Question

错误表示缩进位于＆＃34; if _name＆＃34;部分..我不确定为什么会表明这一点。

class GetFileHandler(tornado.web.RequestHandler):
    def get(self):

        fileid = self.get_argument('fileid', "")

        cur.execute("""SELECT filepath FROM files_table WHERE file_id = %s""", (fileid, ))
        m = cur.fetchall()
        y = m[0]
        x = y[0]

        path = x + "/" + fileid + ".jpg"

        try:
            with open(path, 'rb') as f:
                data = f.read()
                self.write(data)
            self.finish()
if __name__ == "__main__": 
    tornado.options.parse_command_line() 
    app = tornado.web.Application(handlers=[(r"/getit", GetFileHandler)])
    http_server = tornado.httpserver.HTTPServer(app) 
    http_server.listen(options.port) 
    tornado.ioloop.IOLoop.instance().start()

Answer 1

try需要except

try:
    with open(path, 'rb') as f:
        data = f.read()
        self.write(data)
    self.finish()
except IOError:
    print "Failed!!"

为了让它显示为图像，您需要设置内容标题以反映它是图像的mime类型...

Answer 2

匹配get函数的except块在哪里？您应该至少在函数中添加类似以下内容的内容。

except IOError as xcpt:
    # IO error handling
    raise xcpt  # if you want to propagate the exception

Answer 3

如果你有一个单独的图像文件夹，并希望将它们作为带有网址的静态内容提供，例如

http://yourwebsite.com/images/yourimage.jpg

然后您可以使用tornado.web.StaticFileHandler：

handlers = [
        (r"/images/(.*)", tornado.web.StaticFileHandler, {'path': "./images"}),
        (r"/", WebHandler)
]

如何使用Tornado Python服务器在Web浏览器上显示图像？

3 个答案: