will_1990 Posted March 4, 2009 Share Posted March 4, 2009 Why do i get a blank page now i have tried to print some data from the database to a table?! I cant see for the life of me what ive missed! // query $q ="SELECT Loan.BorId, Borrower.BorName, Loan.BcId, BookTitle.BtName, Loan.DateOut, Loan.DateDue, Loan.DateBack FROM Borrower, BookTitle, Loan, BookCopy WHERE Borrower.BorId = Loan.BorId AND BookTitle.BtId = BookCopy.BtId AND BookCopy.BcId = Loan.BcId AND Loan.DateBack > Loan.DateDue ORDER BY Borrower.BorName ASC"; $res1 = mysql_query($q) or die (mysql_error()); echo mysql_num_rows($res1); if ($res1) { //If it ran OK, display the records. while($row = mysql_fetch_array($res1)) { echo "<table border='1' cellpadding=10> <tr> <td class='table'> <center><b> Borrower ID </b></center> </td> <td class='table'> <center><b> Borrower Name </b></center> </td> <td class='table'> <center><b> Book Copy ID </b> </center> </td> <td class='table'> <center><b> Book Title </b></center> </td> <td class='table'> <center><b> Publisher ID </b></center> </td> <td class='table'> <center><b> Date Out </b></center> </td> <td class='table'> <center><b> Date Due </b></center> </td> <td class='table'> <center><b> Date Back </b></center> </td> </tr> <tr> <td class='table'> <center> ".$row['BorId']." </center> </td> <td class='table'> <center> ".$row['BorName']." </center> </td> <td class='table'> <center> ".$row['BcId']." </center> </td> <td class='table'> <center> ".$row['BtName']." </center> </td> <td class='table'> <center> ".$row['PubId']." </center> </td> <td class='table'> <center> ".$row['DateOut']." </center> </td> <td class='table'> <center> ".$row['DateDue']." </center> </td> <td class='table'> <center> ".$row['DateBack']." </center> </td> </tr> </table>"; I feel so stupid many thanks to who ever spots what ive done wrong! Quote Link to comment https://forums.phpfreaks.com/topic/147958-solved-data-output/ Share on other sites More sharing options...
Maq Posted March 4, 2009 Share Posted March 4, 2009 Is this your entire code? Quote Link to comment https://forums.phpfreaks.com/topic/147958-solved-data-output/#findComment-776576 Share on other sites More sharing options...
will_1990 Posted March 4, 2009 Author Share Posted March 4, 2009 No, should i post the entire code, i only posted what i thought would be relivent. Here it is incase its helpful! <?php session_start(); if(!session_is_registered(myusername)){ header("location:index_Staff.php"); } ?> <html> <head> <title> Children's online library -Late Book Return Results </title> </head> <body> <h5 align="left"><?print 'You are Logged in as:';?> <?print $_SESSION['myusername'];?></h5> <h1 align=center > Children's Online Library</h1> <p align = center> <h1 align =center>Borrower Access To Children's Online Library</h1> <table align=center cellpadding="10" border="1"> <tr> <th> <a href="index_Borrower.php"> Homepage</a> </th> <th> <a href="SearchForABook.php">Search For A Book</a> </th> <th> <a href="BorrowABook.php">Borrow A Book</a> </th> <th> <a href="ReturnABook.php">Return A Book</a> </th> <th> <a href="RenewABook.php">Renew A Book</a> </th> <th> <a href="logout.php">Logout</a> </th> </tr> </table> </p> <br> </br> <h2 align=center> Late Book Return Results</h2> <?php #view all late book returns //connect to the database //database information $host = "stocks"; // Host name $username = "njpeddle"; // Mysql username $password = "mysql5"; // Mysql password $db_name = "njpeddle"; // Database name mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); echo '<h1> All Late Book Returns To Current Date</h1>'; // query $q ="SELECT Loan.BorId, Borrower.BorName, Loan.BcId, BookTitle.BtName, Loan.DateOut, Loan.DateDue, Loan.DateBack FROM Borrower, BookTitle, Loan, BookCopy WHERE Borrower.BorId = Loan.BorId AND BookTitle.BtId = BookCopy.BtId AND BookCopy.BcId = Loan.BcId AND Loan.DateBack > Loan.DateDue ORDER BY Borrower.BorName ASC"; $res1 = mysql_query($q) or die (mysql_error()); echo mysql_num_rows($res1); if ($res1) { //If it ran OK, display the records. while($row = mysql_fetch_array($res1)) { echo "<table border='1' cellpadding=10> <tr> <td class='table'> <center><b> Borrower ID </b></center> </td> <td class='table'> <center><b> Borrower Name </b></center> </td> <td class='table'> <center><b> Book Copy ID </b> </center> </td> <td class='table'> <center><b> Book Title </b></center> </td> <td class='table'> <center><b> Publisher ID </b></center> </td> <td class='table'> <center><b> Date Out </b></center> </td> <td class='table'> <center><b> Date Due </b></center> </td> <td class='table'> <center><b> Date Back </b></center> </td> </tr> <tr> <td class='table'> <center> ".