blocks|key|771839|text|r+=+list(qs[:1])
if+r:
++return+r[0]
return+None|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|771840|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>r = list(qs[:1])
if r:
 return r[0]
return None
</code></pre>

blocks|key|771890|text|您可以使用array+slicing|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|771891|Entry.objects.all()[:1].get()|code-block|syntax|javascript|771892|它可以与.filter()一起使用|style|CODE|771893|Entry.objects.filter()[:1].get()|771894|您不希望首先将其转换为列表，因为这将强制对所有记录进行完整的数据库调用。只需执行上面的操作，它只会拉出第一个。你甚至可以使用.order_by()来确保你得到你想要的第一个。|771895|一定要添加.get()，否则将返回一个QuerySet而不是一个对象。|771896|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/reference/expressions.html#slicings|1|https://docs.djangoproject.com/en/dev/ref/models/querysets/#order-by|2|https://docs.djangoproject.com/en/dev/ref/models/querysets/#get|3|https://docs.djangoproject.com/en/dev/ref/models/querysets/^0|5|D|0|0|0|4|9|0|0|1Q|B|1Q|B|1|0|5|6|5|6|2|J|8|3|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@$A|17|B|18|1|19]]|C|$]]|$1|D|3|E|5|F|7|1A|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1B|8|@$A|1C|B|1D|K|L]]|9|@]|C|$]]|$1|M|3|N|5|F|7|1E|8|@]|9|@]|C|$G|H]]|$1|O|3|P|5|6|7|1F|8|@$A|1G|B|1H|K|L]]|9|@$A|1I|B|1J|1|1K]]|C|$]]|$1|Q|3|R|5|6|7|1L|8|@$A|1M|B|1N|K|L]]|9|@$A|1O|B|1P|1|1Q]|$A|1R|B|1S|1|1T]]|C|$]]|$1|S|3|-4|5|6|7|1U|8|@]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]|10|$5|V|W|X|C|$Y|11]]|12|$5|V|W|X|C|$Y|13]]|14|$5|V|W|X|C|$Y|15]]]]

You can use <a href="https://docs.python.org/3/reference/expressions.html#slicings" rel="noreferrer">array slicing</a>:
<pre><code>Entry.objects.all()[:1].get()
</code></pre>
Which can be used with <code>.filter()</code>:
<pre><code>Entry.objects.filter()[:1].get()
</code></pre>
You wouldn't want to first turn it into a list because that would force a full database call of all the records. Just do the above and it will only pull the first. You could even use <a href="https://docs.djangoproject.com/en/dev/ref/models/querysets/#order-by" rel="noreferrer"><code>.order_by()</code></a> to ensure you get the first you want.
Be sure to add the <a href="https://docs.djangoproject.com/en/dev/ref/models/querysets/#get" rel="noreferrer"><code>.get()</code></a> or else you will get a <a href="https://docs.djangoproject.com/en/dev/ref/models/querysets/" rel="noreferrer">QuerySet</a> back and not an object.

blocks|key|771938|text|如果你计划经常获取第一个元素--你可以向这个方向扩展QuerySet：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|771939|class+FirstQuerySet(models.query.QuerySet):
++++def+first(self):
++++++++return+self[0]


class+ManagerWithFirstQuery(models.Manager):
++++def+get_query_set(self):
++++++++return+FirstQuerySet(self.model)|code-block|syntax|javascript|771940|像这样定义模型：|771941|class+MyModel(models.Model):
++++objects+=+ManagerWithFirstQuery()|771942|并像这样使用它：|771943|+first_object+=+MyModel.objects.filter(x=100).first()|771944|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

If you plan to get first element often - you can extend QuerySet in this direction: 

<pre><code>class FirstQuerySet(models.query.QuerySet):
 def first(self):
 return self[0]


class ManagerWithFirstQuery(models.Manager):
 def get_query_set(self):
 return FirstQuerySet(self.model)
</code></pre>

Define model like this: 

<pre><code>class MyModel(models.Model):
 objects = ManagerWithFirstQuery()
</code></pre>

And use it like this:

<pre><code> first_object = MyModel.objects.filter(x=100).first()
</code></pre>

blocks|key|771988|text|你应该使用django方法，比如exists。它就在那里，供你使用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|771989|if+qs.exists():
++++return+qs[0]
return+None|code-block|syntax|javascript|771990|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|$]]

You should use django methods, like exists. Its there for you to use it.

