使用PHP从mysql表中获取多行并描述时间界限

我有一个如下所示的mysql表

ip_add                      mac                     time
10.0.0.97                00 14 2A 2F 72 FE         2013-09-18 16:35:47
10.0.0.97                00 14 2A 2F 72 FE         2013-09-19 08:48:02
10.0.0.98                08 CC 68 71 A1 C0         2013-09-18 16:35:47
10.0.0.98                08 CC 68 71 A1 C0         2013-09-19 08:48:02

我有一个php脚本，我想打印以下几行

The ip address 10.0.0.97 was used by MAC 00 14 2A 2F 72 FE from 2013-09-18 16:35:47 to 2013-09-19 08:48:02
The ip address 10.0.0.97 was used by MAC 00 14 2A 2F 72 FE from  2013-09-19 08:48:02 to present(i.e there has not been a further entry of that ip )
The ip address 10.0.0.98 was used by MAC 00 14 2A 2F 72 FE from 2013-09-18 16:35:47 to 2013-09-19 08:48:02
The ip address 10.0.0.97 was used by MAC 00 14 2A 2F 72 FE from  2013-09-19 08:48:02 to present

我已经写了我的脚本如下

$var = $_POST['IP'](ip which will be obtained from form);
mysql_connect('localhost','root','mysql');
mysql_select_db('logs');
$result=mysql_query("select ip_add,mac from arp_table where ip_add='$var'");
$row=mysql_fetch_row($result);
$count=mysql_query("select COUNT (*)  from arp_table where ip_add='$var'");
$row2 =mysql_fetch_row($count);

echo "The number of instances this ip address was used is {$row2[0]}<br/>"

if($row2==1){
$time=mysql_query("select ip_add, time from arp_table where ip_add='$var'")
$time2=mysql_fetch_row($time);
echo "It was used from {$time2[1]} and it is still in use<br/>";}
else
{$time3=mysql_query("select ip_add,time from arp_table where ip_add='$var'");
while($row3=mysql_fetch_assoc($time3)){
echo "This ip was used by MAC {$row[1]} from (*I`M STUCK HERE*) to (*I`M STUCK HERE*)"}
}

如何让它获得由同一ip注册的上一个时间并将其与下一个在线时间关联

SELECT
    ip_add,
    mac,
    MIN(time) as from_time,
    MAX(time) as to_time
FROM 
    my_table
GROUP BY
    ip_add,
    mac

或者，如果要为每一行执行以下操作：

SELECT
    a.ip_add,
    a.mac,
    a.time as from_time,
    (
        SELECT 
            time 
        FROM 
            my_table b 
        WHERE 
            b.ip_add = a.ip_add 
            AND a.time < b.time 
        ORDER BY 
            b.time 
        LIMIT 1
    ) as to_time
FROM 
    my_table a

试试这个：

SELECT
    ip_add,
    mac,
    MIN(time) as from_time,
    MAX(time) as to_time
FROM 
    my_table
GROUP BY
    ip_add,
    mac

或者，如果要为每一行执行以下操作：

SELECT
    a.ip_add,
    a.mac,
    a.time as from_time,
    (
        SELECT 
            time 
        FROM 
            my_table b 
        WHERE 
            b.ip_add = a.ip_add 
            AND a.time < b.time 
        ORDER BY 
            b.time 
        LIMIT 1
    ) as to_time
FROM 
    my_table a

---

这部分我必须参考上一个ip的下一个相同ip时间，并提及它，因为该ip不时被使用。我得到了你的观点，我发布了一个答案。请检查这部分，我必须参考上一个ip的下一个相同ip时间，并提及它，因为该ip不时被使用。我得到了你的观点，我发布了一个答案，请检查它在这种情况下，我不想找到最短的时间，但我只需要以前的工作time@sosytee为此，您需要使用第二个查询您是否知道在php脚本中放置该查询的位置，因为我现在放置的位置没有返回任何结果，所以您不需要其他查询，只需要这一个查询，然后迭代结果并输出结果。如果我不想找到最短时间，但只需要前一个查询，这是否有效time@sosytee为此，你需要使用第二个问题你知道去哪里吗将它放在php脚本中，因为我现在放置它的地方没有返回任何结果。您不需要其他查询，只需要这一个，然后迭代结果并输出它们