小编PAn*_*cho的帖子

这是我的promise函数,我需要返回rs.rows.item(0)的值;

     public getCustomer()  : any
  {
        let  db =  window.sqlitePlugin.openDatabase({name: 'data.db', location: 'default'});
        return new Promise((resolve, reject) =>
        {
            db.transaction(function(tx)
            {
                tx.executeSql('SELECT * FROM customer ORDER BY customerId DESC LIMIT 1', [], function(tx, rs)
                {
                     return resolve(rs.rows.item(0));
                }, 
                function(tx, error) 
                {
                    console.log('SELECT error: ' + error.message);
                    reject(error);
                });
            });
        });    
  }

返回值我有一个像这个图像的对象

我需要得到这样的例子

var customer = getCustomer();
customer.name;
customer.email;

这是我的JSON字符串: "{'userName' : 'Bachooo'}"

将JSON字符串转换为LoginVO逻辑是:

Gson gson = new GsonBuilder().excludeFieldsWithoutExposeAnnotation().create();
LoginVO loginFrom  = gson.fromJson(jsonInString, LoginVO.class);
System.out.println("userName " + loginFrom.getUserName()); // output null

我的LoginVO.class是:

public class LoginVO {

 private String userName;
 private String password;

 public String getUserName()
 {
    return userName;
 }
 public void setUserName(String userName)
 {
    this.userName = userName;
 }
 public String getPassword()
 {
    return password;
 }
 public void setPassword(String password)
 {
    this.password = password;
 }

}

注意我使用的是jdk 1.8.0_92

loginForm.getUserName()的输出是否有关于此问题NULL的"Bachooo"任何想法？