在尝试在django中创建缩略图时无法识别图像文件

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可能重复：
  resize image on save

我试图在django中创建缩略图，我正在尝试构建一个专门用于生成缩略图的自定义类。如下

from cStringIO import StringIO
from PIL import Image

class Thumbnail(object):

    SIZE = (50, 50)

    def __init__(self, source):
        self.source = source
        self.output = None

    def generate(self, size=None, fit=True):
        if not size:
            size = self.SIZE

        if not isinstance(size, tuple):
            raise TypeError('Thumbnail class: The size parameter must be an instance of a tuple.')

        # resize properties
        box = size
        factor = 1
        image = Image.open(self.source)
        # Convert to RGB if necessary
        if image.mode not in ('L', 'RGB'): 
            image = image.convert('RGB')
        while image.size[0]/factor > 2*box[0] and image.size[1]*2/factor > 2*box[1]:
            factor *=2
        if factor > 1:
            image.thumbnail((image.size[0]/factor, image.size[1]/factor), Image.NEAREST)

        #calculate the cropping box and get the cropped part
        if fit:
            x1 = y1 = 0
            x2, y2 = image.size
            wRatio = 1.0 * x2/box[0]
            hRatio = 1.0 * y2/box[1]
            if hRatio > wRatio:
                y1 = int(y2/2-box[1]*wRatio/2)
                y2 = int(y2/2+box[1]*wRatio/2)
            else:
                x1 = int(x2/2-box[0]*hRatio/2)
                x2 = int(x2/2+box[0]*hRatio/2)
            image = image.crop((x1,y1,x2,y2))

        #Resize the image with best quality algorithm ANTI-ALIAS
        image.thumbnail(box, Image.ANTIALIAS)

        # save image to memory
        temp_handle = StringIO()
        image.save(temp_handle, 'png')
        temp_handle.seek(0)

        self.output = temp_handle

        return self

    def get_output(self):
        self.output.seek(0)
        return self.output.read()

这个类的目的是让我可以在不同的位置使用它来动态生成缩略图。这个类工作得很好，我直接在视图下测试过..我在表单的save方法中实现了缩略图类，以便在保存时调整原始图像的大小。

在我的设计中，我有两个缩略图字段。我能够生成一个缩略图，如果我尝试生成两个崩溃，我已经被困了几个小时不知道是什么问题。

这是我的模特

class Image(models.Model):
    article         = models.ForeignKey(Article)
    title           = models.CharField(max_length=100, null=True, blank=True)
    src             = models.ImageField(upload_to='publication/image/')
    r128            = models.ImageField(upload_to='publication/image/128/', blank=True, null=True)
    r200            = models.ImageField(upload_to='publication/image/200/', blank=True, null=True)

    uploaded_at     = models.DateTimeField(auto_now=True)

这是我的表格

class ImageForm(models.ModelForm):
    """

    """
    class Meta:
        model = Image
        fields = ('src',)


    def save(self, commit=True):
        instance = super(ImageForm, self).save(commit=True)


        instance.r128 = SimpleUploadedFile(
                    instance.src.name,
                    Thumbnail(instance.src).generate((128, 128)).get_output(),
                    content_type='image/png'
                )


        instance.r200 = SimpleUploadedFile(
            instance.src.name,
            Thumbnail(instance.src).generate((200, 200)).get_output(),
            content_type='image/png'
        )

        if commit:
            instance.save()
        return instance

奇怪的是，当我在表单中删除包含instance.r200的行时保存。它工作正常，它完成缩略图并成功存储它。一旦我添加第二个缩略图，它就会失败..

