Re: Still error messages!!
| From: | David OBrien | Date: | Wed, 28 Jan 2004 19:00:16 +0000 |
| Subject: | Re: Still error messages!! | ||
| References: | 1 2 3 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-176098@lists.php.net to get a copy of this message | ||
I believe I'd start here
http://www.php.net/manual/en/function.mssql-fetch-row.php
You get the db result from your script but you do nothing with it in your script
A whole lot more reading and coding you need to accomplish.
Good Luck
-Dave
At 01:54 PM 1/28/2004, Radwan Aladdin wrote:
But I mean by $RightLoginTime = 'LoginTime'; the selected field from the database (LoginTime).. First I selected the LoginTime from the database and now I'm trying to name a variable for it to use it.. So what is the correct code? Regards.. -------------------------------------------------- SASSINC Internet Solutions - Arabic Department raladin@emirates.net.ae http://www.SASSINC.Net ----- Original Message ----- From: "John W. Holmes" <holmes072000@charter.net> To: "Radwan Aladdin" <raladin@emirates.net.ae>; <php-general@lists.php.net> Sent: Wednesday, January 28, 2004 10:46 PM Subject: Re: [PHP] Still error messages!! From: "Radwan Aladdin" <raladin@emirates.net.ae>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 itcontains!!This is what is shown on the screen : UPDATE accounts SET LogoutTime=1075313199SELECTLoginTime,Distance,LessonNumber FROM accounts WHERE UserName='' AND Password=''UPDATE accounts SET LessonNumber=LessonNumber + 1Query error: You have an error in your SQL syntax near '0') WHEREUserName= AND Password=)' at line 1 If you had printed out all the queries, you'd realize that error is coming from $query4.Password="$query4 = "UPDATE accounts SET (Distance='" . $LogoutTime - $RightLoginTime . "') WHERE UserName=" . $UserName . " AND$UserName and $Password have no value. Your "distance" calculation doesn't make any sense, either, as you're subtracting a string from an integer.... $Password . ")";So you're setting distance equal to 1034566 - "LoginTime" (for example), which is just going to come out to $LogoutTime (since a string converted to an integer is zero in this case). ---John Holmes... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php$LogoutTime = date("U"); $RightLoginTime = 'LoginTime';