我对此感到困惑。我搜查了互联网,它仍然令我困惑。我有以下表格:INNER JOIN语句中的歧义列
会议:meeting_id
,title
,chairman
,secretary
,occurances
客房:room_id
,room
,date
,time
,meeting_id
我想在一个表中显示的一切。我的PHP如下:
<?php
$result = mysql_query("SELECT * FROM Meetings INNER JOIN Rooms ON meeting_id = Rooms.meeting_id")
or die(mysql_error()); ;
if (mysql_num_rows($result) == 0) {
echo 'There Arent Any Meetings Setup Yet';
} else {
echo "<table border='0'><table border width=100%><tr><th>Title</th><th>Chairperson</th><th>Secretary</th><th>Terms of Reference</th><th>Occurances</th><th>Room</th><th>Date</th><th>Time</th>";
while($info = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td><br/>" . $info['title']." </td>";
echo "<td><br/>" . $info['chairperson']. "</td>";
echo "<td><br/>" . $info['secretary']."</td>";
echo "<td><br/>" . $info['tof']. "</td>";
echo "<td><br/>" . $info['occurances']. "</td>";
echo "<td><br/>" . $info['room']. "</td>";
echo "<td><br/>" . $info['date']. "</td>";
echo "<td><br/>" . $info['time']. "</td>";
}
}
echo "</tr>";
echo "</table>";
?>
它出来,出现以下错误信息:
列“MEETING_ID”在条款不明确
我想显示所有来自会议桌的字段,显然只有房间,日期和时间。
我该如何解决这个问题?
感谢队友,它工作:) – user1114080 2012-01-03 22:38:44