blocks|key|1015605|text|我不确定它是在Django的什么地方完成的，但你可以这样做。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1015606|通过反射将类名映射到字符串。|1015607|classes+=+[Person,Child,Parent]
def+find_class(name):
+for+clls+in+classes:
++if+clls.__class__.__name__+==+name:
+++return+clls|code-block|syntax|javascript|1015608|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'm not sure where it's done in Django, but you could do this.

Mapping the class name to the string via reflection.

<pre><code>classes = [Person,Child,Parent]
def find_class(name):
 for clls in classes:
 if clls.__class__.__name__ == name:
 return clls
</code></pre>

blocks|key|801206|text|大多数模型“字符串”都以"appname.modelname“的形式出现，因此您可能希望在get_model上使用此变体|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801207|from+django.db.models.loading+import+get_model

your_model+=+get_model+(+*your_string.split('.',1)+)|code-block|syntax|javascript|801208|通常将这些字符串转换为模型的django代码部分要稍微复杂一些，这来自django/db/models/fields/related.py|offset|length|style|CODE|801209|++++try:
++++++++app_label,+model_name+=+relation.split(".")
++++except+ValueError:
++++++++#+If+we+can't+split,+assume+a+model+in+current+app
++++++++app_label+=+cls._meta.app_label
++++++++model_name+=+relation
++++except+AttributeError:
++++++++#+If+it+doesn't+have+a+split+it's+actually+a+model+class
++++++++app_label+=+relation._meta.app_label
++++++++model_name+=+relation._meta.object_name

#+Try+to+look+up+the+related+model,+and+if+it's+already+loaded+resolve+the
#+string+right+away.+If+get_model+returns+None,+it+means+that+the+related
#+model+isn't+loaded+yet,+so+we+need+to+pend+the+relation+until+the+class
#+is+prepared.
model+=+get_model(app_label,+model_name,
++++++++++++++++++seed_cache=False,+only_installed=False)|801210|对我来说，这似乎是在核心代码中将其拆分到单个函数中的一个很好的例子。然而，如果你知道你的字符串是"App.Model“格式的，上面的两行代码就可以工作了。|801211|entityMap^0|0|0|Z|Y|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Most model "strings" appear as the form "appname.modelname" so you might want to use this variation on get_model

<pre><code>from django.db.models.loading import get_model

your_model = get_model ( *your_string.split('.',1) )
</code></pre>

The part of the django code that usually turns such strings into a model is a little more complex This from <code>django/db/models/fields/related.py</code>:

<pre><code> try:
 app_label, model_name = relation.split(".")
 except ValueError:
 # If we can't split, assume a model in current app
 app_label = cls._meta.app_label
 model_name = relation
 except AttributeError:
 # If it doesn't have a split it's actually a model class
 app_label = relation._meta.app_label
 model_name = relation._meta.object_name

# Try to look up the related model, and if it's already loaded resolve the
# string right away. If get_model returns None, it means that the related
# model isn't loaded yet, so we need to pend the relation until the class
# is prepared.
model = get_model(app_label, model_name,
 seed_cache=False, only_installed=False)
</code></pre>

To me, this appears to be an good case for splitting this out into a single function in the core code. However, if you know your strings are in "App.Model" format, the two liner above will work.

blocks|key|801247|text|对于任何遇到困难的人(就像我一样)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801248|from+django.apps+import+apps

model+=+apps.get_model('app_name',+'model_name')|code-block|syntax|javascript|801249|应该像model_name一样使用引号列出app_name+(即不要尝试导入它)|offset|length|style|CODE|801250|get_model接受小写或大写“model_name”|801251|entityMap^0|0|0|3|A|L|8|0|0|9|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]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@$I|Y|J|Z|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

just for anyone getting stuck (like I did):

<pre><code>from django.apps import apps

model = apps.get_model('app_name', 'model_name')
</code></pre>

<code>app_name</code> should be listed using quotes, as should <code>model_name</code> (i.e. don't try to import it)

<code>get_model</code> accepts lower case or upper case 'model_name'

blocks|key|801303|text|django.db.models.loading是deprecated+in+Django+1.7+(removed+in+1.9)，支持新的application+loading+system。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|801304|Django+1.7+docs给我们提供了以下内容：|801305|>>>+from+django.apps+import+apps
>>>+User+=+apps.get_model(app_label='auth',+model_name='User')
>>>+print(User)
<class+'django.contrib.auth.models.User'>|code-block|syntax|javascript|801306|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/django/django/blob/1.7/django/db/models/loading.py|1|https://docs.djangoproject.com/en/1.9/releases/1.9/#features-removed-in-1-9|2|https://docs.djangoproject.com/en/2.2/ref/applications/#initialization-process|3|https://docs.djangoproject.com/en/1.7/ref/applications/#django.apps.apps.get_model^0|0|O|P|O|0|1F|E|1|1Z|Q|2|0|7|8|3|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]|$9|16|A|17|1|18]|$9|19|A|1A|1|1B]]|E|$]]|$1|F|3|G|5|6|7|1C|8|@]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|H|3|I|5|J|7|1G|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]|W|$5|P|Q|R|E|$S|X]]|Y|$5|P|Q|R|E|$S|Z]]]]

<code>django.db.models.loading</code> was <a href="https://github.com/django/django/blob/1.7/django/db/models/loading.py" rel="noreferrer">deprecated in Django 1.7</a> (<a href="https://docs.djangoproject.com/en/1.9/releases/1.9/#features-removed-in-1-9" rel="noreferrer">removed in 1.9</a>) in favor of the the new <a href="https://docs.djangoproject.com/en/2.2/ref/applications/#initialization-process" rel="noreferrer">application loading system</a>.

