Newbie Q

Anon

Hi, I have just started coding PHP tonight, I have an intermediate knowledge of Perl going in to this. But here is my short code:

<body>

<?php
$db = mysql_connect("localhost", "****", "**");
$flg = mysql_select_db("guestbook",$db);

$result = mysql_query("SELECT * FROM Entries",$db);

printf("Name: %s \n", mysql_result($result,0,"Name"));
printf("Email: %s \n", mysql_result($result,0,"Email"));
printf("Page: %s \n", mysql_result($result,0,"Webpage"));
printf("Message: %s \n", mysql_result($result,0,"Message));
?>

</body>

The 'parse error' it gives me occurs at line 17, which is the tag. I basically copied the code right out of a Webmonkey tutorial, and so am lost as to what the problem could be. Any ideas?

Anon

When I posted it didn't print the tags I wanted it to, the error is at the final < /html> tag...

[deleted]

You have left out a closing quotation mark. This:

printf("Message: %s \n", mysql_result($result,0,"Message));

should be this:

printf("Message: %s \n", mysql_result($result,0,"Message"));

As a tip, when you get an error on the last line of your code, normally it is a missing quote....

Matt Wade
http://codewalkers.com/