使用 Django 提供动态生成的图像
[英]Serve a dynamically generated image with Django
问题描述
How do I serve a dynamically generated image in Django?
如何在 Django 中提供动态生成的图像？
I have an html tag我有一个 html 标签
<html>
...
<img src="images/dynamic_chart.png" />
...
</html>
linked up to this request handler, which creates an in-memory image链接到这个请求处理程序，它创建一个内存中的图像
def chart(request):
img = Image.new("RGB", (300,300), "#FFFFFF")
data = [(i,randint(100,200)) for i in range(0,300,10)]
draw = ImageDraw.Draw(img)
draw.polygon(data, fill="#000000")
# now what?
return HttpResponse(output)
I also plan to change the requests to AJAX, and add some sort of caching mechanism, but my understanding is that wouldn't affect this part of the solution.我还计划将请求更改为 AJAX，并添加某种缓存机制，但我的理解是这不会影响解决方案的这一部分。
4 个解决方案
解决方案1
31 已采纳 2009-07-02 13:15:02
I assume you're using PIL (Python Imaging Library).我假设您正在使用 PIL（Python 成像库）。 You need to replace your last line with (for example, if you want to serve a PNG image):您需要将最后一行替换为（例如，如果您想提供 PNG 图像）：
response = HttpResponse(mimetype="image/png")
img.save(response, "PNG")
return response
解决方案2
3 2009-07-02 13:13:36
I'm relatively new to Django myself.我自己对 Django 比较陌生。 I haven't been able to find anything in Django itself, but I have stumbled upon a project on Google Code that may be of some help to you:我没能在 Django 中找到任何东西，但我偶然发现了一个关于 Google Code 的项目，它可能对你有帮助：
解决方案3
1 2020-02-06 23:48:07
I was looking for a solution of the same problem我正在寻找同样问题的解决方案
And for me this simple approach worked fine:对我来说，这种简单的方法效果很好：
from django.http import FileResponse
def dyn_view(request):
response = FileResponse(open("image.png","rb"))
return response
解决方案4
0 2020-08-01 10:12:18
Another way is to use BytesIO.另一种方法是使用 BytesIO。 BytesIO is like a buffer. BytesIO 就像一个缓冲区。 So one can save the image (which is fast enough than writing to disk) in that buffer.因此，可以在该缓冲区中保存图像（这比写入磁盘快得多）。
from PIL import Image, ImageDraw
import io
def chart(request):
img = Image.new('RGB', (240, 240), color=(250,160,170))
draw = ImageDraw.Draw(img)
draw.text((20, 40), 'some_text')
buff = io.BytesIO()
img.save(buff, 'jpeg')
return HttpResponse(buff.getvalue(), content_type='image/jpeg')
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