Display results from multiple tables with php and mysql

[chuck1971]

I have two tables, one table (article_cat) has four fields ("id", "title", "feature1", and "feature2") and the other table has six fields (articles) has ("id", "cat_id", "title", "content", "feature1", and "feature2"). I am looking to display the "title" from the "article_cat" table that has "feature1" with a value of 1. And I want to display beneath that all the "title" results from the "articles" table where "feature1" has a value of 1. I have tried this a thousand ways with no luck. Here is a representation of what I want:

 

TITLE (article_cat)

title (articles)

title (articles)

title (articles)

 

I was hoping someone can show me the err of my ways. Thanks so much in advance to all those who help! Here is the flawed code I have thus far:

 

<?

include("includes/connect.php");

$query = mysql_query("SELECT * FROM articles, article_cat WHERE articles.feature1 = 1 AND article_cat.feature1 = 1 ORDER BY id ASC");

$numrows=@mysql_num_rows($query);

if($numrows != 0) {

while ($result = mysql_fetch_array($query)) {

?>

<tr>

<td align="left" valign="top">

<table width="100%" border="0" cellspacing="0" cellpadding="0">

<tr>

<td align="left" valign="top"><strong><? echo $result['article_cat.title']; ?>:</strong><br>

<strong>></strong><a href="article_details.php?id=<? echo $result['id']; ?>"><? echo $result['articles.title']; ?></a><br>

</td>

</tr>

</table>

</td>

</tr>

<? } } ?>

 

Thanks for the help, All.  I cannot express my appreciation here!

[bri0987]

You have to connect to a database...

ADD this line "mysql_select_db($database, $Db_connection);"

 

Example:

 

include("includes/connect.php");

mysql_select_db($database, $Db_connection);

$query = mysql_query("SELECT * FROM articles, article_cat WHERE articles.feature1 = 1 AND article_cat.feature1 = 1 ORDER BY id ASC");
$numrows=@mysql_num_rows($query);
if($numrows != 0) {
while ($result = mysql_fetch_array($query)) {
?>

 

$database variable will look something like this:

$database = "mysql_database_name_here";

 

$Db_connection will look like:

$Db_connection = mysql_pconnect($hostname, $username, $password) or trigger_error(mysql_error(),E_USER_ERROR);