Member Avatar for whiteyoh

Hi all,

The following code should link 2 tables, showing: student id, surname, firstname, study type, start year.

However, its not :-). its just a white screen. ive got php5.ini with ERROR_REPORTING (E_ALL); in it but im still getting nothing at all.

The tables definitly exist!

have i missed something stupid?

<?php
require_once("myfunctions.php");

 $db = getConnection(); 
 $stmt = $db->query( "SELECT student.studentCode, student.Surname, student.Forename, studyType.typeDesc ".
					" FROM student, studyType ");
 
 	while ( $obj = $stmt->fetchObject()){    	
	
	echo $obj->student.studentCode . ": " . $obj->student.Surname . ": " . $obj->student.Firstname . ": " . $obj->studyType.typeDesc . $obj->studyType.startYear "<br />\n";
	
	}
	}catch( PDOException $e ) {    echo $e->getMessage();}
?>
  • 3 Contributors
  • 5 Replies
  • 115 Views
  • 1 Day Discussion Span
  • Latest Post Latest Post by whiteyoh

Dani AI

Generated

A quick summary and practical checklist that ties the thread together: was right to suggest running the SQL directly, and correctly flagged that a relation/join is required. resolved the immediate problem by adding the missing loop. A blank white page in PHP usually means a fatal parse error or suppressed runtime error — so start by making sure errors are visible and then check for a few common code mistakes illustrated in the snippets above.

Enable visible errors at the top of the script while debugging:

ini_set('display_errors', 1);
error_reporting(E_ALL);

Key things to check (concrete, fast to verify)

  • Syntax problems that cause a parse error: unmatched try/catch, missing concatenation operators, or illegal expressions (for example using dotted-property notation like ->table.field instead of ->field). These stop execution before any output.
  • Suppressed or hidden errors: enable error reporting and check server PHP error logs.
  • Query correctness: run the SQL in your DB client (phpMyAdmin / mysql) as advised.
  • Result handling: ensure you actually iterate the result set (a fetch loop); echoing the raw result resource will not print rows.
  • Mixing APIs: do not use mysql_error() with mysqli — use mysqli_error() or $mysqli->error. Use the correct free_result function name.
  • Column name collisions: alias columns when two tables have the same column name (use AS) and reference the alias in your fetch.

Example PDO pattern (adjust DSN/credentials and column names to your schema):

try {
  $pdo = new PDO('mysql:host=localhost;dbname=yourdb;charset=utf8', 'user', 'pass');
  $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

  $stmt = $pdo->query(
    "SELECT u.id AS uid, u.name, t.label AS type
     FROM users u
     JOIN types t ON u.type_id = t.id"
  );

  while ($row = $stmt->fetch(PDO::FETCH_OBJ)) {
    echo $row->uid . ' ' . $row->name . ' ' . $row->type . "<br>\n";
  }
} catch (PDOException $e) {
  echo 'DB error: ' . $e->getMessage();
}

Final tip: after enabling errors and fixing parse/runtime issues, consider prepared statements and explicit JOINs for clarity and safety.

Recommended Answers

All 5 Replies

Member Avatar for ShawnCplus

If you get no errors and no exception is through but it's still a white page that means the query returned no results. Execute the query directly in MySQL or phpmyadmin or whatever you're using to directly access the database and see if it returns any rows.

Member Avatar for jigarvyas

In addition to this, I believe if you are linking two tables, there needs to be some kind of relation defined between them.
Something like: "Select..... where student.studentCode = studyType.studentCode..."

Although in your case, i agree with ShawnCplus, you definitely ought to try out your query directly first

Member Avatar for whiteyoh

Hi,

I looked at this from an alternative perspective, and i get a little further, but i dont get why no results appear, when there are definitly items in the table.

this will display the grid, but populates nothing........ im baffled, and note, fairly new ish.

I have double checked all the column names and everything matches.

The headers and the grid displays. Im wondering if maybe an sql syntax error perhaps? its linking 2 tables.

<h1>Student Listing</h1>

<?php

// Connect to the mysqldatabase
$link = mysqli_connect("localhost","root","root","university")
or die("ERROR: Could not connect : " . mysql_error());

// Set up a query to select the required fields and records
echo $result = mysqli_query($link, "SELECT studentCode, forename, surname, startYear, typeDesc FROM student, studyType WHERE student.studyTypeID = studyType.studyTypeID")
or die("ERROR:query failed (Student) - " . mysql_error());

$studentResult = mysqli_query($link, $sql);
?>

<!-- Move out of PHP mode and start the table tags (note you
could have stayed in PHP mode and achieved the same
result by placing these tags within echo statements. -->

<table width="70%" border="1">
<tr>
<td width="15%"><strong>Student Code</strong></td>
<td width="25%"><strong>Surname</strong></td>
<td width="25%"><strong>Forename</strong></td>
<td width="20%"><strong>Study Type</strong></td>
<td width="15%"><strong>Start Year</strong></td>
</tr>

<?php

// For every row that was returned by the query, do something.
// In this example, display data from each row within an XHTML table.

// NOTE: The array keys (the field names are the array keys, the values are the field values returned by the SQL query) 

               echo "<tr>\n";
               echo "<td>&nbsp;</td>\n";
               echo "<td>".$studentResult['studentCode']."</td>\n";
               echo "<td>".$studentResult['surname']."</td>\n";
               echo "<td>".$studentResult['forename']."</td>\n";
               echo "<td>".$studentResult['studyType']."</td>\n";
               echo "<td>".$studentResult['startYear']."</td>\n";
               echo "</tr>\n";	
	
mysqli_free_results($studentResult);
// General clean-up of resources (i.e. memory).

mysqli_free_result($studentResult);
mysqli_close($link);

?>

</body>
</html>
Member Avatar for whiteyoh

Hi,

Please excuse my Nativity, ive never thought of ever doing that!!!!! brilliant idea.

I ran

SELECT studentCode, forename, surname, startYear, typeDesc FROM student, studyType WHERE student.studyTypeID = studyType.studyTypeID

directly in mysql and i got the results i expected, so there has to be a common stupid mistake in both scripts.

baffling

Member Avatar for whiteyoh

Hi,

I found the issue, i was missing the fundimental loop for showing every record DOH.

Heres my working code

<?php

      $mysqli = new mysqli('localhost','root','root','university');
   
    // sql query, including the linking of two tables, if it is TRUE
   
   
      if ($result = $mysqli->query("SELECT studentCode, forename, surname, startYear, typeDesc FROM student, studyType WHERE student.studyTypeID = studyType.studyTypeID ORDER BY surname, forename")) {
   
      // if it is true this repeats through all of the records displaying the desired items 
		
   
      echo 'there are '.$result->num_rows.' records.' . "<br>";
  
      // The "fetch_object()" is the new fancy pants OO function of the old mysql_fetch_object() function. 
  
      while ($row = $result->fetch_object()) {
  		
		echo $row->studentCode.' ';
		echo $row->forename.' ';
		echo $row->surname.' ';
		echo $row->startYear.' ';
		echo $row->typeDesc.' ';
		echo "<br>";

 
  
      }
	   
  
      } else {
  
     //error display.  this must be activated in the http conf file error_reporting
  
      // This means the query failed
 
      echo $mysqli->error;
  
      } // if no error, close the connection
  
      $mysqli->close();
  
      ?>
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