Dodgy PHP first attempt off-putting-ness

Bailz

I'm trying to have a basic news retrieval service that can pull stories out of the MySQL database. I have an example story in the database and the following code in the page:

<?php
$db = mysql_connect("localhost","username","password");
mysql_select_db("user_uk_$db);

$request = "SELECT * FROM 'news' WHERE id=".$id;
$result = mysql_query ($request,$db);
$article =mysql_fetch_object($result);
mysql_free_result($result);

?>
</head>
<body>
<?php echo $article->date ?>
<br>
<?php echo $article->author ?>
<br>
<?php echo $article->news ?>
</body>
</html>

with username and pass etc replaced with the relevant data respectively.

When I try to implement the script on the web I get this message:

Parse error: parse error in /data/members/free/tripod/uk/b/o/r/bornrun/htdocs/news.php on line 9

I followed the Lycos tutorial for implementing the database in the page so don't blame me for crappy code.

As you can probably tell i'm not in the best of moods so please be nice!

Regards,
Bailz.

YappyDog

You got a beginning quote, but no ending quote on this line.

mysql_select_db("user_uk_$db);

Bailz

OK...silly mistake number 1 pointed out. Now its time to point out number 2.

My code now looks like this:

<?php

function mysql_die($error = "unknown")
{
echo "<h1>Error:". $error."</h1>";
exit;
}

$db = mysql_connect("localhost","user","pass") or mysql_die("Unable to connect");
mysql_select_db("user_uk_$db");

$request = "SELECT * FROM 'news' WHERE id=1";
$result = mysql_query ($request,$db);
$article =mysql_fetch_object($result) or mysql_die("No record");
mysql_free_result($result);

?>
</head>
<body background="art/bground.JPG">
<?php echo $article->date ?>
<br>
<?php echo $article->author ?>
<br>
<?php echo $article->news ?>
</body>
</html>

But I am getting the error 'No Record'. I have created the database and there is clearly a record that exists and corresponds with the name contained in the .php file.

Please Help!

Regards,
Bailz.

YappyDog

I think the problem is still in your select db statement. You've got:

mysql_select_db("user_uk_$db");

The normal format is mysql_select_db($mydatabase, $db);

Not sure what your db name is:

mysql_select_db("user_uk", $db); ? ? ? ?

Bailz

Right ok, I've changed it as you said, but I am still getting the same error.

I now have:

mysql_select_db("bornrun_uk_db",$db);

Thanks,
Bailz.