任何想法在这里做错了什么？

由于

更新

根据评论请求，附加错误跟踪

IOError at /en/publication/new/

cannot identify image file

Request Method:     POST
Request URL:    http://127.0.0.1:8000/en/publication/new/?image-extra=
Django Version:     1.4.2
Exception Type:     IOError
Exception Value:    

cannot identify image file

Exception Location:     /Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/PIL/Image.py in open, line 1980
Python Executable:  /Users/mo/Projects/pythonic/snowflake-env/bin/python
Python Version:     2.7.2

更新

尝试创建print语句，下面是输出

Source: publication/image/tumblr_m9o7244nZM1rykg1io1_1280_11.jpg
Source: publication/image/tumblr_m9o7244nZM1rykg1io1_1280_11.jpg
ERROR:root:cannot identify image file
ERROR:django.request:Internal Server Error: /en/publication/new/
Traceback (most recent call last):
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/django/core/handlers/base.py", line 111, in get_response
    response = callback(request, *callback_args, **callback_kwargs)
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/django/contrib/auth/decorators.py", line 20, in _wrapped_view
    return view_func(request, *args, **kwargs)
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/django/db/transaction.py", line 209, in inner
    return func(*args, **kwargs)
  File "/Users/mo/Projects/pythonic/snowflake-env/snowflake/snowflake/apps/publication/views.py", line 69, in new
    formset.save()
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/django/forms/models.py", line 497, in save
    return self.save_existing_objects(commit) + self.save_new_objects(commit)
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/django/forms/models.py", line 628, in save_new_objects
    self.new_objects.append(self.save_new(form, commit=commit))
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/django/forms/models.py", line 727, in save_new
    obj = form.save(commit=False)
  File "/Users/mo/Projects/pythonic/snowflake-env/snowflake/snowflake/apps/publication/forms.py", line 113, in save
    Thumbnail(instance.src).generate((200, 200)).get_output(),
  File "/Users/mo/Projects/pythonic/snowflake-env/snowflake/snowflake/apps/core/utils.py", line 23, in generate
    image = Image.open(self.source)
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/PIL/Image.py", line 1980, in open
    raise IOError("cannot identify image file")
IOError: cannot identify image file

如图所示，第二张图像失败后，第一张图像被成功打印并处理。

更新

在缩略图类

中应用copy（）后，

跟踪错误更新

ERROR:root:cannot identify image file
ERROR:django.request:Internal Server Error: /en/publication/new/
Traceback (most recent call last):
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/django/core/handlers/base.py", line 111, in get_response
    response = callback(request, *callback_args, **callback_kwargs)
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/django/contrib/auth/decorators.py", line 20, in _wrapped_view
    return view_func(request, *args, **kwargs)
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/django/db/transaction.py", line 209, in inner
    return func(*args, **kwargs)
  File "/Users/mo/Projects/pythonic/snowflake-env/snowflake/snowflake/apps/publication/views.py", line 69, in new
    formset.save()
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/django/forms/models.py", line 497, in save
    return self.save_existing_objects(commit) + self.save_new_objects(commit)
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/django/forms/models.py", line 628, in save_new_objects
    self.new_objects.append(self.save_new(form, commit=commit))
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/django/forms/models.py", line 727, in save_new
    obj = form.save(commit=False)
  File "/Users/mo/Projects/pythonic/snowflake-env/snowflake/snowflake/apps/publication/forms.py", line 113, in save
    f128.write(Thumbnail(instance.src).generate((128, 128)).get_output())
  File "/Users/mo/Projects/pythonic/snowflake-env/snowflake/snowflake/apps/core/utils.py", line 15, in __init__
    self._pilImage = Image.open(self.source)
  File "/Users/mo/Projects/pythonic/snowflake-env/lib/python2.7/site-packages/PIL/Image.py", line 1980, in open
    raise IOError("cannot identify image file")
IOError: cannot identify image file

更新

最后，我设法让它工作，但我必须将文件流式传输到self.source，因为belo

def __init__(self, source):
    self.source = StringIO(file(source.path, "rb").read())
    self.output = None

    self._pilImage = Image.open(self.source)

是上述理想方法吗？在每次点击时读取文件是个好主意吗？如果不是，我的替代方案是什么？

Answer 1

我看到的问题在于您设计Thumbnail课程的方式。它使用类属性来存储实例变量，这意味着当您尝试多次使用该类时会出现冲突。

不需要静态load方法，因为一旦将属性移动到实例，它就会与类的构造函数完全相同。通过在构造函数中要求source，可以确保在generate中查找空字符串值时不会发生崩溃。

