1

Im trying to update into another table the values blogname, blog description options when the user updated them from General Settings page.

I use this code,

$hook_array = array('update_option_blogname','update_option_admin_email','update_blogdescription','update_option_home','update_option_siteurl');

foreach($hook_array as $hook)
{
    add_action($hook, 'my_functione', 10, 1);

}

function my_functione($arg)
{
    global $wpdb;
    $the_id = get_current_blog_id();
    $blogname =  get_option('blogname'); 
    $blogdescr =  get_option('blogfulldescription');
    $wp_blogs_info = $wpdb->base_prefix . "blogs_info";

    $wpdb->query("UPDATE " . $wp_blogs_info . " SET site_title=".$blogname.", site_description=".$blogdescr." where blog_id = '" . (int)$the_id . "'");

}

Here is the table,

  blog_id int NOT NULL,
  countries_id int(3) NOT NULL,
  site_title varchar(64) NOT NULL,
  site_description varchar(64) NOT NULL,
  site_tags varchar(64) NOT NULL,
  PRIMARY KEY (blog_id)

However, its not updating though there is a value with the blog_id of the current blog.. Please help.

1
  • what is the outcome? do you get any error or message? Commented Jun 19, 2012 at 7:12

1 Answer 1

0

well, i just found that you've error in your query. i've re-written that for you:

$wpdb->query("UPDATE `" . $wp_blogs_info . "` SET `site_title`='".$wpdb->escape($blogname)."', `site_description` = '".$wpdb->escape($blogdescr)."' where blog_id = '" . (int)$the_id . "'");
2
  • Thank you! it works now. I didn't knew the escape one.
    – Ken
    Commented Jun 19, 2012 at 7:56
  • not only escaping, you did not put the quotation sign around the values. Commented Jun 19, 2012 at 8:27

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