blocks|key|2563793|text|为什么不从实例生成路径呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2563794|def+generate_path(instance,+filename):
++++return+os.path.join("hardcoded_prefix",+instance.name,+filename)

class+AFile(models.Model):
++++name+=+models.CharField(max_length=256)
++++file+=+models.FileField(upload_to=generate_path)|code-block|syntax|javascript|2563795|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|$]]

Why don't you generate the path from the instance?

<pre><code>def generate_path(instance, filename):
 return os.path.join("hardcoded_prefix", instance.name, filename)

class AFile(models.Model):
 name = models.CharField(max_length=256)
 file = models.FileField(upload_to=generate_path)
</code></pre>

blocks|key|652496|text|这里的问题是，只有在模型上定义FileField或ImageField时，upload_to才可用。对该字段的任何后续访问都会返回一个FieldFile实例，该实例不能访问已定义的upload_to。长话短说，在最初定义方法之后，没有办法改变它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|652497|但是，您也许能够绕过它进行某种形式的结束。请注意，我还没有真正尝试过这一点，但它应该可以工作：|652498|首先，在模型上定义一个方法，该方法的简单任务是在模型上设置一个实例变量：|652499|def+set_upload_to_info(self,+path,+name):
++++self.upload_to_info+=+(path,+name)|code-block|syntax|javascript|652500|然后，在您的upload_to方法中，您可以测试这些属性是否存在，如果它们存在，则使用它们：|652501|def+my_upload_to(instance,+filename):
++++if+hasattr(instance,+'upload_to_info'):
++++++++path,+name+=+instance.upload_to_info
++++++++#+do+something+and+return+file+path
++++else:
++++++++#+standard+upload_to+bit+here|652502|然后在您的视图中，您只需要在保存模型之前调用您创建的方法：|652503|afile.set_upload_to_info(path,+name)
afile.save()|652504|entityMap^0|F|9|P|A|11|9|1V|9|2I|9|0|0|0|0|6|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|19|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|1A|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|1B|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1F|8|@]|D|@]|E|$M|N]]|$1|S|3|T|5|6|7|1G|8|@]|D|@]|E|$]]|$1|U|3|V|5|L|7|1H|8|@]|D|@]|E|$M|N]]|$1|W|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|X|$]]

The problem here is that <code>upload_to</code> is only available when defining the <code>FileField</code> or <code>ImageField</code> on the model. Any subsequent access to the field returns a <code>FieldFile</code> instance, which doesn't have access to the defined <code>upload_to</code>. Long and short, there's no way to alter the method after it's initially defined.

However, you might be able to do a sort of end-run around it. Note that I haven't actually tried this, but it should work:

First define a method on your model that will have the simple task of setting an instance variable on the model:

<pre><code>def set_upload_to_info(self, path, name):
 self.upload_to_info = (path, name)
</code></pre>

Then, inside your <code>upload_to</code> method, you can test for the presence of these attributes and use them if they are:

<pre><code>def my_upload_to(instance, filename):
 if hasattr(instance, 'upload_to_info'):
 path, name = instance.upload_to_info
 # do something and return file path
 else:
 # standard upload_to bit here
</code></pre>

Then in your view, you just need to call the method you create before you save the model:

<pre><code>afile.set_upload_to_info(path, name)
afile.save()
</code></pre>

blocks|key|652506|text|解决方案是放入一些非持久字段，并仍然在get_path方法中引用它们|type|unstyled|depth|inlineStyleRanges|entityRanges|data|652507|class+AFile(models.Model):
++++name+=+models.CharField(max_length=256)
++++file+=+models.FileField(upload_to=get_path)
++++path+=+''|code-block|syntax|javascript|652508|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|$]]

solution is to put in some non-persistent fields and still refer to them in the get_path method

<pre><code>class AFile(models.Model):
 name = models.CharField(max_length=256)
 file = models.FileField(upload_to=get_path)
 path = ''
</code></pre>

blocks|key|3882375|text|在保存模型之前，可以通过更改FileField的field.upload_to属性在视图中更改upload+to参数。在我的例子中，我使用了基于类的视图和表单。所以在模型实例之前。我使用下面的代码来更改上传到路径。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3882376|with+transaction.atomic():
++++++++model_instance+=+form.save(commit=False)
++++++++model_instance.creator+=+self.request.user
++++++++model_instance.img_field.field.upload_to+=+'directory/'%2Bmodel_instance.name%2B'/logo'
++++++++self.object+=+form.save()|code-block|syntax|javascript|3882377|简而言之，如果您的图像字段命名为imageupload，请相应地更改imageupload.field.upload_to以指向所需的路径。如果这个方法解决了你的问题，请告诉我。|3882378|entityMap^0|E|9|O|F|0|0|Y|R|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

upload to argument can be changed in view by changing <code>field.upload_to</code> attribute of <code>FileField</code> before saving model. In my case I was using class based views together with forms. So before model instance. I used below code to change upload to path.

<pre><code>with transaction.atomic():
 model_instance = form.save(commit=False)
 model_instance.creator = self.request.user
 model_instance.img_field.field.upload_to = 'directory/'+model_instance.name+'/logo'
 self.object = form.save()
</code></pre>

In short if your image field is named as imageupload, change <code>imageupload.field.upload_to</code> accordingly to point to the path you need. Please let know if this approach solved your issue.

I have a model for storing files:

<pre><code>class AFile(models.Model):
 path = models.CharField(max_length=256)
 name = models.CharField(max_length=256)
 file = models.FileField(upload_to=get_path)
</code></pre>

there are many views that save files. I want a seperate path for each. so I put path in the model and use that in the get path function. like so:

<pre><code>afile = AFile(path='blah/foo/', name='filex.jpg')
afile.save()
</code></pre>

so the file is in the right spot. But I don't really want to store the path and name field in my database, its only there to generate a path. any way to achieve this same thing without extra model fields?

django - set the upload_to in the view

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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我有一个存储文件的模型：class AFile(models.Model):    path = models.CharField(max_length=256)    name = models.CharField(max_length=256)    file = models.FileField(upload_t...

问django -在视图中设置upload_to
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