blocks|key|4153228|text|如果您想知道某段时间内每个用户的的最高收益，那么您肯定需要处理来自SQL存储过程的数据。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|4153229|获取一个DataTable中的所有数据，然后找到一个周期的最大值。|4153230|假设你想要找到一周内的最大收益。|4153231|days:+01%7C02%7C03%7C04%7C05%7C06%7C07%7C08%7C09%7C10%7C11%7C12%7C13%7C14%7C15
++++++%7C....................%7C
++++++++%7C....................%7C
+++++++++++%7C....................%7C+
++++++++++++++++++++....+etc+...
++++++++++++++++++++++++++++%7C....................%7C

int+minDay+=+1,+daysPeriod+=+7,+maxDay+=+15,+maxGain+=+0,+startDay+=+0;
for(int+i+=+minDay;+i+<+maxDay+-+daysPeriod;+i%2B%2B)
{
++++++int+value+=+0;
++++++for(int+k+=+0;+k+<+daysPeriod;+i%2B%2B)
++++++++value+%2B=+(int)dataTable.Rows[i+%2B+k]["Gain"];
++++++if(value+>=+maxGain)
++++++{
++++++++maxGain+=+value;
++++++++startDay+=+i;
++++++}
}|code-block|syntax|javascript|4153232|接下来，只需向用户显示开始日期和周期，当然还有最大增益。|4153233|如果您使用一个7+(daysPeriod)数组，并且只迭代一次，直到达到所有15个元素，则此算法可以进行优化。|4153234|entityMap^0|G|S|0|0|X|0|0|G|0|0|0|S|0|0|1J|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|J|3|K|5|L|7|13|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|T|$]]

if you want to know the Highest Gains for a certain period of time per user, you certainly needs to process data out of SQL stored procedures. 
Get all data in a DataTable then find the max value for a period.

let's say you want to find the Max Gain for a period of 1 week. 

<pre><code>days: 01|02|03|04|05|06|07|08|09|10|11|12|13|14|15
 |....................|
 |....................|
 |....................| 
 .... etc ...
 |....................|

int minDay = 1, daysPeriod = 7, maxDay = 15, maxGain = 0, startDay = 0;
for(int i = minDay; i &lt; maxDay - daysPeriod; i++)
{
 int value = 0;
 for(int k = 0; k &lt; daysPeriod; i++)
 value += (int)dataTable.Rows[i + k]["Gain"];
 if(value &gt;= maxGain)
 {
 maxGain = value;
 startDay = i;
 }
}
</code></pre>

 
Next, simply present the start date and period to the user, and the Max Gain of course. 
This algorithm can be optimized if you use an array of 7 (daysPeriod) and iterate just once until reach all 15 elements.

blocks|key|4153260|text|为了显示“谁在给定时间内赢得了最多的XP+"，您需要以至少"X”的粒度存储有多少XP用户获胜。最简单的解决方案几乎肯定是存储XP+won和时间戳，给你一个秒的"X“。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4153261|create+table+xprewarded+(
++track_users_id+int(11)+NOT+NULL,
++xpreward+int(11)+NOT+NULL,
++rewardtime+timestamp+NOT+NULL+DEFAULT+NOW()
);|code-block|syntax|javascript|4153262|要获得在过去一天中赢得最多经验的人：|4153263|select+
++++sum(xpreward),+
++++track_users_id+
from+xprewarded+
where+rewardtime+>+now()-86400
group+by+track_users_id+
order+by+sum(xpreward)+desc+limit+1;|4153264|"+now+()-+86400“是指定”从现在起，回顾86400秒(或'24小时‘)，并返回自那以后创建的任何记录“的过滤器。其他时间段可以用类似的数学方法来完成。|4153265|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Well, in order to display "who won the most XP in a given ", you need to store how many XP users win at a granularity of at LEAST "X". The easiest solution is almost certainly to store the XP won and what the timestamp was, giving you an "X" of seconds.

