<![CDATA[Query sql con variabile]]>Ciao ho un errore eseguendo questa sql

Errore:Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in .....

$sql = mysql_query("SELECT option_name FROM 'wp_'" . $blog_id . "'_options' WHERE option_name='blogname'");
$dati = mysql_fetch_array ($sql);

La variabile $blog_id viene ricavata poco prima, ma come scriverla?

Help!:ciauz:

]]>https://connect.gt/topic/47858/query-sql-con-variabileRSS for NodeSat, 11 Apr 2026 05:34:03 GMTSat, 22 Sep 2007 19:41:57 GMT60<![CDATA[Reply to Query sql con variabile on Mon, 24 Sep 2007 13:41:19 GMT]]>Hai messo qualche apice di troppo. Prova se funziona così
[php]
$sql = mysql_query("SELECT option_name FROM wp_" . $blog_id . "_options WHERE option_name='blogname'");
[/php]

]]>https://connect.gt/post/605854https://connect.gt/post/605854Mon, 24 Sep 2007 13:41:19 GMT