Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Scrapy image download how to use custom filename

Tags:

python

scrapy

For my scrapy project I'm currently using the ImagesPipeline. The downloaded images are stored with a SHA1 hash of their URLs as the file names.

How can I store the files using my own custom file names instead?

What if my custom file name needs to contain another scraped field from the same item? e.g. use the item['desc'] and the filename for the image with item['image_url']. If I understand correctly, that would involve somehow accessing the other item fields from the Image Pipeline.

Any help will be appreciated.

like image 208
fortuneRice Avatar asked May 31 '11 21:05

fortuneRice


4 Answers

This is just actualization of the answer for scrapy 0.24 (EDITED), where the image_key() is deprecated

class MyImagesPipeline(ImagesPipeline):

    #Name download version
    def file_path(self, request, response=None, info=None):
        #item=request.meta['item'] # Like this you can use all from item, not just url.
        image_guid = request.url.split('/')[-1]
        return 'full/%s' % (image_guid)

    #Name thumbnail version
    def thumb_path(self, request, thumb_id, response=None, info=None):
        image_guid = thumb_id + response.url.split('/')[-1]
        return 'thumbs/%s/%s.jpg' % (thumb_id, image_guid)

    def get_media_requests(self, item, info):
        #yield Request(item['images']) # Adding meta. I don't know, how to put it in one line :-)
        for image in item['images']:
            yield Request(image)
like image 192
sumid Avatar answered Nov 17 '22 23:11

sumid


In scrapy 0.12 I solved something like this

class MyImagesPipeline(ImagesPipeline):

    #Name download version
    def image_key(self, url):
        image_guid = url.split('/')[-1]
        return 'full/%s.jpg' % (image_guid)

    #Name thumbnail version
    def thumb_key(self, url, thumb_id):
        image_guid = thumb_id + url.split('/')[-1]
        return 'thumbs/%s/%s.jpg' % (thumb_id, image_guid)

    def get_media_requests(self, item, info):
        yield Request(item['images'])
like image 26
Ivan Saltikov Avatar answered Nov 18 '22 01:11

Ivan Saltikov


I found my way in 2017,scrapy 1.1.3

def file_path(self, request, response=None, info=None):
    return request.meta.get('filename','')

def get_media_requests(self, item, info):
    img_url = item['img_url']
    meta = {'filename': item['name']}
    yield Request(url=img_url, meta=meta)

like the code above,you can add the name you want to a Request meta in get_media_requests(), and get it back in file_path() by request.meta.get('yourname','').

like image 9
Tarjintor Avatar answered Nov 17 '22 23:11

Tarjintor


This was the way I solved the problem in Scrapy 0.10 . Check the method persist_image of FSImagesStoreChangeableDirectory. The filename of the downloaded image is key

class FSImagesStoreChangeableDirectory(FSImagesStore):

    def persist_image(self, key, image, buf, info,append_path):

        absolute_path = self._get_filesystem_path(append_path+'/'+key)
        self._mkdir(os.path.dirname(absolute_path), info)
        image.save(absolute_path)

class ProjectPipeline(ImagesPipeline):

    def __init__(self):
        super(ImagesPipeline, self).__init__()
        store_uri = settings.IMAGES_STORE
        if not store_uri:
            raise NotConfigured
        self.store = FSImagesStoreChangeableDirectory(store_uri)
like image 8
llazzaro Avatar answered Nov 17 '22 23:11

llazzaro