Re: CREATE TABLE QUERY PHP/MySQL

answers inline Ignatius ____________________________________________ ----- Original Message ----- From: "Jonathan" <dbphp@ftbwebsolutions.com> To: <php-db@lists.php.net> Sent: Friday, December 06, 2002 10:39 PM Subject: RE: [PHP-DB] CREATE TABLE QUERY PHP/MySQL > > > Granting access makes sense. > > Now what is the difference between the user information I put into the > mysql_connect function and "PHP user"? > one and same thing. > If there is a difference , how would I grant the access via a query and > how would it affect the new users who will use the application when it > is given to my client? > > Also, a preview reply mentioned mysql_error(). How would I use function > to verify that the table was created? Do I simply write > > If (mysql_error() != "") > Return false > > A typical way to use it is mysql_***( *** ) or die( mysql_error() ) ; > -----Original Message----- > From: Ignatius Reilly [mailto:ignatius.reilly@free.fr] > Sent: Friday, December 06, 2002 3:34 PM > To: php-db@lists.php.net > Subject: Re: [PHP-DB] CREATE TABLE QUERY PHP/MySQL > > Very likely you have to GRANT the CREATE permission to your PHP user > account. > > Also investigate the CREATE TEMPORARY TABLE grant status available since > version 4+. Very convenient. > > Ignatius > ____________________________________________ > ----- Original Message ----- > From: "Jonathan" <dbphp@ftbwebsolutions.com> > To: <php-db@lists.php.net> > Sent: Friday, December 06, 2002 10:00 PM > Subject: [PHP-DB] CREATE TABLE QUERY PHP/MySQL > > > > This is the first time I am trying to create a table in MySQL based on > > user input. > > > > Here is the scenario, > > > > The user will enter in the name of a stage, e.g. Harley Davidson. > > I then look for the first space (to indicate the first word), make it > > into lower case and then do the following: > > > > $new_stage_name = "stage_".$var; > > > > This will give $new_stage_name the value of "stage_harley". > > > > That works fine. > > > > Now I want to run the following query: > > > > $SQL = "CREATE TABLE ".$new_stage_name." (set values and their > > properties)"; > > > > If I run this query using PHPMyAdmin, minus the var of course, the > table > > is created. > > > > When I run it in my code, nothing happens. I do a die($SQL) to see the > > final outcome of the query and it's correct and is what I am using to > > run the query in PHPMyAdmin. > > > > This is what my code is doing: > > > > I have class RecordSet which makes the db connection and its > constructor > > takes the $SQL as an arg. > > > > So, $SQL = "what is listed above"; > > $objRecordSet = new RecordSet($SQL); > > //I used this all the time with all kinds of other Select, Insert, > > Update, and Delete queries and it works perfect. > > > > For one, I don't know how to test whether the table has been created > or > > not. I know I could probably do some sort of mysql_list_tables and > look > > for it or whatever but I was looking for something similar to > > > > If (mysql_affected_rows() == -1) > > Because that is very easy. > > > > And ideas as to why the table is not being created. > > > > =================== > > Jonathan Villa > > Application Developer > > IS Design & Development > > www.isdesigndev.com > > 414.429.0327 > > =================== > > > > > > > > > > -- > > PHP Database Mailing List (http://www.php.net/) > > To unsubscribe, visit: http://www.php.net/unsub.php > > > > > > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > >