chuck1971 Posted November 13, 2007 Share Posted November 13, 2007 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 $result['articles.title']; ?></a><br> </td> </tr> </table> </td> </tr> <? } } ?> Thanks for the help, All. I cannot express my appreciation here! Quote Link to comment Share on other sites More sharing options...
bri0987 Posted November 13, 2007 Share Posted November 13, 2007 You have to connect to a database... ADD this line "mysql_select_db($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); Quote Link to comment Share on other sites More sharing options...
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