo    bar";
print ereg_replace( $pattern, $replacement, $string);

that prints "fubar"