2

我在我的 android 代码中使用安全浏览 api v4。我正在使用 Asynctask 并使用 httpurlconnection 请求安全浏览 api。响应始终为空。

我也使用测试 URL http://malware.testing.google.test/testing/malware/测试了我的连接，然后它也返回空。

class Malicious extends AsyncTask<String, Void, Wrapper> {
    private OnTaskCompleted listener;

    public Malicious(OnTaskCompleted listener){
        this.listener=listener;
    }

    @Override
    protected void onPreExecute() {
        super.onPreExecute();

    }

    protected Wrapper doInBackground(String... args) {
        String postURL = "https://safebrowsing.googleapis.com/v4/threatMatches:find?key=APIKEY";

        String requestBody = "{" +
                "    \"client\": {" +
                "      \"clientId\":      \"twittersentidetector\"," +
                "      \"clientVersion\": \"1.0\"" +
                "    }," +
                "    \"threatInfo\": {" +
                "      \"threatTypes\":      [\"MALWARE\", \"SOCIAL_ENGINEERING\"]," +
                "      \"platformTypes\":    [\"ANY_PLATFORM\"]," +
                "      \"threatEntryTypes\": [\"URL\"]," +
                "      \"threatEntries\": [" +
                "        {\"url\": \"" + args[0] + "\"}," +
                "      ]" +
                "    }" +
                "  }";



        URL url;
        StringBuffer response = new StringBuffer();
        try {
            url = new URL(postURL);
        } catch (MalformedURLException e) {
            throw new IllegalArgumentException("invalid url");
        }

        HttpURLConnection conn = null;
        try {
            conn = (HttpURLConnection) url.openConnection();
            conn.setDoOutput(false);
            conn.setDoInput(true);
            conn.setUseCaches(false);
            conn.setRequestMethod("POST");
            conn.setRequestProperty("Content-Type", "application/json");
            conn.setRequestProperty("User-Agent", USER_AGENT);
            conn.setRequestProperty("Accept-Language", "en-US,en;q=0.5");
            try( DataOutputStream wr = new DataOutputStream( conn.getOutputStream())) {
                byte[] b = requestBody.getBytes();
                wr.write(b);
                wr.flush();
                wr.close();
            }

            // handle the response
            int status = conn.getResponseCode();
            if (status != 200) {
                throw new IOException("Post failed with error code " + status);
            } else {
                BufferedReader in = new BufferedReader(new InputStreamReader(conn.getInputStream()));
                String inputLine;
                while ((inputLine = in.readLine()) != null) {
                    response.append(inputLine);
                }
                in.close();
            }
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            if (conn != null) {
                conn.disconnect();
            }

            //Here is your json in string format
            String responseJSON = response.toString();
            Wrapper w = new Wrapper();
            w.responce = responseJSON;
            w.url = args[0];

            return w;
        }





    }

    @Override
    protected void onPostExecute(Wrapper xml) {

        if(xml.responce.length()==0){
            showtoast("safe");
        }
        else{
            showtoast("not safe");
            listener.onTaskCompleted(xml.url);
        }

    }

}

对于测试 url，它也显示安全。api 面板显示请求已发出，我不知道我在做什么错，我使用的是最新版本的 V4 API，但无论使用什么 URL，始终显示它是安全的.

4

2 回答 2

2

我设置了这个基本的 curl 脚本，它只返回空括号 {} 围绕共识进行搜索似乎是因为我正在查找的网站不是恶意的。只有您提供的示例（http://malware.testing.google.test/testing/malware/）给了我结果。我仍然不相信这能正常工作，但这是脚本。希望能帮助到你。（将 API_KEY 替换为您的 api 密钥）

<\?php

$parameters = array('client' => array('clientId'=>'Evolved Marketing', 'clientVersion'=>'1.5.2'), 'threatInfo' => array('threatTypes'=>array('MALWARE ', 'SOCIAL_ENGINEERING'), 'platformTypes'=>array('ANY_PLATFORM'), 'threatEntryTypes'=>array('URL'), 'threatEntries'=>array(array('url'=>' http://恶意软件.testing.google.test/testing/malware/ ') )), );

$json = json_encode($parameters);

$ch = curl_init(); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);

curl_setopt($ch, CURLOPT_URL, ' https://safebrowsing.googleapis.com/v4/threatMatches:find?key=API_KEY ');

curl_setopt($ch, CURLOPT_POST, TRUE);

curl_setopt($ch, CURLOPT_POSTFIELDS, $json);

curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-type: application/json'));

$response = curl_exec($ch);

print_r($response);

于 2019-04-18T18:35:19.773 回答