winmastergames Posted February 4, 2008 Share Posted February 4, 2008 Well sorry to say but i need a bit more help a deleted the idea of my last script and made a new one i just need 1 thing fixed in it though it comes up with this error Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\swd\ser1\fetch\index.php on line 24 Here is my Code: <html> <head> </head> Your Website ID is: <?php //show website ID, website copyright winmastergames 2008 echo $_GET["website"]; ?> <br> <?php //Show Website, website copyright winmastergames 2008 // connect to db // select db $db_host = "localhost"; $db_username = "root"; $db_password = "com12345"; $db_name = "swd"; $websiteid = $_GET["website"]; $conn = mysql_connect($db_host, $db_username, $db_password) or die(mysql_error()); mysql_select_db($db_name, $conn) or die(mysql_error()); $myquery = mysql_query("select * from ($websiteid)"); while($r=mysql_fetch_array($myquery)) { $rowname=$r["name"]; // put your field's name is where "their_website" is $rowurl=$r["url"]; // put your field's name is where "their_link" is $rowdescription=$r["description"]; // put your field's name is where "their_description" is echo "$rowname | "; echo "<a href=\"$rowurl\">$rowurl</a> | "; echo "$rowdescription<br>"; } ?> </html> I just cant figure it out?? Please help Link to comment https://forums.phpfreaks.com/topic/89426-help-with-my-phpmysql-script/ Share on other sites More sharing options...
revraz Posted February 4, 2008 Share Posted February 4, 2008 $myquery = mysql_query("select * from ($websiteid)") or die (mysql_error()); Link to comment https://forums.phpfreaks.com/topic/89426-help-with-my-phpmysql-script/#findComment-457929 Share on other sites More sharing options...
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