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Il 28/11/2011 20:35, Motaz Abuthiab ha scritto:
<blockquote cite="mid:-8726256970408098379@unknownmsgid" type="cite">
<div>Non devi usare tabella.alias, basta il nome del campo, se hai
due campi con nomi uguali dovresti fare select <a
moz-do-not-send="true" href="http://t1.id">t1.id</a> as t1_id,
<a moz-do-not-send="true" href="http://t2.id">t2.id</a> as t2_id
from t1, t2 where ......</div>
<div><br>
</div>
<div>Ti consiglio di usare mysql_fetch_assoc() invece di
fetch_array()</div>
<br>
</blockquote>
Ad esempio:<br>
<br>
$result = DB_Query("SELECT DISTINCT(t1.sessionId), t2.Status,
t2.addDate, t2.ChgDate, t2.Referrer FROM Answers AS t1 INNER JOIN
SessionsPoll AS t2 ON t1.sessionId=t2.sessionId WHERE
t1.PollID='$pollId' AND t2.PollID='$pollId' ORDER BY t2.addDate DESC
LIMIT 10 OFFSET $startIdx;");<br>
if(mysql_num_rows($result) > 0) {<br>
while($row = mysql_fetch_array($result,MYSQL_ASSOC)) {<br>
$sessionId = $row["sessionId"];<br>
$Status = $row["Status"];<br>
$addDate = $row["addDate"];<br>
$chgDate = $row["ChgDate"];<br>
....<br>
}<br>
}<br>
<br>
Michele
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