blocks|key|3434390|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3434391|会得到所有获得过不止一枚奖牌的选手的名字。|blockquote|3434392|3434393|3434394|(目前还不清楚这意味着什么。获得过一种以上的奖牌吗？或者获得过一枚以上的奖牌？您的示例答案表明是后者。此外，它还将"null“视为另一种奖章，而不是SQL中的特殊奖章。)|3434395|--+rows+where
++++THERE+EXISTS+compId,sport,medal,compId1,compId2,medal2+SUCH+THAT
++++++++in+competition+[compId]+of+sport+[sport]+player+[playerName]+won+[medal]
++++AND+in+competition+[compId2]+of+sport+[sport2]+player+[playerName]+won+[medal2]
++++AND+(compId+<>+compId2+OR+sport+<>+sport2+OR+medal+<>+medal2)|code-block|syntax|javascript|3434396|使用语句速记：|3434397|--+rows+where
++++THERE+EXISTS+compId,sport,medal,compId1,compId2,medal2+SUCH+THAT
++++++++Competition(compId,+sport,+playerName,+medal)
++++AND+Competition(compId2,+sport2,+playerName,+medal2)
++++AND+(compId+<>+compId2+OR+sport+<>+sport2+OR+medal+<>+medal2)|3434398|重新排列(预计每个σ一个比较和每个∪一个属性集的限制)：|3434399|--+rows+where
++++THERE+EXISTS+compId,sport,medal,compId1,compId2,medal2+SUCH+THAT
++++++++(+++Competition(compId,+sport,+playerName,+medal)
++++++++AND+Competition(compId2,+sport2,+playerName,+medal2)
++++++++AND+compId+<>+compId2)
++++OR+(+++Competition(compId,+sport,+playerName,+medal)
++++++++AND+Competition(compId2,+sport2,+playerName,+medal2)
++++++++AND+sport+<>+sport2)
++++OR+(+++Competition(compId,+sport,+playerName,+medal)
++++++++AND+Competition(compId2,+sport2,+playerName,+medal2)
++++++++AND+medal+<>+medal2)|3434400|现在来替换代数：|3434401|通过table/relation|3434402|every+(union)|unordered-list-item|3434403|
|3434404|every++NOT+(必须具有相同的列/属性)+AND+NOT(必须具有相同的列/属性)+AND+comparison+comparison+(select/restrict)|3434405|every+EXISTS+names+to+drop+by+|3434406|+by+names+names+to+to+drop+by∪names+to+keep+(projection)|3434407|every+column/attribute+|3434408|3434409|ρ(rename)。|3434410|3434411|3434412|3434413|竞赛奖牌playerName+(竞赛compId+<>+compId2+(竞赛<>+compID2/compIDρsport2/sportρmedal2+/奖牌竞赛))∪σsport+<>+sport2+(竞赛ρcompID2/compID∪σsport2/sportρMEDADAL2/奖牌竞赛)⋈ρmedal2(∪σ⋈ρcompID2/compIDρsport2/sportρMEDADAL2/+medal+)+)|3434414|(有关详细信息，请参阅this+answer。)|offset|length|3434415|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/24425914/3404097^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|1|1|1|1|0|0|0|0|B|B|0|0^^$0|@$1|2|3|-4|4|5|6|1V|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|1W|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|1X|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|1Y|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|1Z|7|@]|8|@]|9|$]]|$1|H|3|I|4|J|6|20|7|@]|8|@]|9|$K|L]]|$1|M|3|N|4|5|6|21|7|@]|8|@]|9|$]]|$1|O|3|P|4|J|6|22|7|@]|8|@]|9|$K|L]]|$1|Q|3|R|4|5|6|23|7|@]|8|@]|9|$]]|$1|S|3|T|4|J|6|24|7|@]|8|@]|9|$K|L]]|$1|U|3|V|4|5|6|25|7|@]|8|@]|9|$]]|$1|W|3|X|4|5|6|26|7|@]|8|@]|9|$]]|$1|Y|3|Z|4|10|6|27|7|@]|8|@]|9|$]]|$1|11|3|12|4|5|6|28|7|@]|8|@]|9|$]]|$1|13|3|14|4|10|6|29|7|@]|8|@]|9|$]]|$1|15|3|16|4|10|6|2A|7|@]|8|@]|9|$]]|$1|17|3|18|4|10|6|2B|7|@]|8|@]|9|$]]|$1|19|3|1A|4|10|6|2C|7|@]|8|@]|9|$]]|$1|1B|3|-4|4|10|6|2D|7|@]|8|@]|9|$]]|$1|1C|3|1D|4|10|6|2E|7|@]|8|@]|9|$]]|$1|1E|3|-4|4|5|6|2F|7|@]|8|@]|9|$]]|$1|1F|3|-4|4|5|6|2G|7|@]|8|@]|9|$]]|$1|1G|3|-4|4|5|6|2H|7|@]|8|@]|9|$]]|$1|1H|3|1I|4|5|6|2I|7|@]|8|@]|9|$]]|$1|1J|3|1K|4|5|6|2J|7|@]|8|@$1L|2K|1M|2L|1|2M]]|9|$]]|$1|1N|3|-4|4|5|6|2N|7|@]|8|@]|9|$]]]|1O|$1P|$4|1Q|1R|1S|9|$1T|1U]]]]

