To pass a php value in an array in wordpress

Question

I have a function like bellow

if(!function_exists('generate_share_urls')) {
  function generate_share_urls() {
      $title = the_title();
      $content = the_content();
    $share_urls = array (
      'facebook'      => 'http://www.facebook.com/sharer/sharer.php?u=',
      'twitter'       => 'http://twitter.com/share?url=',
      'google_plus'   => 'https://plus.google.com/share?url=',
      'linkedin'      => 'http://www.linkedin.com/shareArticle?mini=true&url=',
      'pinterest'     => 'http://pinterest.com/pin/create/button/?url=',
      'email'         => 'mailto:?subject={'.$title.'}&body='.$content.',
      'permalink'     => ''
   );
   return $share_urls;
  }
}

Now when run the function i cant find the $title and the $content of the page. Is there any fault with my code.

laurent · Accepted Answer · 2013-01-14 08:00:17Z

2

the_title() and the_content() will print the title/content, not return it. What you need instead are the get_ functions:

if(!function_exists('generate_share_urls')) {
  function generate_share_urls() {
      $title = get_the_title();
      $content = get_the_content();
    $share_urls = array (
      'facebook'      => 'http://www.facebook.com/sharer/sharer.php?u=',
      'twitter'       => 'http://twitter.com/share?url=',
      'google_plus'   => 'https://plus.google.com/share?url=',
      'linkedin'      => 'http://www.linkedin.com/shareArticle?mini=true&url=',
      'pinterest'     => 'http://pinterest.com/pin/create/button/?url=',
      'email'         => 'mailto:?subject={'.$title.'}&body='.$content.',
      'permalink'     => ''
   );
   return $share_urls;
  }
}

answered Jan 14, 2013 at 8:00

laurent

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2 Comments

Aesphere Over a year ago

from the syntax highlighting, he may have an unclosed quote in there as well, not sure if he just copied it wrong or whether it's really an error...

Ranjit Over a year ago

Thank you. I have found what i want.

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