Python 图像上传

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2 回答

6192 浏览

提问于 2025-04-18 03:35

我想用Python在我的应用程序中浏览图片并上传到文件夹里。

当我点击提交按钮时，它给我显示了一个链接 http://www.domain.com/store_mp3_view，但是图片没有上传成功。

这是HTML代码：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<form action="/store_mp3_view" method="post" accept-charset="utf-8"
      enctype="multipart/form-data">

    <label for="mp3">Mp3</label>
    <input id="mp3" name="mp3" type="file" value="" />

    <input type="submit" value="submit" />
</form>
</body>
</html>

这是Python文件的代码

import os import uuid from pyramid.response import Response

def store_mp3_view(request):

# filename 变量里存的是文件名，格式是字符串。

# 注意：这个例子没有处理IE浏览器发送的绝对文件路径作为文件名的情况。这个例子比较简单，信任用户输入。

filename = request.POST['mp3'].filename

# ``input_file`` contains the actual file data which needs to be
# stored somewhere.

input_file = request.POST['mp3'].file

# Note that we are generating our own filename instead of trusting
# the incoming filename since that might result in insecure paths.
# Please note that in a real application you would not use /tmp,
# and if you write to an untrusted location you will need to do
# some extra work to prevent symlink attacks.

file_path = os.path.join(/files, '%s.mp3' % uuid.uuid4())

# We first write to a temporary file to prevent incomplete files from
# being used.

temp_file_path = file_path + '~'
output_file = open(temp_file_path, 'wb')

# Finally write the data to a temporary file
input_file.seek(0)
while True:
    data = input_file.read(2<<16)
    if not data:
        break
    output_file.write(data)

# If your data is really critical you may want to force it to disk first
# using output_file.flush(); os.fsync(output_file.fileno())

output_file.close()

# Now that we know the file has been fully saved to disk move it into place.

os.rename(temp_file_path, file_path)

return Response('OK')
return Response('OK')

用户输入浏览器兼容性文件处理文件存储 html表单图像上传响应处理 web应用开发

2 个回答

这段代码可以让你一次上传多个文件。

def insert_file(self):
    for i in request.FILES.getlist('mp3'):
        fileName = i.name
        out_file = open(fileName,'w')
        out_file.write(i.read())
    return HttpResponse('Inserted Successfully')

回答于 2025-04-18 由 Python大师

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你可以在models.py文件里使用文件字段模型。

class Document(models.Model):
    docfile = models.FileField(upload_to='documents/', max_length=5234,blank=True, null=True,)

对应的forms.py文件。

class DocumentForm(forms.Form):
    docfile = forms.FileField(label='', show_hidden_initial='none',required=True,)

在views.py文件里。

if request.FILES.has_key('your_fileName'):
            newdoc = Document(docfile = request.FILES['your_fileName'])
            newdoc.save()

我用上面的代码让它工作了，希望对你有帮助。

回答于 2025-04-18 由 Python大师

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Python 图像上传

2 个回答

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