Paypal付款:如何在webview中加载paypal时获得成功请求

Question

编辑:

在paypal登录后,我可以成功完成transaction.But我需要匹配paypal 中的successUrl来验证两个url是否相同然后显示成功的toast消息.

但是我没有从付款中获得成功.因此我无法与之匹配.我已经发布了相关代码:

WebActivity.java:

public class PaypalWebActivity extends Activity {

    private WebView webView;

    String payUrlStr;

    ProgressDialog dialog;

    String successUrl;


    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);

        setContentView(R.layout.paypal_web_layout);

        successUrl = LOAD_WEBVIEW_PAYMENT_PAYPAL_SUCCESS;

        dialog = ProgressDialog.show(PaypalWebActivity.this, "", "Please wait..", false);

        loadWebViewPaypal();

    }

    private void loadWebViewPaypal() {

        payUrlStr = LOAD_WEBVIEW_PAYMENT_PAYPAL(PAGE_ID);       

        Log.e("payUrlStr", ""+payUrlStr);

        webView = (WebView) findViewById(R.id.webView);
        webView.loadUrl(payUrlStr);
        webView.getSettings().setJavaScriptEnabled(true);

        @SuppressWarnings("unused")
        WebSettings settings= webView.getSettings();
        if (Build.VERSION.SDK_INT >= 21) {
            webView.getSettings().setMixedContentMode( WebSettings.MIXED_CONTENT_ALWAYS_ALLOW );
           }

        webView.setWebViewClient(new WebViewClient() {

            @Override
            public boolean shouldOverrideUrlLoading(WebView view, String url) {
                Log.e("Loading url...", url);
                view.loadUrl(url);

               String loadWebUrl = view.getUrl();

                Log.e("loadWebUrl", ""+loadWebUrl);

                return true;
            }

            @Override
            public void onPageFinished(WebView view, String url) {
                Log.e("Finished url...", url);

                String webUrl = view.getUrl();

                Log.e("webUrl", ""+webUrl);


                if(webUrl.substring(0,95).equals(successUrl)){

                    Log.e("Getting Success Request", "Test");

                }else{

                    Log.e("Failed to get Request", "Test");

                }

                if(dialog.isShowing()){
                    dialog.dismiss();
                }

            }

            @Override
            public void onReceivedError(WebView view, int errorCode,
                    String description, String failingUrl) {

                Log.e("Error in url...", description);
                Log.e("Error in failingUrl...", failingUrl);

            }

        });        

    }



}

表现:

<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
<uses-permission android:name="android.permission.READ_EXTERNAL_STORAGE" />
<uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" />
<uses-permission android:name="com.google.android.providers.gsf.permission.READ_GSERVICES" />
<uses-permission android:name="android.permission.ACCESS_COARSE_LOCATION" />
<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION" />

检查这个讨论.一步一步我把截图和添加的内容清楚地理解.

我收到此错误消息.因此无法匹配成功请求:

06-15 18:12:59.894:I/chromium(3273):[INFO:CONSOLE(0)]"混合内容:' https ://www.sandbox.paypal.com/us/cgi-bin/上的页面webscr？SESSION = LTy9Q59%5fia3wiAdHTQjgQxvUF1BTzLjgXgelCew4AS%2dGAutAfB5WjZXVuX8与调度= 5885d80a13c0db1f8e263663d3faee8dcce3e160f5b9538489e17951d2c62172 "装载通过安全连接,但包含了针对不安全的端点的形式" http://www.myapi-entertainment.com/page_managements/page_featured_subscription_payment_success/4.这些内容应也可通过HTTPS提交. https://www.sandbox.paypal.com/us/cgi-bin/webscr?SESSION=LTy9Q59%5fia3wiAdHTQjgQxvUF1BTzLjgXgelCew4AS%2dGAutAfB5WjZXVuX8&dispatch=5885d80a13c0db1f8e263663d3faee8dcce3e160f5b9538489e17951d2c62172 (0)

这是我的成功回复" http://www.myapi-entertainment.com/page_managements/page_featured_subscription_payment_success/4 " - >更改了服务器名称.

是否可以将成功请求与paypal匹配？如果我有任何建议,对我有帮助.

Answer 1

Paypal在成功付款后返回json响应,如下所示

{
  "client": {
    "environment": "sandbox",
    "paypal_sdk_version": "2.14.1",
    "platform": "Android",
    "product_name": "PayPal-Android-SDK"
  },
  "response": {
    "create_time": "2016-06-15T11:38:04Z",
    "id": "PAY-6CN54299U76194116K5QT4BY",
    "intent": "sale",
    "state": "approved"
  },
  "response_type": "payment"
}

你需要检查响应json对象.如果州获得批准,则意味着PayPal付款成功.然后,您可以在Paypal网站上查看当前的交易.

确保您已按照以下步骤操作:1) private static final int REQUEST_CODE_PAYMENT = 1; private static String CONFIG_ENVIRONMENT=PayPalConfiguration.ENVIRONMENT_SANDBOX; //实时模式的PayPalConfiguration.ENVIRONMENT_PRODUCTION.2)

// note that these credentials will differ between live & sandbox environments.
private static final String CONFIG_CLIENT_ID = "ATBvU5urlaPOhpCrAhFsoG4u63RvNoKUocFPs9yR5q_sbM0yecZawUjoJhIilW8DNg5RrJcRHgRuEP_1";

private static PayPalConfiguration config = new PayPalConfiguration()
                                    .environment(CONFIG_ENVIRONMENT)
                                    .clientId(CONFIG_CLIENT_ID)
                                            // The following are only used in PayPalFuturePaymentActivity.
                                    .merchantName("Example Merchant")
                                    .merchantPrivacyPolicyUri(Uri.parse("https://www.example.com/privacy"))
                                    .merchantUserAgreementUri(Uri.parse("https://www.example.com/legal"));
                            Intent intent = new Intent(getActivity(), PayPalService.class);
                            intent.putExtra(PayPalService.EXTRA_PAYPAL_CONFIGURATION, config);
                            getActivity().startService(intent);

Answer 2

我对您的查询的理解是,您有一个使用Paypal执行付款的网站的URL,而在移动应用程序中,您使用Webview(而不是Paypal SDK)执行相同的操作.

是的,您可以callback在WEB端进行编码.

含义:

每当任何用户导航到付款页面,那么您的服务器必须知道用户是从网站访问还是通过移动Webview访问,服务器可以向additional parameter key任何支付网关发送他们的自定义逻辑.稍后,一旦支付交易完成,支付网关将返回相同additional parameter key的结果(成功或失败).

注意:每个支付网关的设置均为 Redirection url for Success/Failure

一旦Paypal在事务完成后重定向到结果URL(成功/失败),服务器然后再次检查请求是来自网站还是来自移动Webview的帮助additional parameter key; 检查以下案例......

如果来自Mobile Webview

• 成功:重定向到URL www.myserver.com/success
• 失败:重定向到URL www.myserver.com/failure

如果从Webiew那么 Normal flow

现在在Mobile的Webview中

webView.setWebViewClient(new WebViewClient() {

            @Override
            public boolean shouldOverrideUrlLoading(WebView view, String url)
                if(url.equalsIgnoreCase("www.myserver.com/success"))
                  //Success Toast
                else if(url.equalsIgnoreCase("www.myserver.com/failure"))
                   //Failure Toast
                return true;
            }

});

而已.