The get_active_blog_for_user() WordPress PHP function returns one of a user’s active blogs.
Usage
get_active_blog_for_user($user_id);
Example:
Input:
get_active_blog_for_user(1);
Output:
An object containing the active blog information.
Parameters
$user_id (int): The unique ID of the user.
More information
See WordPress Developer Resources: get_active_blog_for_user()
Examples
Display the primary blog of a user
$user_id = 1;
$blog = get_active_blog_for_user($user_id);
if ($blog) {
echo "User's primary blog is: " . $blog->blogname;
} else {
echo "No active blogs found for this user.";
}
Get the blog URL for a user
$user_id = 2;
$blog = get_active_blog_for_user($user_id);
if ($blog) {
echo "User's blog URL: " . $blog->siteurl;
} else {
echo "No active blogs found for this user.";
}
Check if a user’s blog is public
$user_id = 3;
$blog = get_active_blog_for_user($user_id);
if ($blog && $blog->public == 1) {
echo "This user's blog is public.";
} else {
echo "This user's blog is not public.";
}
Print the active blog information for a user
$user_id = 4;
$blog = get_active_blog_for_user($user_id);
if ($blog) {
echo "Active blog information for user $user_id:";
echo "\nBlog ID: " . $blog->blog_id;
echo "\nBlog Name: " . $blog->blogname;
echo "\nSite URL: " . $blog->siteurl;
} else {
echo "No active blogs found for this user.";
}
Get the blog language for a user
$user_id = 5;
$blog = get_active_blog_for_user($user_id);
if ($blog) {
echo "User's blog language: " . $blog->language;
} else {
echo "No active blogs found for this user.";
}