我需要在Web视图中获取必需的内容才能发布登录数据

时间:2019-01-29 12:35:30

标签: android login webview

我需要登录并发布用户ID,并将单词传递到Web视图中。 谁能帮我!预先感谢。

如何在Web视图中显示所需的内容。我需要在Web视图中获取必需的内容才能发布登录数据

 public class MainActivity extends AppCompatActivity {
EditText login_user,login_password;
Button login_button;
String uName,uPwd,data;
WebView webView;
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    login_user=(EditText)findViewById(R.id.userName);
    login_password=(EditText)findViewById(R.id.passWord);
    login_button=(Button)findViewById(R.id.loginButton);

    webView=(WebView)findViewById(R.id.webView);

    login_button.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            try{
                getText();

            }catch(Exception e){
                e.printStackTrace();
            }
        }
    });
}

private void getText() {
    uName = login_user.getText().toString();
    uPwd = login_password.getText().toString();

    WebView webview = new WebView(this);
    setContentView(webview);
    String url = "website url";
    String postData = null;
    try {
        postData = "username=" + URLEncoder.encode(uName, "UTF-8") + "&password=" + URLEncoder.encode(uPwd, "UTF-8");
    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    }
    webview.postUrl(url,postData.getBytes());
}

}

1 个答案:

答案 0 :(得分:0)

您可以尝试以下操作:-

String postData = "username=abc&password=236";

webview.postUrl(
    "website url/exampleURL",
    EncodingUtils.getBytes(postData, "BASE64"));

,并检查此链接:- 1)How to post data for WebView android

2)How to make post requests with webview?