Universal Scraping API

Targeting websites not listed in "Available targets" list

You can use "universal" parameter as your target and supply any URL you want, which will return the HTML of the targeted URL

Code examples for targeting ip.smartproxy.com website (with authorization details changed only):

import requests

url = "https://scrape.smartproxy.com/v1/tasks"

payload = {
    "target": "universal",
    "parse": False,
    "url": "https://ip.smartproxy.com"
}
headers = {
    "Accept": "application/json",
    "Content-Type": "application/json",
    "Authorization": "Base64 encoded user:pass"
}

response = requests.request("POST", url, json=payload, headers=headers)

print(response.text)

<?php
require_once('vendor/autoload.php');

$client = new \GuzzleHttp\Client();

$response = $client->request('POST', 'https://scrape.smartproxy.com/v1/tasks', [
  'body' => '{"target":"universal","parse":false,"url":"https://ip.smartproxy.com"}',
  'headers' => [
    'Accept' => 'application/json',
    'Authorization' => 'Base64 encoded user:pass',
    'Content-Type' => 'application/json',
  ],
]);

echo $response->getBody();

const options = {
  method: 'POST',
  headers: {
    Accept: 'application/json',
    'Content-Type': 'application/json',
    Authorization: 'Base64 encoded user:pass'
  },
  body: JSON.stringify({target: 'universal', parse: false, url: 'https://ip.smartproxy.com'})
};

fetch('https://scrape.smartproxy.com/v1/tasks', options)
  .then(response => response.json())
  .then(response => console.log(response))
  .catch(err => console.error(err));

curl --request POST \
     --url https://scrape.smartproxy.com/v1/tasks \
     --header 'Accept: application/json' \
     --header 'Authorization: Base64 encoded user:pass' \
     --header 'Content-Type: application/json' \
     --data '
{
     "target": "universal",
     "parse": false,
     "url": "https://ip.smartproxy.com"
}
'

Result of the above code: