HI there, I would like to generate a table filled with the result of
an sql command to get a report about stored data.
My query in SQL is like this:
$MyQueryAnreisen="SELECT distinct BP.AnreiseDatum AS AN,
BP.AbreiseDatum AS AB,RQ.name AS ANF, RQ.AnzahlErwachsene AS AnzE,
RQ.AnzahlJugendliche AS AnzJ, RQ.AnzahlKinder AS AnzK, BS.name AS BK,
LS.Zimmernummer AS Zimmer,RQ.meal0_id, RQ.meal1_id, RQ.meal2_id,
concat( CS.name, ': ', CS.Firma, ': ', CS.Vorname, ', ', CS.Nachname )
AS Kunde FROM `booking_positions` AS BP, `bookings` AS BS, `lodgings`
AS LS, `customers` AS CS, `requests` AS RQ, meals as MS WHERE 1 AND
bs.id = bp.`booking_id` AND `lodging_id` = LS.id AND BS.customer_id =
CS.id AND RQ.id = BS.request_id ORDER BY AN";
current I do:
if (!mysql_connect("localhost", "xxx", "pwdpwdpwd")) {
die ("Verbindung zum Server fehlgeschlagen.");
}
if (!mysql_select_db("mydatabase")) {
die ("Datenbank nicht gefunden.");
}
$read_cursor = mysql_query($MyQuery);
while($result = mysql_fetch_array($read_cursor)) { echo
$result[xx] ... }
Now I want a less dumb way to get the data from my db and output it in
a view.
Is there a way to pass my $result to the $html->tableCells() helper or
even better, how to get the result using the find() and then pass it
to $html->tableCells?
Thanks in advance
Michael
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