Question

以下是查看哪个网址可以安全浏览的代码。我使用了google api。我面临的问题是我无法获得SafeBrowsing类对象来命中给定的URL。亲切地看看是否有人有解决方案。

if ($scope.model.pageState === condition) {
    $scope.model.disableDate = true;
}

Answer 1

试试这个。

public final NetHttpTransport httpTransport = GoogleNetHttpTransport.newTrustedTransport();
public static final JacksonFactory GOOGLE_JSON_FACTORY = JacksonFactory.getDefaultInstance();

Safebrowsing.Builder safebrowsingBuilder = new Safebrowsing.Builder(httpTransport, GOOGLE_JSON_FACTORY, null);
Safebrowsing safebrowsing = safebrowsingBuilder.build();
FindThreatMatchesResponse findThreatMatchesResponse = safebrowsing.threatMatches().find(findThreatMatchesRequest).setKey(GOOGLE_API_KEY).execute();

List<ThreatMatch> threatMatches = findThreatMatchesResponse.getMatches();

if (threatMatches != null && threatMatches.size() > 0) {
    for (ThreatMatch threatMatch : threatMatches) {
        threatUrls.add(threatMatch.getThreat().getUrl());
    }
}

完整的示例代码：https://github.com/kalinchih/java_google_safebrowsing_v4

Google安全浏览v4 API java

1 个答案: