cayley wrote:
>
> $id = array( 1, 2, 3, 6, 10 );
> $sql = "select name from resource where id in ( ? )";
>
Try this:
$sql = $db->quoteInto("select name from resource where id in ( ? )", $id);
The quoteInto() method recognizes an array argument as a special case. It
joins the values of the array together in a comma-separated list, and
substitutes that for the placeholder character. This is useful in cases
like the one you describe.
Regards,
Bill Karwin
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