因此,我正在尝试为Google云端存储上托管的图片动态创建公共网址。我是否可以调用文件名的值并将其附加到结构化URL,例如?
import os
filenames = os.listdir()
our_filename = "foo"
cur = 0
cur_filename = "foo"
extension = ".txt"
while(True):
if (cur_filename) in filenames:
cur += 1
cur_filename = our_filename + str(cur) + extension
else:
# found a filename that doesn't exist
f = open(cur_filename,'w')
f.write(stuff)
f.close()
<%= img_tag(https://storage.googleapis.com/cods-images/uploads/<% @latestpost.thumbImg %>) %>
与文件名完全匹配的位置?
尝试此操作时,出现以下错误:
@latestpost.thumbImg
如果我执行此操作keyword argument must be followed by space after: https:
,则只会将<img src="https://storage.googleapis.com/cods-images/uploads/@latestpost.thumbImg)" />
的值反对添加以创建完整的URL字符串。
答案 0 :(得分:2)
您应该能够执行以下操作:
<img src="https://storage.googleapis.com/cods-images/uploads/<%= @latestpost.thumbImg %>" />