<blockquote>
 Get the names of all players who have won more than one medal.
</blockquote>

(It's not clear what this means. Have won than one kind of medal? Or have received more than one medal? Your example answer suggests the latter. Also, it treats "null" as just another kind of medal, not specially as in SQL.)

<pre><code>-- rows where
 THERE EXISTS compId,sport,medal,compId1,compId2,medal2 SUCH THAT
 in competition [compId] of sport [sport] player [playerName] won [medal]
 AND in competition [compId2] of sport [sport2] player [playerName] won [medal2]
 AND (compId &lt;&gt; compId2 OR sport &lt;&gt; sport2 OR medal &lt;&gt; medal2)
</code></pre>

Using statement shorthand:

<pre><code>-- rows where
 THERE EXISTS compId,sport,medal,compId1,compId2,medal2 SUCH THAT
 Competition(compId, sport, playerName, medal)
 AND Competition(compId2, sport2, playerName, medal2)
 AND (compId &lt;&gt; compId2 OR sport &lt;&gt; sport2 OR medal &lt;&gt; medal2)
</code></pre>

Rearranging (anticipating the limitations of one comparison per σ and one attribute set per ∪):

<pre><code>-- rows where
 THERE EXISTS compId,sport,medal,compId1,compId2,medal2 SUCH THAT
 ( Competition(compId, sport, playerName, medal)
 AND Competition(compId2, sport2, playerName, medal2)
 AND compId &lt;&gt; compId2)
 OR ( Competition(compId, sport, playerName, medal)
 AND Competition(compId2, sport2, playerName, medal2)
 AND sport &lt;&gt; sport2)
 OR ( Competition(compId, sport, playerName, medal)
 AND Competition(compId2, sport2, playerName, medal2)
 AND medal &lt;&gt; medal2)
</code></pre>

Now to get the algebra replace:

<ul>
<li>every statement by its table/relation</li>
<li>every AND of table/relation by ⋈ (natural join)</li>
<li>every OR of table/relation (which must have the same columns/attributes) by ∪ (union)</li>
<li>every AND NOT (which must have the same columns/attributes) by \ (difference)</li>
<li>every AND comparison by σ comparison (select/restrict)</li>
<li>every EXISTS names to drop by π names to keep (projection)</li>
<li>every column/attribute renaming by ρ (rename).

<pre><code>π playerName (
 σ compId &lt;&gt; compId2 (Competition
 ⋈ ρ compID2/compID ρ sport2/sport ρ medal2/medal Competition)
∪ σ sport &lt;&gt; sport2 (Competition
 ⋈ ρ compID2/compID ρ sport2/sport ρ medal2/medal Competition)
∪ σ medal &lt;&gt; medal2 (Competition
 ⋈ ρ compID2/compID ρ sport2/sport ρ medal2/medal Competition)
)
</code></pre></li>
</ul>

(For more see <a href="https://stackoverflow.com/a/24425914/3404097">this answer</a>.)

blocks|key|3434486|text|在我看来，有一种更简单的方法来解决这个问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3434487|​|3434488|📷|atomic|offset|length|3434489|本质上，您找到medal不为空的记录之间的关系，然后将此记录与其自身连接在一起，按名称连接。生成的记录将是名称重复的记录。|3434490|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/a35bde96ed82494b2f025cbf03144a95.png|imageAlt^0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|V|8|@]|9|@$G|W|H|X|1|Y]]|A|$]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R|S|-4]]]]

There is a much simpler way to solve this problem, in my opinion:

<a href="https://i.stack.imgur.com/eIqcA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eIqcA.png" alt="enter image description here"></a> 

Essentially, you find the relation of records where medal isn't null, and then join this record with itself, joining on name. The resulting records will be the ones where there are duplicate names.

