creating a table via mysql

[Fire Juggler]

I have this:
$mysql_link = mysql_connect($hostname, $username, $password);
mysql_select_db($databasename);
$create = "CREATE TABLE 'photosgigs' ('id' INT(4) NOT NULL AUTO_INCREMENT,
'table_gal' VARCHAR(100) NOT NULL, 'gallery' VARCHAR(100) NOT NULL, PRIMARY
KEY ('id'))";
mysql_query($create,$mysql_link) or die(failed);

with $hostname, $username & $password set.

This won't create a table, can you give me any ideas y?
Thanks

--
Kathryn (Fire Juggler)
http://www.firejugglers.34sp.com
http://www.cornwalljugglers.co.uk

[Gordon Burditt]

>$mysql_link = mysql_connect($hostname, $username, $password);

mysql_select_db($databasename);
$create = "CREATE TABLE 'photosgigs' ('id' INT(4) NOT NULL AUTO_INCREMENT,
'table_gal' VARCHAR(100) NOT NULL, 'gallery' VARCHAR(100) NOT NULL, PRIMARY
KEY ('id'))";
mysql_query($create,$mysql_link) or die(failed);

with $hostname, $username & $password set.

This won't create a table, can you give me any ideas y?

Use print mysql_error() to find out why.

My guess is that the quotes should be backquotes. Look at the error
message to be sure. You also might not have sufficient privileges
to create tables.

Gordon L. Burditt

[Marcin Dobrucki]

Fire Juggler wrote:

I have this:
$mysql_link = mysql_connect($hostname, $username, $password);
mysql_select_db($databasename);
$create = "CREATE TABLE 'photosgigs' ('id' INT(4) NOT NULL AUTO_INCREMENT,
'table_gal' VARCHAR(100) NOT NULL, 'gallery' VARCHAR(100) NOT NULL, PRIMARY
KEY ('id'))";
mysql_query($create,$mysql_link) or die(failed);

Giving you have sufficient privileges to create the table, I'd
completely remove the single quotes.

I think you also need to provide default values for both table_gal
and gallery since you are declaring them "not null".

/Marcin