Can google vision API accept external image URL?

Question

I am reading documentation on vision API request schema. In image source, I only see option of using url of GCS image paths. Is it possible to use external image url like http://example.com/images/image01.jpg ?

Answer 1

Yes, this is working perfectly fine:

{
  "requests": [
    {
      "features": [
        {
          "type": "WEB_DETECTION"
        }
      ],
      "image": {
        "source": {
          "imageUri": "http://i3.kym-cdn.com/photos/images/facebook/000/242/592/1c8.jpg"
        }
      }
    }
  ]
}

Answer 2

Yes it can, but ONLY using google cloud storage urls. Try this:

{
  "requests": [
    {
      "image": {
        "source": {
          gcsImageUri: 'gs://something.com/image',
        },
      },
      "features": ...
      "imageContext": ...
    },
  ]
}

Answer 3

Yes, you can do this with any image as long as it is smaller than 4Mb. It does not have to be on Google Cloud Storage.

Here is an example using the Golang client library:

// Copyright 2016 Google Inc. All rights reserved.
// Use of this source code is governed by the Apache 2.0
// license that can be found in the LICENSE file.

// [START vision_quickstart]
// Sample vision-quickstart uses the Google Cloud Vision API to label an image.
package main

import (
    "fmt"
    "log"

    // Imports the Google Cloud Vision API client package.
    vision "cloud.google.com/go/vision/apiv1"
    "golang.org/x/net/context"
)

func main() {
    ctx := context.Background()

    // Creates a client.
    client, err := vision.NewImageAnnotatorClient(ctx)
    if err != nil {
        log.Fatalf("Failed to create client: %v", err)
    }

    image := vision.NewImageFromURI("https://www.denaligrizzlybear.com/assets/images/scenic-denali.jpg")

    labels, err := client.DetectLabels(ctx, image, nil, 10)
    if err != nil {
        log.Fatalf("Failed to detect labels: %v", err)
    }

    fmt.Println("Labels:")
    for _, label := range labels {
        fmt.Println(label.Description)
    }
}

Here is the function on the Godoc:https://godoc.org/cloud.google.com/go/vision/apiv1#NewImageFromURI

The docs state:

NewImageFromURI returns an image that refers to an object in Google Cloud Storage (when the uri is of the form "gs://BUCKET/OBJECT") or at a public URL.

Answer 4

Answer for python users.

def detect_labels_uri(uri):
    """Detects labels in the file located in Google Cloud Storage or on the
    Web."""
    from google.cloud import vision
    client = vision.ImageAnnotatorClient()
    image = vision.types.Image()
    image.source.image_uri = uri

    response = client.label_detection(image=image)
    labels = response.label_annotations
    print('Labels:')

    for label in labels:
        print(label.description)
# [END vision_label_detection_gcs]

Answer 5

Yes you can use external image URL and even can use BASE64 image content also.

You can find the same in official documentation also.

Documentation URL : https://cloud.google.com/vision/docs/request#json_request_format

I have checked it with all available options and it is working fine for me.

Answer 6

Yes, Google Cloud Vision API DOES accept external URL images. I just used this image to get labels:

python label_sample.py -u "https://upload.wikimedia.org/wikipedia/commons/e/e6/Bleeding_finger.jpg"

Found label: red with 96.47 percent probability!!!!!!!!!!!
Found label: color with 95.46 percent probability!!!!!!!!!!!
Found label: pink with 92.15 percent probability!!!!!!!!!!!
Found label: finger with 91.06 percent probability!!!!!!!!!!!
Found label: hand with 90.45 percent probability!!!!!!!!!!!
Found label: nail with 73.23 percent probability!!!!!!!!!!!
Found label: lip with 73.09 percent probability!!!!!!!!!!!
Found label: jewellery with 68.84 percent probability!!!!!!!!!!!
Found label: produce with 68.39 percent probability!!!!!!!!!!!
Found label: macro photography with 67.86 percent probability!!!!!!!!!!!

You have to give it the url in proper format using urllib library and comment the part of opening an image like this:

url = url_file
opener = urllib.urlopen(url)

#    with open(photo_file, 'rb') as image:
image_content = base64.b64encode(opener.read())