Django <a href="https://docs.djangoproject.com/en/1.7/ref/applications/#django.apps.apps.get_model" rel="noreferrer">1.7 docs</a> give us the following instead:

<pre><code>&gt;&gt;&gt; from django.apps import apps
&gt;&gt;&gt; User = apps.get_model(app_label='auth', model_name='User')
&gt;&gt;&gt; print(User)
&lt;class 'django.contrib.auth.models.User'&gt;
</code></pre>

blocks|key|801355|text|在Django+1.7%2B中实现这一点的幸运方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801356|import+django
model_cls+=+django.apps.apps.get_model('app_name',+'model_name')|code-block|syntax|javascript|801357|因此，在所有框架教程的规范示例中：|801358|import+django
entry_cls+=+django.apps.apps.get_model('blog',+'entry')++#+Case+insensitive|801359|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|$]]

The blessed way to do this in Django 1.7+ is:

<pre class="lang-python prettyprint-override"><code>import django
model_cls = django.apps.apps.get_model('app_name', 'model_name')
</code></pre>

So, in the canonical example of all framework tutorials:

<pre class="lang-python prettyprint-override"><code>import django
entry_cls = django.apps.apps.get_model('blog', 'entry') # Case insensitive
</code></pre>

blocks|key|1015886|text|如果你不知道你的模型存在于哪个应用中，你可以这样搜索：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1015887|from+django.contrib.contenttypes.models+import+ContentType+
ct+=+ContentType.objects.get(model='your_model_name')+
model+=+ct.model_class()|code-block|syntax|javascript|1015888|请记住，your_model_name必须为小写。|1015889|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|$]]

In case you don't know in which app your model exists, you can search it this way:

<pre><code>from django.contrib.contenttypes.models import ContentType 
ct = ContentType.objects.get(model='your_model_name') 
model = ct.model_class()
</code></pre>

Remember that your_model_name must be lowercase.

blocks|key|801475|text|下面是一种不太特定于django的方法，用于从string中获取类：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801476|mymodels+=+['ModelA',+'ModelB']
model_list+=+__import__('<appname>.models',+fromlist=mymodels)
model_a+=+getattr(model_list,+'ModelA')|code-block|syntax|javascript|801477|或者，您可以使用here所示的importlib|offset|length|801478|import+importlib
myapp_models+=+importlib.import_module('<appname>.models')
model_a+=+getattr(myapp_models,+'ModelA')|801479|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/10773699^0|0|0|8|4|0|0|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]]|$1|G|3|H|5|6|7|W|8|@]|9|@$I|X|J|Y|1|Z]]|A|$]]|$1|K|3|L|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

Here is a less django-specific approach to get a class from string:

<pre><code>mymodels = ['ModelA', 'ModelB']
model_list = __import__('&lt;appname&gt;.models', fromlist=mymodels)
model_a = getattr(model_list, 'ModelA')
</code></pre>

or you can use importlib as shown <a href="https://stackoverflow.com/a/10773699">here</a>:

<pre><code>import importlib
myapp_models = importlib.import_module('&lt;appname&gt;.models')
model_a = getattr(myapp_models, 'ModelA')
</code></pre>

blocks|key|1016002|text|另一种具有较少代码的延迟呈现。在Django+2%2B中测试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1016003|from+django.apps+import+apps
model+=+apps.get_model("appname.ModelName")+#+e.g+"accounts.User"|code-block|syntax|javascript|1016004|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|$]]

Another rendition with less code for the lazy. Tested in Django 2+

<pre><code>from django.apps import apps
model = apps.get_model("appname.ModelName") # e.g "accounts.User"
</code></pre>

blocks|key|801625|text|2020年解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|801626|from+django.apps+import+apps

apps.get_model('app_name',+'Model')|code-block|syntax|javascript|801627|根据您的示例：|801628|apps.get_model('people',+'Person')|801629|per：Import+Error+:cannot+import+name+get_model|offset|length|801630|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/38443628/import-error-cannot-import-name-get-model^0|0|0|0|0|4|16|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@$M|11|N|12|1|13]]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

2020 solution:
<pre><code>from django.apps import apps

apps.get_model('app_name', 'Model')
</code></pre>
per your eg:
<pre><code>apps.get_model('people', 'Person')
</code></pre>
per:
<a href="https://stackoverflow.com/questions/38443628/import-error-cannot-import-name-get-model">Import Error :cannot import name get_model</a>

In Django, you can specify relationships like:

<pre><code>author = ForeignKey('Person')
</code></pre>

And then internally it has to convert the string "Person" into the model <code>Person</code>. 

Where's the function that does this? I want to use it, but I can't find it.

Django: Get model from string?

Django

在Django中，您可以指定如下关系：author = ForeignKey('Person')然后在内部，它必须将字符串"Person“转换为模型Person。执行此操作的函数在哪里？我想用它，但我找不到。

问Django:从字符串中获取模型？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Django:从字符串中获取模型？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Django:从字符串中获取模型？
EN