sidorak95 Posted May 21, 2011 Share Posted May 21, 2011 Hi, I've been trying to get a simple select statement to work in my code, and I just don't know why it returns an error. Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /adoptnow.php on line 13 And the code is: <?php $con = mysql_connect("localhost","***","***"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("a8784hos_pets2", $con) or die ("Couldn't select database because: ".mysql_error()); $query = "SELECT * FROM species WHERE available = 1"; $petcrap = mysql_query($query) or die ("Error in number one query: $query. ".mysql_error()); $petinfo = mysql_fetch_array($petcrap); while ($row = mysql_fetch_array($petinfo)){ echo $row['petid']; } ?> I've read that this is usually because you haven't selected the database or it can't find the table, but I've run the sql query in phpmyadmin, and it comes back perfectly. I honestly have no idea what I'm doing wrong right now and would appreciate some help. Thanks. Link to comment https://forums.phpfreaks.com/topic/237057-simple-select-issues/ Share on other sites More sharing options...
wildteen88 Posted May 21, 2011 Share Posted May 21, 2011 Remove this line. $petinfo = mysql_fetch_array($petcrap); And change $petinfo to $petcrap on this line while ($row = mysql_fetch_array($petinfo)){ Link to comment https://forums.phpfreaks.com/topic/237057-simple-select-issues/#findComment-1218461 Share on other sites More sharing options...
sidorak95 Posted May 21, 2011 Author Share Posted May 21, 2011 /facepalm I don't know how I missed that. I guess that's what I deserve for copying and pasting code snippets. Thanks! Link to comment https://forums.phpfreaks.com/topic/237057-simple-select-issues/#findComment-1218463 Share on other sites More sharing options...
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