Home  >  Article  >  Backend Development  >  Android如何把数据转换成List<Map<string,string>>

Android如何把数据转换成List<Map<string,string>>

PHPz
PHPzOriginal
2016-06-06 20:42:562247browse

Android如何把数据转换成List<Map<string,string>>

Android如何用Gson把数据转换成List03736eea88e2f3e7c93f615ae27fda5c>?

具体问题:

我现在有任意长度的json数据,用php的json_encode()生成,如下格式

[{"rowID":"1","Name":"tqtqwet","Comment":"qewrtqwe"},{"rowID":"2","Name":"waf","Comment":"agsadga"},{"rowID":"3","Name":"android","Comment":"fafdadsf"},{"rowID":"4","Name":"android222","Comment":"fasdfas"},{"rowID":"5","Name":"fadfa","Comment":"adgasd"}]

现在想用把它转换成Listb9ce0bbc1c208c036cee4db71a23ea6a>的格式,并用listView显示出来。其中hashmap的第一个键对应“Name”,第二个键对应“Comment”,“rowID”无视掉。请问怎么能最快捷地完成这种转换?如果有办法可以绕过也可以!谢谢

解决办法:

这个json数据格式很简单,不用你想象的那么复杂就能解析;

你需要一个JavaBean对象:

CustomClass.java:

    public class CustomClass {
     private String Name;
     private String Comment;
    public String getName() {
      return Name;
    }
    public void setName(String name) {
      Name = name;
    }
    public String getComment() {
      return Comment;
    }
    public void setComment(String comment) {
      Comment = comment;
    }
    }

MainActivity.java:

    public class MainActivity extends Activity {
    //你的json数据
    String jsonString="     [{\"rowID\":\"1\",\"Name\":\"tqtqwet\",\"Comment\":\"qewrtqwe\"},{\"rowID\":\"2\",\"Name\":\"waf\",\"Comment\":\"agsadga\"},{\"rowID\":\"3\",\"Name\":\"android\",\"Comment\":\"fafdadsf\"},{\"rowID\":\"4\",\"Name\":\"android222\",\"Comment\":\"fasdfas\"},{\"rowID\":\"5\",\"Name\":\"fadfa\",\"Comment\":\"adgasd\"}]";
        @Override
    protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    Type listType = new TypeToken>(){}.getType();
    ArrayList customList = new Gson().fromJson(jsonString, listType);
    }
    }

customList就是你想要的那个带有json数据的集合,你可以通过customList.get(index).getName() or getXXX()得到你想要的值

更多相关技术文章,请访问PHP中文网

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn