Hyperlink result in php

[hanna1]

Hi,

 

I have a php program where the query result will turn to hyperlink. But my problem is when i click the hyperlink search result it returns to blank page, I have a hard time working on this around. I'm newbie in php please help me thanks.  :'(

 

Below is my code:

 

//hyperlink_result.php (this code is the hyperlink query result)

 

//echo out construct

$sql = "SELECT * FROM search WHERE $construct";

$run = mysql_query($sql);

 

 

$foundnum = mysql_num_rows($run);

if ($foundnum==0)

echo "No results found.";

else

{

echo "$foundnum results found!<p>";

 

while ($runrows= mysql_fetch_array($run))

{

//hyperlink search result

 

 

$title = $runrows['title'];

 

//echo '<p><a href="hyperlink_results.php?id='.$runrows['id'].'">'.$runrows['title'].'</a></p>';

echo "<a href=\"hyperlink_results.php?id=".$runrows['id']."\">".$title."</a>";

 

 

}

}

 

 

 

 

// hyperlink_results.php (this code will fetch hyperlink query)

 

if (isset($_GET['id']))

{

include ("module/conn.php");

$select=mysql_query("SELECT * FROM search WHERE id='".$_GET['id']."'");

$r = mysql_num_rows($select);

 

while($row= mysql_fetch_array($r)){

 

$des = $row['description'];

$title = $row['title'];

 

echo "$des";

echo "$title";

 

 

}

 

}

}

 

 

[wildteen88]

mysql_fetch_array expects a result resource, not the number of rows your query returns. A result resource is returnedby mysql_query. So you need to pass thee $select variable to mysql_fetch_array, not the $r variable

$select=mysql_query("SELECT * FROM search WHERE id='".$_GET['id']."'");
   $r = mysql_num_rows($select);
   
            while($row= mysql_fetch_array($r)){

 

Change $row= mysql_fetch_array($r) to $row= mysql_fetch_array($select) in hyperlink_results.php

 

Also when developing your scripts you should configure PHP so error_reporting is set to E_ALL and set  display_errors to on. You can do this either within your php.ini or within your scripts using

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

Enabling these two directives will display any PHP errors on screen rather than a blank screen.