如何在Django中获取上传文件的名称 [英] How to get the name of uploaded file in Django

问题描述

我已经看到一些有关此问题的信息，但这些问题无法解决我的问题，这就是为什么我要提出一个新问题.请不要将此标记为重复项！

I have seen some question about it already but these couldn't solve my issue, that's why I'm asking a new question.So, don't mark this as duplicate, please!

使用Python(3.6)&Django(1.10)我正在尝试获取上传文件的名称，但返回

Using Python(3.6) & Django(1.10) I'm trying to get the name of uploaded file, but it returns

AttributeError:"NoneType"对象没有属性名称"

这是我尝试过的:来自 models.py

Here's what I have tried: From models.py

sourceFile = models.FileField(upload_to='archives/', name='sourceFile', blank=True)

通过 HTML模板:

<div class="form-group" hidden id="zipCode">
                    <label class="control-label" style="font-size: 1.5rem; color: black;">Select File</label>
                    <input id="sourceFile" name="sourceFile" type="file" class="file" multiple
                           data-allowed-file-extensions='["zip"]'>
                    <small id="fileHelp" class="form-text control-label" style="color:black; font-size: .9rem;">
                        Upload a Tar or Zip
                        archive which include Dockerfile, otherwise your deployment will fail.
                    </small>
                </div>

来自 views.py :

if form.is_valid():
               func_obj = form
               func_obj.sourceFile = form.cleaned_data['sourceFile']
               func_obj.save()
               print(func_obj.sourceFile.name)

这是怎么了?

请帮帮我！

提前谢谢！

推荐答案

要获取文件名，只需使用request.FILES字典(我假设只有1个文件正在上传)

To get the filename, you simply use the request.FILES dictionary (I assume that there is only 1 file being uploaded)

示例:

try:
    print(next(iter(request.FILES))) # this will print the name of the file
except StopIteration:
    print("No file was uploaded!")

请注意，这要求文件通过POST方法作为表单的一部分发送.

Note that this requires that the files were sent as a part of a form by the POST method.

要将其名称更改为随机字符串，建议使用 uuid.uuid4 ，因为这会生成一个随机字符串，该字符串非常不可能与已经存在的任何内容发生冲突.另外，您需要通过提供生成名称的功能来编辑 sourceFile 模型的 upload_to = 部分:

To change their name to a random string, I recommend uuid.uuid4, as this generates a random string that is VERY unlikely to collide with anything already there. Also, you need to edit your upload_to= section of your sourceFile model by providing a function to generate the name:

# In models.py

def content_file_name(instance, filename):
    filename = "{}.zip".format(str(uuid.uuid4().hex))
    return os.path.join('archives', filename)

# later....

sourceFile = models.FileField(upload_to=content_file_name, name='sourceFile', blank=True)

希望这会有所帮助！

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