How to display average number of posts per day?

According to the plugin’s source code, it directly outputs the post count, rather than return it as a variable for you to use.

So edit the plugin to look like this, instead:
<?php
/*
Plugin Name: Post Count
Plugin URI: https://mtdewvirus.com/code/wordpress-plugins/
Description: Outputs the total number of posts.
Version: 1.01
Author: Nick Momrik
Author URI: https://mtdewvirus.com/
modified by Scott Merrill (https://www.skippy.net/)
*/
function mdv_new_post_count() {
global $wpdb;
$now = gmdate("Y-m-d H:i:s",time());
$request = "SELECT COUNT(*) FROM $wpdb->posts WHERE post_status = 'publish' AND post_date < '$now'";
return $wpdb->get_var($request);
}
function mdv_post_count() {
echo mdv_new_post_count();
}
?>

Now, in your template file, you can call:
<?php
$post_count = mdv_new_post_count();
$from = strtotime("Apr 19, 2005");
$to = time();
echo $post_count / ($to - $from);
?>

Note: totally untested.

Thank you, skippy. Apparently, I haven’t been working very hard since Apr 19. This calculation says that I’ve made an average of 0.00052437762500377 posts per day!

OK, I’ve got a PHP text open here and I’m trying to understand this. I think strtotime() and time() both return timestamps expressed as seconds_since_unix_epoch. So wouldn’t that value have to be converted back into number_of_days before "echo $post_count / ($to - $from);?

Also, if that’s true, how would I make the result of that calculation an integer?

Thanks for your help.

This is what I’ve got now:

<?php $post_count = mdv_new_post_count();
$from = strtotime("Apr 19, 2005");
$to = time();
$days = ($to - $from) * 86400;
echo $post_count / $days; ?>

The only part I haven’t yet figured out is how to format the result (“6.0663586632747E-09”) as an integer.

Try echo (int) $post_count / $days; ?>

If that doesn’t work, try:
$foo = (int) $post_count / $days;
echo $foo;
?>

This is what I’m doing now:
<?php $post_count = mdv_new_post_count();
$from = strtotime("Apr 19, 2005");
$to = time();
$days = ($to - $from) * 86400;
$avg = (int) $post_count / $days;
echo $avg; ?>

But I’m still getting the result in scientific notation. (Same with ‘echo (int) $post_count / $days’.

All set now. Here’s what worked:

<?php
$post_count = mdv_new_post_count();
$from = mktime(0, 0, 0, 4, 19, 2005);
$to = time();
$days = ($to - $from) / 86400;
$avg = (int) $post_count / $days;
printf('%u', $avg);
?>

Please pretend you didn’t notice I was multiplying by 86400 instead of dividing. Shhhh… Do not go there.