Question

我想从android app执行java web服务方法。我的Web服务方法签名如下：

public String getData(String category) throws Exception

该方法返回string并接受一个字符串作为参数。

我已经从google chrome的地址栏执行了该方法：

http://localhost:8080/JsonWebService/services/JsonWebService/getData?category=Marketing

这里getData是方法的名称，Marketing是该方法的参数。从资源管理器中它运行良好。

但是当我向Android应用程序的httppost请求添加相同的url时，它无法说出错误的参数数量。我的Android应用程序代码是：

HttpPost post = new HttpPost("http://192.168.1.7:8080/JsonWebService/services/JsonWebService/getData?category=Marketing");
HttpResponse httpres = httpClient.execute(post);

请注意：这里需要192.168.1.7，因为我直接在设备上执行应用程序，因此我没有使用localhost。

非参数方法也可以从Android应用程序正确执行。但是当在url中插入参数时，为什么它在android应用程序中错误地说错误的参数数量以及它如何在PC上的谷歌浏览器中正确执行。请帮忙......谢谢...

我在这里添加代码。我的java Web服务代码如下：

public class JsonWebService {
    @POST
    @Path("getData")
    public String getData(String category) throws Exception {
        JSONObject jsonData = new JSONObject();
        String Email = "";
        String Name = "";
        String receivedCat = "";
        boolean status = false;
        try {
            Class.forName("com.mysql.jdbc.Driver");
            Connection con = DriverManager.getConnection("jdbc:mysql://localhost/admindb","root","");
            java.sql.PreparedStatement query = con.prepareStatement("SELECT * FROM sample WHERE Category =" + "'" + category + "'" + ";");
            ResultSet result = query.executeQuery();

            while(result.next()){
                receivedCat = result.getString("Category");
                Name = result.getString("Name");
                Email = result.getString("Email");
            }
            if(receivedCat.equals(category)){
                status = true;
                jsonData.put("Name",Name);
                jsonData.put("Email", Email);
                jsonData.put("status", status);
            }
        }
        catch(Exception e) {
            e.printStackTrace();
        }
    return jsonData.toString();
}

我的Android客户端代码如下所示：

btnCategory = (Button)findViewById(R.id.button1);
txtCategory = (EditText)findViewById(R.id.editText1);
gridV = (GridView)findViewById(R.id.gridView1);
txtName = (EditText)findViewById(R.id.editText3);
txtEmail = (EditText)findViewById(R.id.editText2);

btnCategory.setOnClickListener(new View.OnClickListener() {

    @Override
    public void onClick(View v) {
    // TODO Auto-generated method stub
        Thread netThread = new Thread() {
            public void run() {
                try {
                    final JSONObject receivedJson;// = new JSONObject();
                    String URL = "http://192.168.1.7:8080/JsonWebService/services/JsonWebService/getData?category=Marketing?";
                    DefaultHttpClient httpClient = new DefaultHttpClient();
                    HttpPost post = new HttpPost(URL);
                    HttpResponse httpres = httpClient.execute(post);
                    HttpEntity entity = httpres.getEntity();                
                    String json = EntityUtils.toString(entity).toString();
                    String parts[] = json.split("<ns:return>");
                    parts = parts[1].split("</ns:return>");
                    String jsonPart = parts[0];
                    receivedJson = new JSONObject(jsonPart);
                    runOnUiThread(new Runnable() {

                    @Override
                    public void run() {
                        try {
                            txtName.setText(receivedJson.getString("Name"));
                            txtEmail.setText(receivedJson.getString("Email"));
                        }
                        catch(Exception e){                     
                        }
                    }
                };
            netThread.start();
            }
        });
}

问题出在String Client中的java客户端，它是调用java Web服务方法的URL字符串。请帮帮我......

从android app传递参数到java web服务方法

0 个答案: