Question

这是我的 json编码代码

{"detail":
       {"name":["The name field is required."],
        "password":["The password field is required."],
        "email":["The email field is required."],
        "phone":["The phone field is required."],
        "address":["The address field is required."]},"success":0}

如何将其转换为jsonobject。例如"name":["The name field is required."]我想向用户显示名称的价值，任何人都可以告诉我如何做到这一点。

这是我的Android代码

public class AccountRegister extends Activity {
private ProgressDialog pDialog;
String password;
Button bnt_Submit;
EditText edt_email,edt_password,edt_name,edt_phone,edt_address;

JSONParser jsonParser = new JSONParser();

private static String url_create_product = "http://192.168.1.2/laravel/public/registerUser";


private static final String TAG_SUCCESS = "success";
    @Override


protected void onCreate(Bundle savedInstanceState) {
    // TODO Auto-generated method stub
    super.onCreate(savedInstanceState);
    setContentView(R.layout.register);


bnt_Submit =(Button)findViewById(R.id.btnsubmit);
edt_email = (EditText)findViewById(R.id.Edt_Email);
edt_password = (EditText)findViewById(R.id.Edt_Password);
edt_name = (EditText)findViewById(R.id.Edt_Name);
edt_phone = (EditText)findViewById(R.id.Edt_Phone);
edt_address = (EditText)findViewById(R.id.Edt_Address);



bnt_Submit.setOnClickListener(new OnClickListener() {


@Override
        public void onClick(View v) {
        // TODO Auto-generated method stub

        password = edt_password.getText().toString();
        if (!isValidPassword(password)) {
        edt_password.setError("The Passward must be at least 8 character ");
        }
        new RegisterUser().execute();



        }
        });



}


    private boolean isValidPassword(String pass) {
    if (pass != null && pass.length() > 7) {
    return true;
    }
    return false;
    }

    class RegisterUser extends AsyncTask<String, String, String> {



    @Override
    protected void onPreExecute() {
    // TODO Auto-generated method stub
    super.onPreExecute();

    pDialog = new ProgressDialog(AccountRegister.this);
    pDialog.setMessage("Registering...");
    pDialog.setIndeterminate(false);
    pDialog.setCancelable(true);
    pDialog.show();


    }

@Override
protected String doInBackground(String... params) {

// TODO Auto-generated method stub

String name =edt_name.getText().toString();

String email =edt_email.getText().toString();
String phone =edt_phone.getText().toString();
String address =edt_address.getText().toString();




List<NameValuePair> params1 = new ArrayList<NameValuePair>();
params1.add(new BasicNameValuePair("name", name));
params1.add(new BasicNameValuePair("password", password));
params1.add(new BasicNameValuePair("email", email));
params1.add(new BasicNameValuePair("phone", phone));
params1.add(new BasicNameValuePair("address", address));



JSONObject json = jsonParser.makeHttpRequest(url_create_product,
"POST", params1);


Log.d("Create Response", json.toString());


try {
int success = json.getInt(TAG_SUCCESS);

if (success == 1) {


} else if(success == 0) {



 JSONArray arr = new JSONArray("detail");

    for(int i = 0; i < arr.length(); i++){

            JSONObject c = arr.getJSONObject(i);        
            JSONArray ar_in = c.getJSONArray("name");

        for(int j = 0; j < ar_in.length(); j++){    
            Log.v("result--", ar_in.getString(j));
        }
   }





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

return null;

  }

protected void onPostExecute(String file_url) {
    // dismiss the dialog once done

pDialog.dismiss();
}

}
}

Answer 1

你的Json必须是这样的：

{
"detail": [
    {
        "name": "The name field is required.",
        "password": "The password field is required.",
        "email": "The email field is required.",
        "phone": "The phone field is required.",
        "address": "The address field is required."
    },
    {
        "name": "The name field is required.",
        "password": "The password field is required.",
        "email": "The email field is required.",
        "phone": "The phone field is required.",
        "address": "The address field is required."
    }
],
"success": 0
}

Answer 2

试试这样：

   JSONObject c = json.getJSONObject("detail");

   JSONArray ar_in = c.getJSONArray("name");

    for(int j = 0; j < ar_in.length(); j++){    
        Log.v("result--", ar_in.getString(j));
    }

Answer 3

如何将其转换为jsonobject。例如"name":["The name field is required."]

您不需要将这些JSONarrays转换为JSONobjects。

当您发现"[]"表示其数组时，"{}"表示其对象。

在您的情况下，首先将您的总json转换为JSONobject，然后从其JSONarray中检索其值。

    JSONObject jsonObject = json.getJSONObject("detail");

   JSONArray jsonArray = c.getJSONArray("name");

    //you dont need to use for loop you have only one value in the array.  
        Log.v("Name is", jsonArray.getString(0));
    //like this change the array name and retrieve other array values.
    }

试试这个。希望这会帮助你。：）

Answer 4

    String detail = "{\"detail\":{\"name\":[\"The name field is required.\"],\"password\":[\"The password field is required.\"]," + 
        "\"email\":[\"The email field is required.\"]," + 
        "\"phone\":[\"The phone field is required.\"]," + 
        "\"address\":[\"The address field is required.\"]},\"success\":0}";

    JSONObject contentObj = new JSONObject(detail);
    JSONObject detObj = contentObj.optJSONObject("detail");
    if (detObj != null) {
     //if you need directly name
        JSONArray nameArr = detObj.optJSONArray("name");
        String nameStr = nameArr.optString(0, "");

     //or iterate by all keys
        Iterator<String> keys = detObj.keys();
        while (keys.hasNext()) {
            JSONArray errorsArr = detObj.optJSONArray(keys.next());
            String errorStr = nameArr.optString(0, "");
        }
    }

如何在android中显示jsonarray值

4 个答案: