Here is the lean, corrected code:
<?php
$link = mysql_connect( $site, $id, $pass ) or die ("error connecting");
echo 'connection ok<br>';
mysql_select_db( $dbname, $link) or die ("eror selecting db");
echo 'selected db ok'<br>;
$result = mysql_query( 'Select * From newsletter_subscribers',$link) or die
("error query");
// some here code to disply result
?>
It is easier to evaluate each time the return of the functions using the
expression "or die...".
Also some sintax error:
resource mysql_query(string query [, connection id], int result mode])
you used the mysql_select_db() return instead of mysql_connect() return
value (altough it is optional), also it is not necessary to write ";" at the
end of the query string inside the function.
Bye
"Joshuah Goldstein" <mrbiglesworth3@xxxxxxxxx> ha scritto nel messaggio
news:20040615070106.51066.qmail@xxxxxxxxxxxxxxxxxxxxxxxxxx
>
> To: php-db-digest@xxxxxxxxxxxxx
> MIME-Version: 1.0
> Content-Type: text/plain; charset=us-ascii
>
> I'm trying this query:
>
> $link = mysql_connect( $site, $id, $pass );
> if (!$link) {
> die('Could not connect: ' . mysql_error());
> }
> echo 'Connected successfully<br>';
>
> $db = mysql_select_db( $dbname, $link);
> if( !$db ) {
> echo "DB false<br>";
> exit;
> }
>
> $result = mysql_query( 'Select * From
> newsletter_subscribers;', $db );
>
> if (!$result) {
> echo "DB Error, could not list tables<br>";
> echo 'MySQL Error: ' . mysql_error();
> exit;
> }
>
>
> with this output:
>
> Connected successfully
> DB Error, could not list tables
> MySQL Error:
>
> I dont understand why there is an error, but it prints
> nothing for the error. Any ideas? Thanks, Josh
>
>
>
>
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