此外，我认为您面临的一个主要问题是当您使用类似文件的对象包装器时，您的django模型将返回ImageField。虽然如果传入字符串路径，则不会看到此内容，但是当您传入文件对象时，generate方法会将其读取到最后。然后，您使用相同的源对象再次调用generate，但它最后会得到IOError。现在，一种方法是确保在再次使用0再次调用Thumbnail之前寻找源代码，但是您可以省去麻烦并让您的Thumbnail课程打开，在构造函数中缓存PIL图像一次。然后generate不需要每次都重新读取它。

# Example from your code #
def generate(self, size=None, fit=True):
    ...
    # The first time you do this, it will read
    # self.source to the end, because in Django, you
    # are passing a file-like object.
    image = Image.open(self.source)

# this will work the first time
generate()
# uh oh. self.source was a file object that is at the end
generate() # crash

重写缩略图类

from cStringIO import StringIO
from PIL import Image

class Thumbnail(object):

    SIZE = (50, 50)

    def __init__(self, source):
        self.source = source
        self.output = None

        self._pilImage = Image.open(self.source)

    def generate(self, size=None, fit=True):
        if not size:
            size = self.SIZE

        if not isinstance(size, tuple):
            raise TypeError('Thumbnail class: The size parameter must be an instance of a tuple.')

        # resize properties
        box = size
        factor = 1
        image = self._pilImage.copy()

        # Convert to RGB if necessary
        if image.mode not in ('L', 'RGB'): 
            image = image.convert('RGB')
        while image.size[0]/factor > 2*box[0] and image.size[1]*2/factor > 2*box[1]:
            factor *=2
        if factor > 1:
            image.thumbnail((image.size[0]/factor, image.size[1]/factor), Image.NEAREST)

        #calculate the cropping box and get the cropped part
        if fit:
            x1 = y1 = 0
            x2, y2 = image.size
            wRatio = 1.0 * x2/box[0]
            hRatio = 1.0 * y2/box[1]
            if hRatio > wRatio:
                y1 = int(y2/2-box[1]*wRatio/2)
                y2 = int(y2/2+box[1]*wRatio/2)
            else:
                x1 = int(x2/2-box[0]*hRatio/2)
                x2 = int(x2/2+box[0]*hRatio/2)
            image = image.crop((x1,y1,x2,y2))

        #Resize the image with best quality algorithm ANTI-ALIAS
        image.thumbnail(box, Image.ANTIALIAS)

        # save image to memory
        temp_handle = StringIO()
        image.save(temp_handle, 'png')
        temp_handle.seek(0)

        self.output = temp_handle

        return self

    def get_output(self):
        self.output.seek(0)
        return self.output.read()

用法：Thumbnail(src).generate((200, 200)).get_output()

source和output对每个实例都必须是唯一的。但是在您的版本中，您可以将output设置为类级别，这意味着Thumbnail的两个实例使用最新版本的共享output。

# your code #
    # this is assigning the most recently processed
    # object to the class level. shared among all.
    self.output = temp_handle

    return self

def get_output(self):
    # always read the shared class level
    return self.output.read()

此外，我觉得有一种更简单的方法来执行调整大小/适合/裁剪。如果你解释想要为图像做的确切转换，我也可以简化它。

<强>更新

我忘了特别提到我保存源图像一次的建议，你的用法应该是这样的：

def save(self, commit=True):
    instance = super(ImageForm, self).save(commit=True)

    thumb = Thumbnail(instance.src)

    instance.r128 = SimpleUploadedFile(
        instance.src.name,
        thumb.generate((128, 128)).get_output(),
        content_type='image/png'
    )

    instance.r200 = SimpleUploadedFile(
        instance.src.name,
        thumb.generate((200, 200)).get_output(),
        content_type='image/png'
    )

请注意，我们只使用源创建一个Thumbnail实例，它只在PIL中打开一次。然后，您可以根据需要生成任意数量的图像。

Answer 2

PIL.Image.open(...)的参数可以是文件名或文件对象。如果使用类似对象的文件，则读取位置应位于文件的开头。您使用文件对象。 （这是肯定的，因为您使用instance.src.name然后传递Thumbnail(instance.src)。）

解决方案：在创建第二个缩略图之前将文件回到instance.src.seek(0)开头，或者只传递文件名，而不是文件对象：Thumbnail(instance.src.name)。