weemee500 Posted October 20, 2009 Share Posted October 20, 2009 Hi, I am currently trying to make a part for my user driven website where one user can subscribe to another and whoever they have subscribed to is echoed back on there profile page. my users table structure is: id username password and my subscription table is: id user_id follow_id i am currently trying to query the logged in users subscriptions with this sql_query $subsQuery = mysql_query("SELECT U.username FROM users AS U INNER JOIN subscription AS Sub.follow_id=U.id WHERE Sub.user_id=".$ID); $ID being the logged in users id. However when trying to echo this on the page using this bit of php: <? while ($subs = mysql_fetch_assoc($subsQuery)) { ?><br /><br /> <? echo $subs['username'] ?> <? } ?> it throws back this error: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/avince/public_html/users/user.php on line 53 Any help would be much appreciated! Quote Link to comment https://forums.phpfreaks.com/topic/178328-querying-info-from-one-table-based-on-info-in-another/ Share on other sites More sharing options...
ILMV Posted October 20, 2009 Share Posted October 20, 2009 Simple, your MySQL query is incorrect, could you please echo out your query and post it here. ILMV Quote Link to comment https://forums.phpfreaks.com/topic/178328-querying-info-from-one-table-based-on-info-in-another/#findComment-940308 Share on other sites More sharing options...
Mark Baker Posted October 20, 2009 Share Posted October 20, 2009 Sub isn't automatically an alias for the subscription table unless you define it as such SELECT U.username FROM users U INNER JOIN subscription Sub AS Sub.follow_id=U.id WHERE Sub.user_id=1 Quote Link to comment https://forums.phpfreaks.com/topic/178328-querying-info-from-one-table-based-on-info-in-another/#findComment-940314 Share on other sites More sharing options...
ILMV Posted October 20, 2009 Share Posted October 20, 2009 Bugger, could you also find the mysql_error description too, after your mysql_query function do this: $subsQuery = mysql_query("SELECT U.username FROM users AS U INNER JOIN subscription AS Sub.follow_id=U.id WHERE Sub.user_id=".$ID) or die ("Query '$query' failed with error message: \"" . mysql_error () . '"'); Quote Link to comment https://forums.phpfreaks.com/topic/178328-querying-info-from-one-table-based-on-info-in-another/#findComment-940322 Share on other sites More sharing options...
sasa Posted October 20, 2009 Share Posted October 20, 2009 changr to $subsQuery = mysql_query("SELECT U.username FROM users AS U INNER JOIN subscription AS Sub ON Sub.follow_id=U.id WHERE Sub.user_id=".$ID); Quote Link to comment https://forums.phpfreaks.com/topic/178328-querying-info-from-one-table-based-on-info-in-another/#findComment-940531 Share on other sites More sharing options...
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