Re: Shot in the dark

If you added some error-checking you'd know right away. ie. after your mysql_query() call. if(!$news) die(mysql_error()); -Rasmus On Sat, 6 Jul 2002, jtjohnston wrote: > MySQL4 installation. I don't get this. One mysql_query works, the other > 2 don't. What did I mess up in my PHP syntax? > The SQL queries all work. > > $au ="Atwood, Margaret"; > $id ="1624"; > > THIS WORKS: > $news = mysql_query('select * from '.$table.' where id like '.$id.' > order by AU desc'); > > This gives "Warning: Supplied argument is not a valid MySQL result > resource ... on line 48": > $news = mysql_query('select * from '.$table.' where AU like '.$au.' > order by AU desc'); > #SQL-Query works: select * from ccl_main where AU like 'Atwood, > Margaret' order by AU desc > > This gives "Warning: Supplied argument is not a valid MySQL result > resource ... on line 48": > $news = mysql_query('SELECT * FROM '.$table.' WHERE MATCH (AU) AGAINST > ('.$au.' IN BOOLEAN MODE) order by AU desc`'); > #SQL-Query works: SELECT * FROM ccl_main WHERE MATCH (AU) AGAINST > ('Atwood, Margaret' IN BOOLEAN MODE) order by AU desc > > while ($mydata = mysql_fetch_object($news)) <----- this is line > 48 > { > $mycounter++; > echo $mydata->id." - ".$mydata->AU."<hr>"; > } > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php >