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Ander_3

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Posts posted by Ander_3

  1. index.php: $selectSql='select '. 'Transactions.TransID, '. 'Transactions.OrderID, '. 'Transactions.DVDID, '. 'Transactions.DateOut, '. 'Transactions.DateDue, '. 'Transactions.DateIn, '. 'Customers.CustFN, '. 'Customers.CustLN, '. 'DVDs.DVDName '. 'from Transactions, Orders, Customers, DVDs '. 'where Orders.OrderID=Transactions.OrderID '. 'and Customers.CustID=Orders.CustID '. 'and DVDs.DVDID=Transactions.DVDID '. 'order by Transactions.OrderID desc, Customers.CustLN asc, '. 'Customers.CustFN asc, '. 'Transactions.DateDue desc, DVDs.DVDName asc;';//Execute the SQL query$result=mysql_query($selectSql, $link);

  2. I forgot to put this code in. This is from update.php. The code looks very similiar to the code I posted previously but this one does not have any errors: <?php//Build the update statement$sqlString='update Transactions set OrderID='.$orderId.', DVDID='.$dvdId.', DateOut="'.$dateOut.'", DateDue="'.$dateDue.'", DateIn="'.$dateIn.'" where TransID='.$transactionId.'';//Execute the update statementmysql_query($sqlString) or die('Error in query $sqlString' .mysql_error());?>

  3. I keep getting this error and can't figure out how to fix it... Error in query $sqlString Cannot add or update a child row: a foreign key constraint fails (`dvdrentals`.`transactions`, CONSTRAINT `transactions_ibfk_2` FOREIGN KEY (`DVDID`) REFERENCES `dvds` (`DVDID`)) This is the code causing the error: <?php//Build the insert statement$sqlString='insert into Transactions(OrderID, DVDID, DateOut, DateDue) values('.$orderId.', '.$dvdId.', "'.$dateOut.'", "'.$dateDue.'")';//Execute the insert statementmysql_query($sqlString) or die('Error in query $sqlString '.mysql_error());?>Thank you!

  4. I've just begun trying to learn php and MySQL and am having problems running some example code in a beginner book.I have tried everything I know to get the program working but keep getting an 'undefined variable: dateIn' error.The code is included from the beginning of program down to the error line. The line throwing the error is typed in red. Could someone please take a look and see if I have missed something?Thank you! <?php//connect to server or return an error$link=mysql_connect('localhost', 'mysqlapp', 'pw1')or die('Could not connect: '.mysql_error());//Select a database or return an errormysql_select_db('DVDRentals', $link);//Initialize an error-related variable$error='';//initialize variables with parameters retrieved from the posted formif(isset($_POST['command'])){ $command=$_POST['command'];}if(isset($_POST['transaction_id'])){ $transactionId=$_POST['transaction_id'];}if(isset($_POST['date_due'])){ $dateDue=$_POST['date_due'];}if(isset($_POST['order_id'])){ $orderId=$_POST['order_id'];}if(isset($_POST['dvd_id'])){ $dvdId=$_POST['dvd_id'];}if(isset($_POST['date_out'])){ $dateOut=$_POST['date_out'];}if(isset($_POST['date_in'])){ $dateIn=$_POST['date_in'];}//Process the save and savenew commandsif((strcmp('save', $command)==0)||(strcmp('savenew', $command)==0)){ //Check for missing parameters if($orderId==-1){ $error.='Please select an \'Order\'<br />'; } if($dvdId==-1){ $error.='Please select a \'DVD\'<br />'; } if(($dateOut==null)||(strlen($dateOut)==0)){ $error.='Please enter a \'Date Out\' Value<br />'; } if(($dateDue==null)||(strlen($dateDue)==0)){ $error.='Please enter a \'Date Due\' Value<br />'; } if(strlen($error)==0){ //Reformat dates so that they are compatible with the MySQL format if($dateOut!=null){ $dateOut=substr($dateOut, 6, 4).'-'.substr($dateOut, 0, 2).'-'.substr($dateOut, 3, 2); } if($dateDue!=null){ $dateDue=substr($dateDue, 6, 4).'-'.substr($dateDue, 0, 2).'-'.substr($dateDue, 3, 2); } if($dateIn!=null){ $dateIn=substr($dateIn, 6, 4).'-'.substr($dateIn, 0, 2).'-'.substr($dateIn, 3, 2); }

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