$row['BorId']." </center> </td> <td class='table'> <center> ".$row['BorName']." </center> </td> <td class='table'> <center> ".$row['BcId']." </center> </td> <td class='table'> <center> ".$row['BtName']." </center> </td> <td class='table'> <center> ".$row['PubId']." </center> </td> <td class='table'> <center> ".$row['DateOut']." </center> </td> <td class='table'> <center> ".$row['DateDue']." </center> </td> <td class='table'> <center> ".$row['DateBack']." </center> </td> </tr> </table>"; //echo mysql_result($res1, 5); //mysql_free_result ($res1); //Free up the resources. } else { // If it did not run OK. //Public message: echo '<p> Unfortunately we could not retrieve the results. We apologise for any inconvience.</p>'; // Debugging message: echo '<p>' . mysqli_error($db_name) . '<br /><br />Querey: ' .$q. '</p>'; } //End of it ($res1) IF. //mysqli_close($db_name); // Close the database connection. ?> Thanks! Quote Link to comment https://forums.phpfreaks.com/topic/147958-solved-data-output/#findComment-776577 Share on other sites More sharing options...
wildteen88 Posted March 4, 2009 Share Posted March 4, 2009 Why are using mysqli_error here / Debugging message: echo '<p>' . mysqli_error($db_name) . '<br /><br />Querey: ' .$q. '</p>'; } //End of it ($res1) IF. when in the rest of your code you're using the standard mysql library functions? mysqli_* and mysql_* functions are not cross compatible. You should also realise session_is_registered or session_register are depreciated and should only be used if register_globals is enabled. You should change if(!session_is_registered(myusername)){ header("location:index_Staff.php"); } to if(!isset($_SESSION['myusername'])){ header("location:index_Staff.php"); } Quote Link to comment https://forums.phpfreaks.com/topic/147958-solved-data-output/#findComment-776583 Share on other sites More sharing options...
Maq Posted March 4, 2009 Share Posted March 4, 2009 For starters: if(!session_is_registered(myusername)){ has been deprecated as of 5.3 so you have to use: if(!isset($_SESSION['myusername'])){ Put this at the top of your page as your first PHP statements and see if there are any errors. ini_set ("display_errors", "1"); error_reporting(E_ALL); Quote Link to comment https://forums.phpfreaks.com/topic/147958-solved-data-output/#findComment-776587 Share on other sites More sharing options...
will_1990 Posted March 4, 2009 Author Share Posted March 4, 2009 Thanks for the help, i have made those changes but still have no output on screen what so ever! Its like ive missed a curly brace or a semi colon somewhere! Thanks anyway Quote Link to comment https://forums.phpfreaks.com/topic/147958-solved-data-output/#findComment-776611 Share on other sites More sharing options...
will_1990 Posted March 4, 2009 Author Share Posted March 4, 2009 found it, it was a curly brace! all sorted now, thanks for your help everyone!! Quote Link to comment https://forums.phpfreaks.com/topic/147958-solved-data-output/#findComment-776628 Share on other sites More sharing options...
Maq Posted March 4, 2009 Share Posted March 4, 2009 And you didn't get any errors when you put: ini_set ("display_errors", "1"); error_reporting(E_ALL); at the top of your script? Hmm, weird... Quote Link to comment https://forums.phpfreaks.com/topic/147958-solved-data-output/#findComment-776633 Share on other sites More sharing options...
will_1990 Posted March 4, 2009 Author Share Posted March 4, 2009 No i didnt, i know it is wierd! Quote Link to comment https://forums.phpfreaks.com/topic/147958-solved-data-output/#findComment-776691 Share on other sites More sharing options...
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