1

可能重复：
保存时调整图像大小

我正在尝试在 django 中创建缩略图，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()

该课程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。该类完美运行，我直接在视图下对其进行了测试。我在表单的保存方法中实现了缩略图类，以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它会崩溃并且我已经卡了几个小时不知道是什么问题。

这是我的模型

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，当我在保存表单中删除包含 instance.r200 的行时。它工作正常，它会生成缩略图并成功存储它。一旦我添加了第二个缩略图，它就会失败..

有什么想法在这里做错了吗？

谢谢

更新：

根据评论请求，我附加了错误跟踪

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

试图创建打印语句，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，第一张图像打印并成功处理，第二张图像失败。

更新

在缩略图类中应用 copy() 后追溯错误更新

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，但我不得不将文件流式传输到 self.source 作为 belo

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

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

以上是理想的方法吗？每次点击时读取文件是个好主意吗？如果没有，我的替代方案是什么？

4

2 回答 2

4

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发生冲突。

不需要静态load方法，因为一旦将属性移动到实例，它的作用就与类的构造函数完全相同。通过在构造函数中要求 a ，您可以确保稍后在查找空字符串值时source不会发生崩溃。generate

此外，我认为您面临的主要问题之一是当您使用 django 模型为ImageField's. 如果您传入字符串路径，则不会看到这一点，但当您传入文件对象时，该generate方法会将其读取到最后。然后你generate用相同的源对象第二次调用，但它在最后，你得到一个IOError. 现在一种方法是确保在再次0调用之前先找到源代码Thumbnail，但是您可以省去麻烦，只需Thumbnail打开您的类并在构造函数中缓存一次 PIL 图像。然后generate就不需要每次都不断地重读一遍。

# 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

重写的缩略图类

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需要对于每个实例都是唯一的。但是在您的版本中，您将设置output为类级别，这意味着两个实例Thumbnail使用共享的最新版本的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()

另外，我觉得有一种更简单的方法来执行调整大小/适合/裁剪。如果您解释要对图像进行的确切转换，我也可以将其简化。

更新

我忘了特别提到我建议保存一次源图像，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，我们只创建了一个使用源的实例，Thumbnail它只会在 PIL 中打开一次。然后，您可以从中生成任意数量的图像。

于 2012-12-01T17:00:58.430 回答
2

的参数PIL.Image.open(...)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，因为你使用instance.src.name然后你通过了Thumbnail(instance.src)。）

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

于 2012-12-02T01:16:19.880 回答