blocks|key|499862|text|我知道这是一个非常古老的帖子，但我正在学习这个主题，我确实找到了另一种方法来表达练习的要求。也许这对搜索它的人很有用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|499863|答案是：|499864|π+Lc.playerName+(
++++ρ+Lc+σ+medal+≠+'null'+Competition
++++++++++++⨝+Lc.playerName+=+Rc.playerName+AND+Lc.sport+≠+Rc.sport
++++ρ+Rc+σ+medal+≠+'null'+Competition)|code-block|syntax|javascript|499865|我假设没有人可以在同一个项目中获得两枚奖牌，如果一个人拥有不止一枚奖牌，那么他必须在不同的项目上获得奖牌。我还假设关系中的每个元组代表一个奖励，所以...|499866|上面的表达式成功地获得了多次出现在不同运动中的playerName。|499867|499868|编辑的：我处理了null...您可以使用此工具https://dbis-uibk.github.io/relax/calc.htm查看它的实际效果|offset|length|style|BOLD|499869|要设置数据，请在组编辑器选项卡中使用此代码|499870|group:Competition

Competition+=+{
++++compId:number,+sport:string+,+playerName:string,+medal:string
++++193++++++++++,+'Tennis'+++++,+'John+Doe'+++++++,+'Gold'++++++
++++931++++++++++,+'Skiing'+++++,+'Mary+White'+++++,+'Bronze'++++
++++193++++++++++,+'Tennis'+++++,+'Arnold+Black'+++,+'null'++++++
++++182++++++++++,+'Bobsledding',+'John+Doe'+++++++,+'Gold'++++++
++++901++++++++++,+'Ping-Pong'++,+'Adam+Brown'+++++,+'Silver'++++
++++248++++++++++,+'Bobsledding',+'Mary+White'+++++,+'Silver'++++
}|499871|entityMap|0|LINK|mutability|MUTABLE|url|https://dbis-uibk.github.io/relax/calc.htm^0|0|0|0|0|0|0|0|3|N|16|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|16|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|17|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|18|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|19|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1B|8|@$P|1C|Q|1D|R|S]]|9|@$P|1E|Q|1F|1|1G]]|A|$]]|$1|T|3|U|5|6|7|1H|8|@]|9|@]|A|$]]|$1|V|3|W|5|F|7|1I|8|@]|9|@]|A|$G|H]]|$1|X|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|Y|$Z|$5|10|11|12|A|$13|14]]]]

I know this is a very old post, but I'm learning this subject, and I did find another way to represent what the exercise asks. Maybe it can be usefull for someone searching for this.

The answer:

<pre><code>π Lc.playerName (
 ρ Lc σ medal ≠ 'null' Competition
 ⨝ Lc.playerName = Rc.playerName AND Lc.sport ≠ Rc.sport
 ρ Rc σ medal ≠ 'null' Competition)
</code></pre>

I'm assuming that nobody can have two medals in the same sport, if somone have more than one medal, it has to be in different sports. I'm also assuming that each tuple in the relation represents an awarding, so ...

The expression above manages to get the playerName that appears more than once for different sports.

<hr>

Edited: I treated null...
you can see it in action with this tool <a href="https://dbis-uibk.github.io/relax/calc.htm" rel="nofollow noreferrer">https://dbis-uibk.github.io/relax/calc.htm</a>

to set the data, use this this code in the Group Editor tab

<pre><code>group:Competition

Competition = {
 compId:number, sport:string , playerName:string, medal:string
 193 , 'Tennis' , 'John Doe' , 'Gold' 
 931 , 'Skiing' , 'Mary White' , 'Bronze' 
 193 , 'Tennis' , 'Arnold Black' , 'null' 
 182 , 'Bobsledding', 'John Doe' , 'Gold' 
 901 , 'Ping-Pong' , 'Adam Brown' , 'Silver' 
 248 , 'Bobsledding', 'Mary White' , 'Silver' 
}
</code></pre>

I wasn't sure quite what to call this problem but it's not exactly counting rows. Let's say we have the relation:

<pre><code>Competition(compId, sport, playerName, medal)
</code></pre>

And let's say the attribute medal can be either gold, silver, bronze, or null. So we have the following data:

<pre><code>(193, Tennis, John Doe, Gold)
(931, Skiing, Mary White, Bronze)
(193, Tennis, Arnold Black, null)
(182, Bobsledding, John Doe, Gold)
(901, Ping-Pong, Adam Brown, Silver)
(248, Bobsledding, Mary White, Silver)
</code></pre>

I am having a very hard time figuring out how to answer this question: Get the names of all players who have won more than one medal. In this data the answers would be John Doe and Mary White. How could I get that answer on arbitrary data for this relation using relational algebra?

(This is a simplified version of the actual homework problem, and this simplification represents (I hope) the part of that problem I'm struggling with. There are an arbitrary and unknown number of competitions, sports, and players, but only 4 possibilities for medal)

Relational algebra to count rows

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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我不太确定这个问题是什么，但它并不完全是在计算行数。假设我们有这样的关系：Competition(compId, sport, playerName, medal)假设medal属性可以是金、银、铜或null。因此，我们有以下数据：(193, Tennis, John Doe, Gold)(931, Skiing, M...

问用于计算行的关系代数
EN

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问用于计算行的关系代数EN