buchberg Posted February 18, 2012 Share Posted February 18, 2012 Windows 7, Apache 2.2.21 / PHP 5.3.5. What does the below error mean? I am new and fighting through this particular piece of code. Thank you in advance for your help. Warning: mysql() expects parameter 3 to be resource, string given in C:\website\do_createtable.php on line 3 <? $db_name="booster"; $connection=mysql("localhost", "USERNAME", "PASSWORD") or die (mysql_error()); $db=mysql_select_db($db_name, $connection) or die (mysql_error()); $sql="CREATE TABLE $_POST[table_name] ("; for ($i=0; $i < count($_POST[field_name]); $i++) { $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i]; if ($_POST[field_length][$i] != "") { $sql .= " (".$_POST[field_length][$i]."),"; } else { $sql .= ","; } } $sql=substr($sql,0,-1); $sql .= ")"; $result=mysql_query($sql,$connection) or die (mysql_error()); if ($result) { $msg="<P>".$_POST[table_name]." has been created!</P>"; } ?> <HTML> <HEAD> <TITLE>Create a Database Table: Step 3</TITLE> </HEAD> <BODY> <h1>Adding table to <? echo "$db_name"; ?>....</h1> <? echo "$msg"; ?> </BODY> </HTML> Quote Link to comment https://forums.phpfreaks.com/topic/257228-php-form-for-creating-table-in-mysql/ Share on other sites More sharing options...
joel24 Posted February 18, 2012 Share Posted February 18, 2012 you want mysql_connect(), not mysql() Quote Link to comment https://forums.phpfreaks.com/topic/257228-php-form-for-creating-table-in-mysql/#findComment-1318540 Share on other sites More sharing options...
buchberg Posted February 18, 2012 Author Share Posted February 18, 2012 joel24--YOU are the man! Thank you so much and have a good day....buchberg Quote Link to comment https://forums.phpfreaks.com/topic/257228-php-form-for-creating-table-in-mysql/#findComment-1318595 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.