I would do:
def getImage(url,db):
user_agent = 'Mozilla/4.0 (compatible; MSIE 5.5; Windows NT)'
headers = {'User-Agent': user_agent}
req = urllib2.Request(url,"",headers)
try:
response = urllib2.urlopen(req)
except urllib2.URLError, e:
if hasattr(e, 'reason'):
print 'We failed to reach a server.'
print 'Reason: ', e.reason
elif hasattr(e, 'code'):
print 'The server couldn\'t fulfill the request.'
print 'Error code: ', e.code
else:
pass # everything is fine
print response.geturl()
print response.info()
# print response.read() # <<< do not read it
# f = open(response.read(),'rb') # << response is already a stream
# stream = open(response,'rb')
id =
db.image.insert(url=url,image=db.image.image_upload.store(response,filename=url))
db.commit() # not sure if needed, depends on where executed
return id
then return it with the provided download action or
def imageLookup():
id = request.vars.id
#response.headers['Content-Type']='image/jpeg'
path = os.path.join(request.folder,'uploads',db.image[id].image_file)
return response.stream(open(path,'rb'))
On Thursday, 5 July 2012 06:23:31 UTC-5, RCTYCO wrote:
>
>
> I am trying to save an image from another database into my
> database, essentially i am caching the a copy into my database so I don't
> have to query the external site.
>
> I have a link
> http://eandata.com/image/products/004/900/000/0049000000443.jpg.
>
> I want to save the jpg into a blob in my database.
>
> Model
> db.define_table('image',
> Field('url','string'),
> Field('image_upload','upload',uploadfield='image_file'),
> Field('image_file','blob'))
>
> Function. I have this running a module and called when I need to download
> the external image.
>
> def getImage(url,db):
>
> user_agent = 'Mozilla/4.0 (compatible; MSIE 5.5; Windows NT)'
> headers = {'User-Agent': user_agent}
>
> req = urllib2.Request(url,"",headers)
>
> try:
> response = urllib2.urlopen(req)
> except urllib2.URLError, e:
> if hasattr(e, 'reason'):
> print 'We failed to reach a server.'
> print 'Reason: ', e.reason
> elif hasattr(e, 'code'):
> print 'The server couldn\'t fulfill the request.'
> print 'Error code: ', e.code
> else:
> pass # everything is fine
>
> print response.geturl()
> print response.info()
> print response.read()
> f = open(response.read(),'rb')
> #stream = open(response,'rb')
>
> return
> db.image.insert(url=url,image=db.image.image_upload.store(f,url))
>
>
> *I am having a problem with my insert statement. I can't seem to figure
> out how to save the image into database.* urllib2.urlopen(req) will
> download the file and provide a file object. But I don't know if i can just
> save the file or do i need to read the file object within
> the urllib2.urlopen(req).
>
> I would like the image to be available to be via
> http://mydomain.com/myapp/imageLookup?id=2
> <http://mydomain.com/myapp/imageLookup?id=2>
>
> Controller
> def imageLookup():
> id = request.vars.id
> #response.headers['Content-Type']='image/jpeg'
>
> return db.image[id].image_file
>
>
> Do you have any suggestions? The code abo
>
