[PHP] Re: syntax help
If you want to actually want the variables $flyertotal, $emailtotal, and
$phonetotal, use the line:
${$array[0].total} = $row[0];
But as you defined $array, that will actually create the variables
$Flyertotal, $Emailtotal, and $Phonetotal (note the caps).
On 2/13/2004 6:10 AM, Bob pilly wrote:
$array[$i].total=$row[0];-- problem here
so for the above code i would like to create three
variable $flyertotal,$emailtotal $phonetotal
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[PHP] Re: syntax help~~~
Hi..
I just wanna laugh really want to laugh WUWUAHAHAHAHAHA
I have figure out the solution amazingly! here will be the code I am running
LoL
?
mysql_connect('localhost','coconut','tkming') or die (Unable to connect to SQL
Server);
mysql_select_db('helpwatch') or die (Unable to select database);
$temp = $username.watch
?
?
$watchlist_query = mysql_query( Create Table .$temp. ( WId int auto_increment
not null, QId int not null, Primary Key (WId) ) ) or die (Error! Cannot
create table ! . mysql_error() );
?
I separated the query code with the database connection coding LoL after that..
its work!
Thanks for everyone who help me :)
Regards
Kok Ming
Arcadius A. wrote:
Hello !
We could keep things simpler ...
Why not give a try to this ?:
$watchlist_query = mysql_query( Create Table $temp ( WId int Not
Nullauto_increment, QId int not null, Primary Key (WId) ) ) or die (Error
! Cannot create table ! . mysql_error() );
Hope it would work :o)
(At least the mysql_error() would show you the error code )
Arcad
Coconut Ming [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi
I am having the problem in the coding below
?
mysql_connect('localhost','123','123') or die (Unable to connect to SQL
Server);
mysql_select_db('Helpwatch') or die (Unable to select database);
$temp = $username.watch;
$watchlist_query = mysql_query(Create Table $temp(WId int Not Null
auto_increment, QId int not null, Primary Key (WId)););
?
So.. in the above coding. I need to create a table. The table name
should be a user-define name + watch
I acquire the $username correctly and the variable $temp is working fine
when I try to echo the value of it.
Just say. I enter my username as coconut so I wish the mysql query to
create a table called coconutwatch
but I can't do that.. Because of the syntax error, the coding I have
underline is where the parser told me that is an error there.
I have playing around with it for 2 hours and more... but I cant solve
it.. Anyway help is greatly appreciated.
Thanks in advance.
Sincerely
Kok Ming
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Re: [PHP] Re: syntax help~~~
At 06.08.2001 10:31, you wrote:
Hi..
I just wanna laugh really want to laugh WUWUAHAHAHAHAHA
I have figure out the solution amazingly! here will be the code I am
running
LoL
it works without separating the code if you add the ; at the end of the
$temp = $username.watch line...
?
mysql_connect('localhost','coconut','tkming') or die (Unable to connect
to SQL
Server);
mysql_select_db('helpwatch') or die (Unable to select database);
$temp = $username.watch
?
?
$watchlist_query = mysql_query( Create Table .$temp. ( WId int
auto_increment
not null, QId int not null, Primary Key (WId) ) ) or die (Error! Cannot
create table ! . mysql_error() );
?
--
Andreas D Landmark / noXtension
Real Time, adj.:
Here and now, as opposed to fake time, which only occurs there
and then.
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To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] Re: syntax help~~~
Hello !
We could keep things simpler ...
Why not give a try to this ?:
$watchlist_query = mysql_query( Create Table $temp ( WId int Not
Nullauto_increment, QId int not null, Primary Key (WId) ) ) or die (Error
! Cannot create table ! . mysql_error() );
Hope it would work :o)
(At least the mysql_error() would show you the error code )
Arcad
Coconut Ming [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi
I am having the problem in the coding below
?
mysql_connect('localhost','123','123') or die (Unable to connect to SQL
Server);
mysql_select_db('Helpwatch') or die (Unable to select database);
$temp = $username.watch;
$watchlist_query = mysql_query(Create Table $temp(WId int Not Null
auto_increment, QId int not null, Primary Key (WId)););
?
So.. in the above coding. I need to create a table. The table name
should be a user-define name + watch
I acquire the $username correctly and the variable $temp is working fine
when I try to echo the value of it.
Just say. I enter my username as coconut so I wish the mysql query to
create a table called coconutwatch
but I can't do that.. Because of the syntax error, the coding I have
underline is where the parser told me that is an error there.
I have playing around with it for 2 hours and more... but I cant solve
it.. Anyway help is greatly appreciated.
Thanks in advance.
Sincerely
Kok Ming
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
