Mysql 求和日期时间和分组

这是我的一张有日期时间进出的桌子。我想对每个p_ID的时间差求和，并按p_ID对它们进行分组

t_ID p_ID   t_IN                     t_OUT
1    1      2011-07-13 18:54:56      2011-07-13 20:16:12
2    1      2011-07-14 09:26:56      2011-07-14 09:46:02
3    1      2011-07-14 10:06:39      2011-07-14 10:56:31
4    3      2011-07-14 13:07:04      2011-07-14 13:58:35

我尝试了一些MySQL命令，但都没有用。。。我上次的尝试是这样的：

SELECT p_ID, TIME_FORMAT(SUM(TIMEDIFF(t_OUT,t_IN)),'%H:%i') AS time FROM timeclock GROUP BY p_ID

我猜我的命令不是对每个p_ID的每个记录求和。。。有什么帮助吗？

试试这个：

SELECT p_ID,
       SUM(TIMESTAMPDIFF(HOUR, t_OUT, t_IN)) AS time
FROM timeclock
GROUP BY p_ID;

SELECT p_ID, TIME_FORMAT(minutes,'%H:%i') as time
FROM (SELECT p_ID, SUM(TIMESTAMPDIFF(MINUTE, t_OUT, t_IN)) AS minutes
    FROM timeclock
    GROUP BY p_ID) x;

现在只需要几个小时，不过你可以用几分钟来替换它。

试试这个：

SELECT p_ID,
       SUM(TIMESTAMPDIFF(HOUR, t_OUT, t_IN)) AS time
FROM timeclock
GROUP BY p_ID;

SELECT p_ID, TIME_FORMAT(minutes,'%H:%i') as time
FROM (SELECT p_ID, SUM(TIMESTAMPDIFF(MINUTE, t_OUT, t_IN)) AS minutes
    FROM timeclock
    GROUP BY p_ID) x;

---

现在只需要几个小时，但你可以用几分钟来替换它。

你很接近，你只需要两步的方法，因为格式化需要在获得

总和之后完成，否则它会妨碍聚合时间差。

试试这个：

SELECT p_ID, TIME_FORMAT(minutes,'%H:%i') as time
FROM (SELECT p_ID, SUM(TIMESTAMPDIFF(MINUTE, t_OUT, t_IN)) AS minutes
    FROM timeclock
    GROUP BY p_ID) x;

不确定时间\u格式的格式，但此查询应该是关闭的。
如果您已经关闭，您只需要两步方法，因为格式设置需要在获得
总和之后完成，否则它会妨碍聚合时间差。

试试这个：

SELECT p_ID, TIME_FORMAT(minutes,'%H:%i') as time
FROM (SELECT p_ID, SUM(TIMESTAMPDIFF(MINUTE, t_OUT, t_IN)) AS minutes
    FROM timeclock
    GROUP BY p_ID) x;

不确定时间\u格式的格式，但此查询应已关闭。
已解决此问题：

$timeDiffReq = mysql_query("CREATE VIEW totals AS SELECT p_ID, TIME_TO_SEC(TIMEDIFF(t_OUT,t_IN)) AS time FROM timeclock");
$timeTotalReq = mysql_query("SELECT p_ID, SEC_TO_TIME(SUM(time)) AS timetotals FROM totals GROUP BY p_ID");
while($row = mysql_fetch_array($timeTotalReq)) {
echo "<div class=\"clockInOut\"><div id=\"name\">" . $row['p_ID'] . " - " . $row['timetotals'] . "</div></div>";}
$timeTotalDrop = mysql_query("DROP VIEW totals");

$timeDiffReq=mysql\u查询（“创建视图总计作为选择p\u ID，时间到秒（TIMEDIFF（t\u OUT，t\u IN））作为时间时钟的时间”）；
$timeTotalReq=mysql_查询（“按p_ID从总计组中选择p_ID、秒到时间（总和（时间））作为时间总计”）；
while（$row=mysql\u fetch\u数组（$timeTotalReq））{
echo“$row['p_ID']”-“$row['timetotals']”；}
$timeTotalDrop=mysql_查询（“下拉视图总计”）；
找到了答案：

$timeDiffReq = mysql_query("CREATE VIEW totals AS SELECT p_ID, TIME_TO_SEC(TIMEDIFF(t_OUT,t_IN)) AS time FROM timeclock");
$timeTotalReq = mysql_query("SELECT p_ID, SEC_TO_TIME(SUM(time)) AS timetotals FROM totals GROUP BY p_ID");
while($row = mysql_fetch_array($timeTotalReq)) {
echo "<div class=\"clockInOut\"><div id=\"name\">" . $row['p_ID'] . " - " . $row['timetotals'] . "</div></div>";}
$timeTotalDrop = mysql_query("DROP VIEW totals");

$timeDiffReq=mysql\u查询（“创建视图总计作为选择p\u ID，时间到秒（TIMEDIFF（t\u OUT，t\u IN））作为时间时钟的时间”）；
$timeTotalReq=mysql_查询（“按p_ID从总计组中选择p_ID、秒到时间（总和（时间））作为时间总计”）；
while（$row=mysql\u fetch\u数组（$timeTotalReq））{
echo“$row['p_ID']”-“$row['timetotals']”；}
$timeTotalDrop=mysql_查询（“下拉视图总计”）；
还有谁能帮我解决这个问题吗？还有谁能帮我解决这个问题吗？