<pre><code>if qs.exists():
 return qs[0]
return None
</code></pre>

blocks|key|986834|text|它可以是这样的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|986835|obj+=+model.objects.filter(id=emp_id)[0]|code-block|syntax|javascript|986836|或|986837|obj+=+model.objects.latest('id')|986838|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|$]]

It can be like this

<pre><code>obj = model.objects.filter(id=emp_id)[0]
</code></pre>

or

<pre><code>obj = model.objects.latest('id')
</code></pre>

blocks|key|772059|text|使用方便的方法.first()和.last()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|772060|MyModel.objects.filter(blah=blah).first()|code-block|syntax|javascript|772061|如果查询集不返回任何对象，则它们都接受结果异常并返回None。|772062|这些是在Django1.6中添加的，也就是released+in+Nov+2013。|772063|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.djangoproject.com/en/dev/ref/models/querysets/#first|1|https://docs.djangoproject.com/en/dev/ref/models/querysets/#last|2|https://www.djangoproject.com/weblog/2013/nov/06/django-16-released/^0|7|8|G|7|7|8|0|G|7|1|0|0|Q|4|0|L|K|2|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]|$9|18|A|19|1|1A]]|E|$]]|$1|F|3|G|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1F|8|@]|D|@$9|1G|A|1H|1|1I]]|E|$]]|$1|O|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]|Y|$5|R|S|T|E|$U|Z]]]]

Use the convenience methods <a href="https://docs.djangoproject.com/en/dev/ref/models/querysets/#first" rel="noreferrer"><code>.first()</code></a> and <a href="https://docs.djangoproject.com/en/dev/ref/models/querysets/#last" rel="noreferrer"><code>.last()</code></a>:
<pre><code>MyModel.objects.filter(blah=blah).first()
</code></pre>
They both swallow the resulting exception and return <code>None</code> if the queryset returns no objects.
These were added in Django 1.6, which was <a href="https://www.djangoproject.com/weblog/2013/nov/06/django-16-released/" rel="noreferrer">released in Nov 2013</a>.

blocks|key|986943|text|现在，在Django1.9中，您可以使用first()方法来处理查询集。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|986944|YourModel.objects.all().first()|code-block|syntax|javascript|986945|这是一种比.get()或[0]更好的方法，因为如果查询集为空，它不会抛出异常，因此不需要使用exists()进行检查|986946|entityMap^0|K|7|0|0|5|6|C|3|1A|8|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|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

Now, in Django 1.9 you have <code>first()</code> method for querysets.

<pre><code>YourModel.objects.all().first()
</code></pre>

This is a better way than <code>.get()</code> or <code>[0]</code> because it does not throw an exception if queryset is empty, Therafore, you don't need to check using <code>exists()</code>

blocks|key|772171|text|这也可以起作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|772172|def+get_first_element(MyModel):
++++my_query+=+MyModel.objects.all()
++++return+my_query[:1]|code-block|syntax|javascript|772173|如果为空，则返回一个空列表，否则返回列表中的第一个元素。|772174|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This could work as well:

<pre><code>def get_first_element(MyModel):
 my_query = MyModel.objects.all()
 return my_query[:1]
</code></pre>

if it's empty, then returns an empty list, otherwise it returns the first element inside a list.

Often I find myself wanting to get the first object from a queryset in Django, or return <code>None</code> if there aren't any. There are lots of ways to do this which all work. But I'm wondering which is the most performant.

<pre><code>qs = MyModel.objects.filter(blah = blah)
if qs.count() &gt; 0:
 return qs[0]
else:
 return None
</code></pre>

Does this result in two database calls? That seems wasteful. Is this any faster? 

<pre><code>qs = MyModel.objects.filter(blah = blah)
if len(qs) &gt; 0:
 return qs[0]
else:
 return None
</code></pre>

Another option would be:

<pre><code>qs = MyModel.objects.filter(blah = blah)
try:
 return qs[0]
except IndexError:
 return None
</code></pre>

This generates a single database call, which is good. But requires creating an exception object a lot of the time, which is a very memory-intensive thing to do when all you really need is a trivial if-test.

How can I do this with just a single database call and without churning memory with exception objects?

Fastest way to get the first object from a queryset in django?

数据库

Django

我经常发现自己想从Django中的查询集中获取第一个对象，如果没有对象，则返回None。有很多方法可以做到这一点，而且都是有效的。但是我想知道哪一个是最好的。qs = MyModel.objects.filter(blah = blah)if qs.count() > 0:    return qs[0]else:  ...

问在django中从查询集中获取第一个对象的最快方法？
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