blocks|key|1441777|text|这是Rails+3的方式|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1441778|SomeModel.last(5)+#+last+5+records+in+ascending+order

SomeModel.last(5).reverse+#+last+5+records+in+descending+order|code-block|syntax|javascript|1441779|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is the Rails 3 way

<pre><code>SomeModel.last(5) # last 5 records in ascending order

SomeModel.last(5).reverse # last 5 records in descending order
</code></pre>

blocks|key|1441755|text|在Rails3.1中实现这一点的新方法是SomeModel.limit(5).order('id+desc')|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1441756|entityMap^0|K|Z|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

new way to do it in rails 3.1 is <code>SomeModel.limit(5).order('id desc')</code>

blocks|key|1656834|text|对于Rails+5(可能还有Rails+4)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1656835|不好：|1656836|Something.last(5)|code-block|syntax|javascript|1656837|因为：|1656838|Something.last(5).class
=>+Array|1656839|所以：|1656840|Something.last(50000).count|1656841|很可能会毁了你的记忆，或者永远都不会。|1656842|好的方法：|1656843|Something.limit(5).order('id+desc')|1656844|1656845|Something.limit(5).order('id+desc').class
=>+Image::ActiveRecord_Relation

Something.limit(5).order('id+desc').to_sql
=>+"SELECT++\"somethings\".*+FROM+\"somethings\"+ORDER+BY+id+desc+LIMIT+5"|1656846|后者是未评估的作用域。您可以将其链接，或通过.to_a将其转换为数组。所以：|offset|length|style|CODE|1656847|Something.limit(50000).order('id+desc').count|1656848|..。只需要一秒钟。|1656849|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|M|5|0|0|0^^$0|@$1|2|3|4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1C|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1D|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|1E|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|1F|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|1G|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|1H|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|1I|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1J|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|1K|8|@]|9|@]|A|$G|H]]|$1|W|3|J|5|6|7|1L|8|@]|9|@]|A|$]]|$1|X|3|Y|5|F|7|1M|8|@]|9|@]|A|$G|H]]|$1|Z|3|10|5|6|7|1N|8|@$11|1O|12|1P|13|14]]|9|@]|A|$]]|$1|15|3|16|5|F|7|1Q|8|@]|9|@]|A|$G|H]]|$1|17|3|18|5|6|7|1R|8|@]|9|@]|A|$]]|$1|19|3|-4|5|6|7|1S|8|@]|9|@]|A|$]]]|1A|$]]

<h1>For Rails 5 (and likely Rails 4)</h1>

<h3>Bad:</h3>

<pre><code>Something.last(5)
</code></pre>

because:

<pre><code>Something.last(5).class
=&gt; Array
</code></pre>

so:

<pre><code>Something.last(50000).count
</code></pre>

will likely blow up your memory or take forever.

<h3>Good approach:</h3>

<pre><code>Something.limit(5).order('id desc')
</code></pre>

because:

<pre><code>Something.limit(5).order('id desc').class
=&gt; Image::ActiveRecord_Relation

Something.limit(5).order('id desc').to_sql
=&gt; "SELECT \"somethings\".* FROM \"somethings\" ORDER BY id desc LIMIT 5"
</code></pre>

The latter is an unevaluated scope. You can chain it, or convert it to an array via <code>.to_a</code>. So:

<pre><code>Something.limit(50000).order('id desc').count
</code></pre>

... takes a second.

blocks|key|1441723|text|解决方案在这里：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1441724|SomeModel.last(5).reverse|code-block|syntax|javascript|1441725|由于rails是懒惰的，它最终会用SQL命中数据库，比如："SELECT+table.*+FROM+table+ORDER+BY+table.id+DESC+LIMIT+5“。|offset|length|style|CODE|1441726|entityMap^0|0|0|11|5|1E|5|1T|5|1Z|2|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|N|$]]

Solution is here:

<pre><code>SomeModel.last(5).reverse
</code></pre>

Since rails is lazy, it will eventually hit the database with SQL like: "SELECT <code>table</code>.* FROM <code>table</code> ORDER BY <code>table</code>.<code>id</code> DESC LIMIT 5".

blocks|key|1656619|text|如果您需要对结果设置一些排序，则使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1656620|Model.order('name+desc').limit(n)+#+n=+number|code-block|syntax|javascript|1656621|如果您不需要任何排序，只需要将记录保存在表中，则使用：|1656622|Model.last(n)+#+n=+any+number|1656623|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If you need to set some ordering on results then use:

<pre><code>Model.order('name desc').limit(n) # n= number
</code></pre>

if you do not need any ordering, and just need records saved in the table then use:

<pre><code>Model.last(n) # n= any number
</code></pre>

blocks|key|1656630|text|试试看：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1656631|Model.order("field_for_sort+desc").limit(5)|code-block|syntax|javascript|1656632|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Just try:

<pre><code>Model.order("field_for_sort desc").limit(5)
</code></pre>

blocks|key|1656670|text|在我的rails+(rails+4.2)项目中，我使用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1656671|Model.last(10)+#+get+the+last+10+record+order+by+id|code-block|syntax|javascript|1656672|而且它是有效的。|1656673|entityMap^0|9|B|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

