private int ID=0;
.
.
Integer field2=0;
field2 = Integer.parseInt(item.get("credentialID").toString());
ID=field2.intValue();
UName.setText(Integer.toString(ID));
上面的內容:根據預期,field2的值爲'9',UName在文本字段顯示值'9'。Integer.toString()更改int的值
HashMap<String,String> paramage= new HashMap<String, String>();
paramage.put("credential", Integer.toString(ID));
現在,當我調用使用paramage的方法時,我沒有得到任何結果(意味着方法中的ID比較導致false)。
但是,如果我這樣做,並調用方法現在,它完美的作品(但我無法提供靜態輸出方法,應該從用戶採取)
HashMap<String,String> paramage= new HashMap<String, String>();
paramage.put("credential", "9");
什麼問題?如何解決它?
順便說一句我打電話給kumulos的方法,我正在爲android編程。
編輯:我的確切代碼按要求。
package lcukerd.com.logintest;
import android.os.Bundle;
import android.support.annotation.BoolRes;
import android.support.v7.app.AppCompatActivity;
import android.support.v7.widget.Toolbar;
import android.text.Editable;
import android.view.View;
import android.view.Menu;
import android.view.MenuItem;
import android.app.Application;
import android.widget.Button;
import android.widget.EditText;
import com.kumulos.android.Kumulos;
import com.kumulos.android.ResponseHandler;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.LinkedHashMap;
public class MainActivity extends AppCompatActivity {
private int ID=0;
private EditText UName,UPass,UAge;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
Kumulos.initWithAPIKeyAndSecretKey("removed", "removed", this);
UName =(EditText) findViewById(R.id.name);
UPass = (EditText) findViewById(R.id.password);
UAge = (EditText) findViewById(R.id.age);
Button login = (Button) findViewById(R.id.login);
login.setOnClickListener(new View.OnClickListener()
{
@Override
public void onClick(View view) {
String username = UName.getText().toString();
String password = UPass.getText().toString();
LinkedHashMap<String, String> params = new LinkedHashMap<String, String>();
params.put("accountName", username);
params.put("password",password);
Kumulos.call("login", params, new ResponseHandler() {
@Override
public void didCompleteWithResult(Object result) {
Integer field2=0;
ArrayList<LinkedHashMap<String, Object>> objects = (ArrayList<LinkedHashMap<String,Object>>) result;
LinkedHashMap<String, Object> item= objects.get(0);
field2 = Integer.parseInt(item.get("credentialID").toString());
ID=field2.intValue();
UName.setText(Integer.toString(ID));
}
});
params.clear();
HashMap<String,String> paramage= new HashMap<String, String>();
paramage.put("credential", Integer.toString(ID)); //up here
Kumulos.call("getage", paramage, new ResponseHandler() {
@Override
public void didCompleteWithResult(Object result) {
Integer field2=0;
ArrayList<LinkedHashMap<String, Object>> objects = (ArrayList<LinkedHashMap<String,Object>>) result;
LinkedHashMap<String, Object> item= objects.get(0);
//Boolean check = item.containsKey("age");
field2 = Integer.parseInt(item.get("age").toString());
int age=field2.intValue();
UAge.setText(Integer.toString(age));
}
});
}
});
Button signup = (Button) findViewById(R.id.signup);
signup.setOnClickListener(new View.OnClickListener()
{
@Override
public void onClick(View view) {
String username = UName.getText().toString();
String password = UPass.getText().toString();
HashMap<String, String> params = new HashMap<String, String>();
params.put(username, password);
try {
Kumulos.call("signup", params, new ResponseHandler() {
@Override
public void didCompleteWithResult(Object result) {
ID = (int) result;
}
});
UName.setText("");
UPass.setText("");
params.put(UAge.getText().toString(), Integer.toString(ID));
Kumulos.call("setAge", params, new ResponseHandler() {
@Override
public void didCompleteWithResult(Object result) {
}
});
UAge.setText(Integer.toString(ID));
}
catch (Exception e)
{
UAge.setText("5");
e.printStackTrace();
}
}
});
}
}
註冊後的代碼按鈕聲明沒有測試和編輯不看那裏。
檢查你的鑰匙..憑證和憑證ID。您是否使用該鍵訪問相同的數據集? –
您已將'item.get(「credentialID」)。toString()'作爲'String'。你爲什麼要把它轉換成一個整數,然後再返回? – EJP
我發現有點奇怪,你如何獲得一個字符串轉換爲整數重新將其轉換爲字符串。有必要嗎 ?另外,請提供[mcve]。 – AxelH