<pre><code>create table xprewarded (
 track_users_id int(11) NOT NULL,
 xpreward int(11) NOT NULL,
 rewardtime timestamp NOT NULL DEFAULT NOW()
);
</code></pre>

To get the person who won the most XP in the past day:

<pre><code>select 
 sum(xpreward), 
 track_users_id 
from xprewarded 
where rewardtime &gt; now()-86400
group by track_users_id 
order by sum(xpreward) desc limit 1;
</code></pre>

"now()-86400" being the filter that specifies "from now, look back 86400 seconds (or '24 hours'), and return any record that was created since that point." Other time periods can be accomplished with similar math.

blocks|key|4153322|text|@TML:我需要得到每个人的XP，Rank和Level的区别，而不仅仅是XP。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4153323|@pixel3cs：|4153324|你的帖子让我开始思考。如果我遍历所有用户，如果差值大于已经存在的值，我可以将差值放入一个数组中。所以，我认为这是可行的：http://pastebin.com/m26b60996|offset|length|4153325|entityMap|0|LINK|mutability|MUTABLE|url|http://pastebin.com/m26b60996^0|0|0|1O|T|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@$F|S|G|T|1|U]]|A|$]]|$1|H|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]]]

@TML: I need to get the differences of XP, Rank, and Level for everyone though, not just XP.

@pixel3cs:

Your post got me thinking. If I loop through all the users, I could put the differences into an array if they are bigger than the already existing value. So, I think this would work: <a href="http://pastebin.com/m26b60996" rel="nofollow noreferrer">http://pastebin.com/m26b60996</a>

blocks|key|1837087|text|我假设只要其中一个统计数据发生更改，就会向track_users表中写入一条新记录。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1837088|要获得两个给定日期($START_DATE和$END_DATE)之间统计信息的变化，您可以使用以下查询：|1837089|SELECT+a.id,+a.date,+b.date,+b.skill1rank+-+b.skill1rank+AS+skill1rank_gain
++++FROM+track_users+AS+a+
++++INNER+JOIN+track_users+AS+b
++++++++++ON+a.date+=+$START_DATE+AND+
+++++++++++++b.date+=+$END_DATE+AND+
+++++++++++++a.id+=+b.id
++++ORDER+BY+skill1rank_gain+DESC|code-block|syntax|javascript|1837090|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I assume you write a new record to the track_users table whenever one of the stats change.

To get the change of a stat between two given dates ($START_DATE and $END_DATE), for all user you can use the following query:

<pre><code>SELECT a.id, a.date, b.date, b.skill1rank - b.skill1rank AS skill1rank_gain
 FROM track_users AS a 
 INNER JOIN track_users AS b
 ON a.date = $START_DATE AND 
 b.date = $END_DATE AND 
 a.id = b.id
 ORDER BY skill1rank_gain DESC
</code></pre>

I have a table that has the columns skill1xp, skill1lvl, and skill1rank for 25 skills. Each day I update it for all users, so there's a datetime column. I'm making a high-scores progress tracker for a game, for many users, and I've got just one feature left, which is Highest Gains for a certain period of time (day/week/month).

I guess this is a hefty question, but...how would I go about doing that? =/.

Table structure: <a href="http://pastebin.com/m5903aa17" rel="nofollow noreferrer">http://pastebin.com/m5903aa17</a>

How to get the biggest differences between two fields comparing all users?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个表，其中包含25个技能的列skill1xp、skill1lvl和skill1rank。每天我都为所有用户更新它，所以有一个datetime列。我正在为一款游戏做一个高分的进度跟踪器，对许多用户来说，我只剩下一个功能，那就是某段时间(天/周/月)的最高收益。我想这是一个沉重的问题，but...how我会这么做吗？...

问如何获取比较所有用户的两个字段之间的最大差异？
EN

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问如何获取比较所有用户的两个字段之间的最大差异？EN