[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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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() );
>?>
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Real Time, adj.:
Here and now, as opposed to fake time, which only occurs there
and then.
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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
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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[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
>
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