From: "Radwan Aladdin" <[EMAIL PROTECTED]>
> I made that.. but there is something strange!!
It's strange you can't follow instructions...
> I SELECTED field from the database.. but it is not showing what it
contains!!
>
> This is what is shown on the screen :
>
> UPDATE accounts SET LogoutTime=1075313199SELECT
LoginTime,Distance,LessonNumber FROM accounts WHERE UserName='' AND
Password=''UPDATE accounts SET LessonNumber=LessonNumber + 1
> Query error: You have an error in your SQL syntax near '0') WHERE
UserName= AND Password=)' at line 1
If you had printed out all the queries, you'd realize that error is coming
from $query4.
> > > > $query4 = "UPDATE accounts SET (Distance='" . $LogoutTime -
> > > > $RightLoginTime . "') WHERE UserName=" . $UserName . " AND
Password="
> > > >. $Password . ")";
$UserName and $Password have no value. Your "distance" calculation doesn't
make any sense, either, as you're subtracting a string from an integer...
> > > $LogoutTime = date("U");
> > > $RightLoginTime = 'LoginTime';
So you're setting distance equal to 1034566 - "LoginTime" (for example),
which is just going to come out to $LogoutTime (since a string converted to
an integer is zero in this case).
---John Holmes...
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