In my rails <code>(rails 4.2)</code> project, I use

<pre><code>Model.last(10) # get the last 10 record order by id
</code></pre>

and it works.

blocks|key|1442224|text|我们可以在Rails5.2中使用Model.last(5)或Model.limit(5).order(id:+:desc)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1442225|entityMap^0|G|D|U|V|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

we can use <code>Model.last(5)</code> or <code>Model.limit(5).order(id: :desc)</code> in rails 5.2

blocks|key|1656562|text|我发现这个查询使用"pluck“方法更好/更快，我喜欢这个方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1656563|Challenge.limit(5).order('id+desc')|code-block|syntax|javascript|1656564|这将给出一个ActiveRecord作为输出；因此您可以对其使用.pluck，如下所示：|1656565|Challenge.limit(5).order('id+desc').pluck(:id)|1656566|在使用最佳SQL代码的同时，它可以快速地将ids作为数组给出。|1656567|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|$]]

I find that this query is better/faster for using the "pluck" method, which I love:

<pre><code>Challenge.limit(5).order('id desc')
</code></pre>

This gives an ActiveRecord as the output; so you can use .pluck on it like this:

<pre><code>Challenge.limit(5).order('id desc').pluck(:id)
</code></pre>

which quickly gives the ids as an array while using optimal SQL code.

blocks|key|1442086|text|假设N=5，你的模型是Message，你可以这样做：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1442087|Message.order(id:+:asc).from(Message.all.order(id:+:desc).limit(5),+:messages)|code-block|syntax|javascript|1442088|看一下sql：|1442089|SELECT+"messages".*+FROM+(
++SELECT++"messages".*+FROM+"messages"++ORDER+BY+"messages"."created_at"+DESC+LIMIT+5
)+messages++ORDER+BY+"messages"."created_at"+ASC|1442090|关键是子选择。首先，我们需要定义我们想要的最后一条消息是什么，然后我们必须按升序对它们进行排序。|1442091|entityMap^0|B|7|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

Let's say N = 5 and your model is <code>Message</code>, you can do something like this:

<pre class="lang-ruby prettyprint-override"><code>Message.order(id: :asc).from(Message.all.order(id: :desc).limit(5), :messages)
</code></pre>

Look at the sql:

<pre class="lang-sql prettyprint-override"><code>SELECT "messages".* FROM (
 SELECT "messages".* FROM "messages" ORDER BY "messages"."created_at" DESC LIMIT 5
) messages ORDER BY "messages"."created_at" ASC
</code></pre>

The key is the subselect. First we need to define what are the last messages we want and then we have to order them in ascending order.

blocks|key|1656510|text|如果你在你的模型中有一个默认的作用域，在Rails+3中指定了升序，那么你需要使用reorder而不是上面Arthur+Neve指定的order：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1656511|Something.limit(5).reorder('id+desc')|code-block|syntax|javascript|1656512|或|1656513|Something.reorder('id+desc').limit(5)|1656514|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If you have a default scope in your model that specifies an ascending order in Rails 3 you'll need to use reorder rather than order as specified by Arthur Neves above:

<pre><code>Something.limit(5).reorder('id desc')
</code></pre>

or 

<pre><code>Something.reorder('id desc').limit(5)
</code></pre>

blocks|key|1442278|text|一个简单的答案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1442279|Model.limit(5).order(id:+:desc)|code-block|syntax|javascript|1442280|这种解决方案存在一个问题，因为id不能成为记录创建时间的唯一决定因素。|offset|length|style|CODE|1442281|更可靠的解决方案是：|1442282|Model.order(created_at:+:desc).limit(5)|1442283|正如其他人所指出的，我们也可以使用Model.last(5)。唯一的问题是它返回Array，而不是Model::ActiveRecord_Relation。|1442284|entityMap^0|0|0|F|2|0|0|0|H|D|14|5|1D|S|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|11|8|@$I|12|J|13|K|L]|$I|14|J|15|K|L]|$I|16|J|17|K|L]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|T|$]]

A simple answer would be:
<pre><code>Model.limit(5).order(id: :desc)
</code></pre>
There is a problem with this solution, as <code>id</code> can't be the sole determiner of when a record was created in the time.
A more reliable solution would be:
<pre><code>Model.order(created_at: :desc).limit(5)
</code></pre>
As others have pointed out, one can also use <code>Model.last(5)</code>. The only gotcha with this is that it returns <code>Array</code>, and not <code>Model::ActiveRecord_Relation</code>.

blocks|key|1656364|text|将:order参数添加到查询中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1656365|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Add an :order parameter to the query

With <code>:limit</code> in query, I will get first N records. What is the easiest way to get last N records?

How to get last N records with activerecord?

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在查询中使用:limit，我将获得前N条记录。获取最后N条记录的最简单方法是什么？

问如何使用activerecord获取最后N条记录？
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回答 13

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问如何使用activerecord获取最后N条记录？
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