Need Help With Displaying Data

blah

<html>
<head>
<title>Site Title</title>

<?php
function mysql_die($error = "unknown")
{
echo "<H1>Error:". $error."</H1>";
exit;
}
$db = mysql_connect("localhost","username","password") or mysql_die("Unable to connect");
mysql_select_db("databasename",$db);

$requete = "SELECT * FROM article WHERE id=".$id;
$result = mysql_query ($requete,$db)or mysql_die("No record");
$article =mysql_fetch_object($result);
mysql_free_result($result);

</head>
<body>
<?php echo $article->id ?>
<BR>
<?php echo $article->author ?>
<BR>
<?php echo $article->size ?>
<BR>
<?php echo $article->date ?>
</body>
</html>

Does anyone know why this code won't work when I go to:
http://domain.com/site/index.php?id=1

cgraz

Try changing your select statment too

$requete = "SELECT * FROM article WHERE id='$id'";

blah

Thanks for your help 🙂
but for some reason, it still doesn't work

cgraz

I've never used mysql_fetch_object(), but I have two suggestions. You don't have any semi-colons after your echo's, or, try using mysql_fetch_array() instead.

<? 

$requete = mysql_query("SELECT * FROM article WHERE id='$id'");
$a_row = mysql_fetch_array($requete);

?> 

</head> 

<body> 
<?php echo $a_row["id"]; ?> 
<BR> 
<?php echo $a_row["author"]; ?> 
<BR> 
<?php echo $a_row["size"]; ?> 
<BR> 
<?php echo $a_row["date"]; ?> 


</body> 
</html>

Let me know how that works.

Cgraz

blah

I copied the code and replaced my code with the one that you posted but it still doesn't work.
I wonder what I'm doing wrong...
It seems to work when I replaced

id='$id'"
with
id=1"

but when I don't do it, it doesn't work

cgraz

What happens if you echo $id on your page.

<? echo "ID " . $id . ""; ?>

That'll let you know if there is a value assigned to it. If there's no value, you can try

<? 

$id = $_GET['id']; 
echo "ID " . $id . ""; 

?>

blah

Wow!
It works! Thank you so much!
You're a genius! ^_

blah

One thing I noticed about the code that I was using was that it only allowed one entry to be displayed at a time. Is there a way to modify the code so that more than one can be shown?
Instead of $id, I'll use $type so more than one could be displayed
It still works but it only displays the first entry.

I heard that I'm supposed to add "while"
I'm not too sure where though, I'm a beginner in this.

cgraz

</head> 

<body> 

<? 

$requete = mysql_query("SELECT * FROM article WHERE id='$id'");
while ($a_row = mysql_fetch_array($requete)) {

?> 

<?php echo $a_row["id"]; ?> 
<BR> 
<?php echo $a_row["author"]; ?> 
<BR> 
<?php echo $a_row["size"]; ?> 
<BR> 
<?php echo $a_row["date"]; ?> 

<?

}

?>


</body> 
</html>

use $_get['id'] if that returned a value and $id didn't. same for the other variables.

blah

Thank you for all your help!
Is there a way to make it work when I search the same file three ways?

http://domain.com/site/index.php?id=1
http://domain.com/site/index.php?type=1
http://domain.com/site/index.php?author=1

Because I find that I have to do change
type=$type
to
id=$id
or to
author=$author

in the file so I would need three different files for 3 different searches.

I also found that
http://domain.com/site/index.php?type=1&id=4
would just give me a list of the same files.

How can I fix this problem?