python - PIL 值错误 : not enough image data?

Question

当我尝试从 URL 获取图像并将其响应中的字符串转换为 App Engine 中的 Image 时，我收到了上述消息的错误。

from google.appengine.api import urlfetch

def fetch_img(url):
  try:
    result = urlfetch.fetch(url=url)
    if result.status_code == 200:
      return result.content
  except Exception, e:
    logging.error(e)

url = "http://maps.googleapis.com/maps/api/staticmap?center=Narita+International+Airport,Narita,Chiba+Prefecture,+Japan&zoom=18&size=512x512&maptype=roadmap&markers=color:blue|label:S|40.702147,-74.015794&markers=color:green|label:G|40.711614,-74.012318&markers=color:red|color:red|label:C|40.718217,-73.998284&sensor=false"

img = fetch_img(url)
# As the URL above tells, its size is 512x512 
img = Image.fromstring('RGBA', (512, 512), img)

根据 PIL ，大小选项假设是一个像素元组。这是我指定的。谁能指出我的误解？

Answer 1

image 返回的数据是图像本身而不是 RAW RGB 数据，因此您不需要将其作为原始数据加载，而是只需将该数据保存到文件中，这将是一个有效图像或使用 PIL 来打开它，例如(我已将您的代码转换为不使用 appengine api，以便任何安装了普通 python 的人都可以运行该示例)

from urllib2 import urlopen
import Image
import sys
import StringIO

url = "http://maps.googleapis.com/maps/api/staticmap?center=Narita+International+Airport,Narita,Chiba+Prefecture,+Japan&zoom=18&size=512x512&maptype=roadmap&markers=color:blue|label:S|40.702147,-74.015794&markers=color:green|label:G|40.711614,-74.012318&markers=color:red|color:red|label:C|40.718217,-73.998284&sensor=false"
result = urlopen(url=url)
if result.getcode() != 200:
  print "errrrrr"
  sys.exit(1)

imgdata = result.read()
# As the URL above tells, its size is 512x512 
img = Image.open(StringIO.StringIO(imgdata))
print img.size

输出:

(512, 512)