Hi Will,
$result = $wpdb->get_results("select * from wp_posts");
isn’t really ‘something simple’.
You are basically returning a very big result set; more so if your blog has a big number of posts.
Try something like:
$result = $wpdb->get_results("select * from wp_posts where ID='1'");
Better-still, for an even faster query, don’t use the ‘*’ syntax but specify exactly which field(s) you want from wp_posts.
$first_post_author = $wpdb->get_results("select post_author from wp_posts where ID='1'");
I have same problem. how to show the $result? is it work via echo?
and another question is about join 3 table. how to do it?i know how to join in sql. i need to learn the way that in wordpress. tnx
It is not working… Strange!!!
In ‘bpModBackend.php’, when i paste the below code on line 156:
$mod_status = $wpdb->get_var( $wpdb->prepare( “SELECT status FROM $wpdb->bp_mod_contents;” ) );
echo $mod_status;
Even echo anything is working very well!
I am trying to do something that falls very much within this discussion. I am not at all literate with PHP though.
I have two WP sites. I want to reach from one of them, into the other, and retrieve specific featured images for posts, and display them with a link to that post.
I am using this line (with the correct info for my target database):
$mydb = new wpdb(‘username’,’password’,’database’,’localhost’);
What would I follow that with to retrieve and display a post thumbnail?
I am tinkering with a line of code I found elsewhere that looks like this:
$result = $mydb->get_results(“select * from wp_posts where ID=’1885′”);
where ‘1885’ is a sample post ID number.
In an ideal world, I would develop this to the point where it functions as a widget. The client could simply enter a post ID number from the external WP site, and the widget would retrieve the appropriate thumbnail, and display it as a link back to the appropriate post on the other site.
