如何在PHP/MySQL中显示选定列
我正在用PHP运行此查询:如何在PHP/MySQL中显示选定列,php,mysql,Php,Mysql,我正在用PHP运行此查询: $sql="SELECT t.* FROM tickets t join (select ticketnumber, max(datetime) as maxdt from ticket_updates tu GROUP BY ticketnumber) tu ON t.ticketnumber = tu.ticketnumber WHERE t.status <> 'Completed' AND tu.maxdt < now() - inte
$sql="SELECT t.* FROM tickets t join
(select ticketnumber, max(datetime) as maxdt from ticket_updates tu GROUP BY ticketnumber) tu
ON t.ticketnumber = tu.ticketnumber
WHERE t.status <> 'Completed' AND tu.maxdt < now() - interval 1 hour; ";
$rs=mysql_query($sql,$conn);
while($result=mysql_fetch_array($rs)) {
echo $result["tu.maxdt"];
}
但它不显示datetime列
我做错了什么?该列被称为maxdt而不是tu.maxdt,因此它需要是echo$result['maxdt']。您还需要在select语句中包含该列:
SELECT t.*, tu.*
FROM tickets t join
(SELECT ticketnumber, max(datetime) AS maxdt
FROM ticket_updates tu
GROUP BY ticketnumber) tu
...
该列被称为maxdt而不是tu.maxdt,因此它需要是echo$result['maxdt']。您还需要在select语句中包含该列:
SELECT t.*, tu.*
FROM tickets t join
(SELECT ticketnumber, max(datetime) AS maxdt
FROM ticket_updates tu
GROUP BY ticketnumber) tu
...
您正在子选择中选择maxdatetime作为maxdt。这不会显示为tu.maxtd,它需要包含在主选择中
SELECT t.*, max(tu.datetime) as maxdt FROM tickets
echo $result["maxdt"];
由于mysql_uu3;函数已被弃用,您应该切换到mysqli_3; 您正在子选择中选择maxdatetime作为maxdt。这不会显示为tu.maxtd,它需要包含在主选择中
SELECT t.*, max(tu.datetime) as maxdt FROM tickets
echo $result["maxdt"];
由于mysql_uu3;函数已被弃用,您应该切换到mysqli_3; 您必须在查询中包含此列:
$sql="SELECT t.*, tu.* FROM tickets t join
(select ticketnumber, max(datetime) as maxdt from ticket_updates tu GROUP BY ticketnumber) tu
ON t.ticketnumber = tu.ticketnumber
WHERE t.status <> 'Completed' AND tu.maxdt < now() - interval 1 hour; ";
您必须在查询中包含此列:
$sql="SELECT t.*, tu.* FROM tickets t join
(select ticketnumber, max(datetime) as maxdt from ticket_updates tu GROUP BY ticketnumber) tu
ON t.ticketnumber = tu.ticketnumber
WHERE t.status <> 'Completed' AND tu.maxdt < now() - interval 1 hour; ";
尝试echo$result['maxdt']您需要将其包含在select中。不再支持mysql_*函数,它们已不再维护,将来也将继续使用。您应该使用或更新代码,以确保将来项目的功能。请尝试echo$result['maxdt'],您需要将其包含在select中。不再支持mysql_*函数,它们已不再维护,并且将在将来提供。您应该使用或更新代码,以确保将来项目的功能。请参阅上面的robin答案。请参阅上面的robin答案。