Why can't I display data from four joined tables? - php

Question

I'm using the following query to join two tables. Table gz_topics features basic information, such as page Title and Subtitle, while the articles for each page are stored in table gz_articles_topics_intro. GZ.URL is actually a variable that matches each page's URL, but I'm using "Birds" here as an example.
$Zext = mysql_fetch_assoc(mysql_query("SELECT GZ.N, GZ.URL, GZ.Title, GZ.Live,
AI.URL, AI.Article, AI.Pagedex
FROM gz_topics GZ
LEFT JOIN gz_articles_topics_intro AI ON AI.URL = GZ.URL
WHERE GZ.URL LIKE 'Birds' AND GZ.Live = 1"));
It works fine. The problems begin when I put my articles in three separate tables and try to join them...
$Zext = mysql_fetch_assoc(mysql_query("SELECT GZ.N, GZ.URL, GZ.Title, GZ.Live,
Art.URL, Art.Article, Art.Pagedex,
AI.URL, AI.Article, AI.Pagedex, AN.URL, AN.Article, AN.Pagedex
FROM gz_topics GZ
LEFT JOIN gz_articles_topics Art ON Art.URL = GZ.URL
LEFT JOIN gz_articles_topics_intro AI ON AI.URL = GZ.URL
LEFT JOIN gz_articles_topics_names AN ON AN.URL = GZ.URL
WHERE GZ.URL LIKE '$MyURL' AND GZ.Live = 1"));
When I paste it into SQL, it works just fine, displaying all the data, including the article. But no values display on my page.
This is the code I use to display the article:
$Article = $Zext['Article'];
echo $Article;
I've tried inner joins and outer joins, while loops, etc., but nothing seems to work. My PHP/MySQL skills are intermediate, and I don't have a clue what the problem is because, as I said, it works fine when I paste it into SQL, and I don't see any error messages.
The problem must be right under my nose, because this looks pretty simple - even for me. ;)
Thanks for any tips.
On edit: I fixed one mistake, and my query is now displaying items from the table gz_topics, like the page title. However, it still doesn't display the article.

Answer 1

Problem 1) You have three columns with the same name "Article"
Art.Article, AI.Article, AN.Article
You may rename them like this:
Art.Article AS Article1, AI.Article AS Article2, AN.Article AS Article3
And then you can use the appropriate column (1, 2 or 3):
$Article = $Zext['Article1'];
Problem 2) If the contents are plain text (not HTML), you should escape the output:
echo htmlentities($Article);