2muchspam Posted May 26, 2003 Share Posted May 26, 2003 Hello again! I\'ve got a total hits script that I kinda got from an old nuke site. The script does incriment the database just fine. But when I call the result, I get \"Resource id #4\" What the heck am I doing wrong? Database: # Table structure for table `hits` # CREATE TABLE hits ( type varchar(80) NOT NULL default \'\', var varchar(80) NOT NULL default \'\', count int(10) unsigned NOT NULL default \'0\' ) TYPE=MyISAM; # # Dumping data for table `hits` # INSERT INTO hits VALUES (\'total\', \'hits\', 45612); Here is the header script: (this works fine. stolen from nuke) include_once \'db.php\'; mysql_query("UPDATE hits SET count=count+1 WHERE (type=\'total\' AND var=\'hits\')"); Here is the footer scriipt: $total_hits = mysql_query("SELECT count FROM hits WHERE type=\'total\' AND var=\'hits\'"); echo "$total_hits"; Thanks in advance! Link to comment https://forums.phpfreaks.com/topic/513-query-error-and-im-going-nuts/ Share on other sites More sharing options...
2muchspam Posted May 26, 2003 Author Share Posted May 26, 2003 Got it!! The result returned by a mysql query is not the data, it\'s a resource that contains the data. To get the data out, you need to \'fetch\' it out. Try it with mysql_fetch_array(). $query_th = mysql_query("SELECT count FROM hits"); $fetched_result = mysql_fetch_array($query_th); $total_hits = $fetched_result[\'count\']; Link to comment https://forums.phpfreaks.com/topic/513-query-error-and-im-going-nuts/#findComment-1715 Share on other sites More sharing options...
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