Query Isn't Working like others

I for the life of me cannot figure out why this is not outputting a 0 when i try to echo it. It works on the previous query but not for the last one.

// This one works and if there are no movies above 0 then it outpus 0 fine
$tag_movies_result = mysql_query("SELECT * FROM tags WHERE tagname='$n' AND movie > 0");
$total_times_used_movies = mysql_num_rows($tag_movies_result);

// Where as this query returns an error 
$tag_shows_result = mysql_query("SELECT * FROM tags WHERE tagname='$n' AND show > 0"); <-- `show`
$total_times_used_shows = mysql_num_rows($tag_shows_result); <-- Line 12

// the error
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/a5410474/public_html/tags/tag.php on line 12

this is what i use to call it

<li><a href="#shows" rel="shows">Shows (<?php echo $total_times_used_shows; ?>)</a></li>

demo: http://mydb.host56.com/tags.php?n=the

the column name SHOWS needed to be enclosed in `.开发者_高级运维 Topic Closed.

SHOW is reserved keyword wrap tilde (`) around show, like below:

`show` 

Reference

If there is an error in a query, the value returned from mysql_query will be FALSE rather than a resource. Try this:

if(!$tag_show_result) {
 echo mysql_error();
}

This should help you to identify the error in the query

Actually, I see what the problem might be. You should still add echo mysql_error(); to the line right after the query that is failing, however there is an error in your query. the word show is a mysql reserved word. Show is used to display information about the database. You can wrap it in backticks like this:

`show`

There's clearly an error in your query; find that, and you will resolve the problem.

More to the point, you should be checking the result of your mysql_query for FALSE to make sure that the query was successful, and if it wasn't, output the error using mysql_error().