[SOLVED] Data Output

will_1990 · 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!

Maq · March 4, 2009

Is this your entire code?

will_1990 · 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!

wildteen88 · 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");

}

Maq · 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);

will_1990 · 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

will_1990 · March 4, 2009

found it, it was a curly brace! all sorted now, thanks for your help everyone!!

Maq · 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...

will_1990 · March 4, 2009

No i didnt, i know it is wierd!

Sign In

[SOLVED] Data Output

Recommended Posts

will_1990

Link to comment

Share on other sites

Maq

Link to comment

Share on other sites

will_1990

Link to comment

Share on other sites

wildteen88

Link to comment

Share on other sites

Maq

Link to comment

Share on other sites

will_1990

Link to comment

Share on other sites

will_1990

Link to comment

Share on other sites

Maq

Link to comment

Share on other sites

will_1990

Link to comment

Share on other sites

Archived

Browse

Activity

Important Information