SQL页面视图Php

Question

So i want to have page view on my php pages and googled this... i dont want to make a new database for it so ive found a available spot in my database (so cant use any solutions that requires more than 1 cell in the database.

I got this code wich counts the visit so that part works fine, but i cant seem to make the number of views to be shown (printed i guess its the right word)

<?php  //Adds one to the counter 
  mysql_query("UPDATE news SET post = post + 1, published=published, last_edit=last_edit WHERE id=$id");

//Retreives the current count 
$count1 = mysql_fetch_row(mysql_query("SELECT post FROM post"));  

// Displays the count on your site print

             echo $count1; ?>

Any idea why i wont get the result to show?

Answer 1

$count is an array, not a scalar. Try echo $count[];

There is a set of techniques called debugging when things don't work as expected. The very first step you make is "divide and conquer" by separating complex expressions into simple ones:

$resource = mysql_query("SELECT count = count(post) FROM post");
if ($resource) {
    $count = mysql_fetch_row();
    if ($count > 0) {
       echo $count[0];
    } else {
       echo 'it seems that post table is empty';
    }
} else {
    echo 'I need to figure out what went wrong by looking in PHP documentation for mysql error query functions';
}

Secondly, when debugging, instead of echo use var_dump($resource) or var_dump($count). That way you will see the real type and value of a variables.

PS. mysql_query is deprecated and might be removed from PHP in any future version