2

我正在使用 Android Web 服务器。当我在模拟器浏览器上访问 localhost:8080 时，它会提供一个带有密码字段的页面/表单。在成功验证密码后，我想将用户重定向到成功/失败页面。读取传入的 http post 请求并解析密码字段以进行验证的最佳方法是什么？任何正确方向的指针都是非常感激。我有一个处理表单提交到的网址的处理程序。处理程序的代码是：

import java.io.IOException;
import java.io.OutputStream;
import java.io.OutputStreamWriter;

import org.apache.http.HttpEntity;
import org.apache.http.HttpException;
import org.apache.http.HttpRequest;
import org.apache.http.HttpResponse;
import org.apache.http.entity.ContentProducer;
import org.apache.http.entity.EntityTemplate;
import org.apache.http.protocol.HttpContext;
import org.apache.http.protocol.HttpRequestHandler;

import android.content.Context;

public class LoginHandler implements HttpRequestHandler {
private Context context = null;
public LoginHandler(Context context) {
    this.context = context;
}

@Override
public void handle(final HttpRequest request, HttpResponse response,
        HttpContext httpcontext) throws HttpException, IOException {
       HttpEntity entity = new EntityTemplate(new ContentProducer() {
        public void writeTo(final OutputStream outstream) throws IOException {
            String resp = null;
            OutputStreamWriter writer = new OutputStreamWriter(outstream, "UTF-8");
            if(validatePassword()==true){
             resp ="<html><head></head><body><h1>Home<h1><p>Success.</p></body></html>";
            }
            else{resp="<html><head></head><body><h1>Home<h1><p>Login Failed.</p></body></html>";}
            writer.write(resp);
            writer.flush();
        }


    });
    response.setHeader("Content-Type", "text/html");
    response.setEntity(entity);

}
boolean validatePassword(){
boolean pass=false;
//parse request body here and check for the password if true return true/else false
 return pass;
 }


 }
4

2 回答 2

5

环顾四周后，我找到了解决方案。在句柄方法中添加以下内容就可以了。感谢原始海报。http://www.androiddevblog.net/android/a-bare-minimum-web-server-for-android-platform

        if (request instanceof HttpEntityEnclosingRequest) {
HttpEntity entity = ((HttpEntityEnclosingRequest) request).getEntity();
if (entity != null) {
Log.v("RequestBody", EntityUtils.toString(entity, "UTF-8"));
entity.consumeContent();
}
}
于 2012-08-15T18:40:33.473 回答
2

如果这不是您所要求的，我深表歉意，所以如果不是，请告诉我。

您可以使用JSONObject返回该密码是否被验证为正确。

例如，如果密码正确，您可以将 HTTP 结果存储为：

{"status":200,"confirmed":"true"}

否则为“假”。

当您从 HTTP 发布请求返回时，您可以将此结果存储为字符串，然后从中JSONObject提取。例如：

// Send the URL to a postRequest function and return the result as a String
String output = makePostRequest(url);

// Parse the String as a JSONObject and receive whether or not the login was confirmed
JSONObject o = new JSONObject(output);
String confirmed = o.getString("confirmed");
if (confirmed.equals("true")) {
    // Password confirmed - redirect user to success page
} else {
    // Password incorrect - redirect user to failure page
}

注意：如果您需要了解如何从发布请求中接收响应代码，这里有一些示例代码：

String output = {};

// Use bufferedreader and stringbuilder to build an output string (where conn is your HTTPUrlConnection object you used to make the post request    
BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream()));
StringBuilder sb = new StringBuilder();
String line;

// Loop through response to build JSON String
while((line = br.readLine()) != null) {
    sb.append(line + "\n");
}

// Set output from response
output = sb.toString();

现在output是可以转换为 JSONObject 的字符串。

这有什么帮助吗？

编辑：
好的，所以您将获得的字符串格式为{"password":"somepassword"}. 要解析这个，试试这个：

String s = /* the string in the format {"password":"somepassword"} */
JSONObject o = new JSONObject(s);
String password = o.getString("password");
if (password.equals(random_password_at_beginning_of_webservice) {
    // Password confirmed - redirect user to success page
} else {
    // Password incorrect - redirect user to failure page
}
于 2012-08-08T12:04:52.453 回答