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我正在做一个“天气预报”项目,信息是从这个静态 URL中给出的。我必须获得城市列表及其条件。例如,当我输入 Tokyo 时,它应该显示如下:

Tokyo, Japan#14#Cloudy ##

我还创建了ArrayList城市列表。我已经完成了布局 xml 文件。

输出格式为纯文本:

  • 每个输出行都以换行符结束(\n)
  • 流程总是由包含两个符号的行终止(##)

它给了我错误:

Invalid layout of java.lang.String at value

A fatal error has been detected by the Java Runtime Environment:

Internal Error (javaClasses.cpp:129), pid=6464, tid=7052
fatal error: Invalid layout of preloaded class

这是我的代码:

import android.app.Activity;
import android.os.Bundle;

public class Meteo extends Activity{

        @Override
    protected void onCreate(Bundle savedInstanceState) { {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.meteo);

        String url = `http://myexample.info/?cities`;
        // Weather information
        String weather = "Tokyo, Japan#14#Cloudy ##";
        // Get the city name
        String city = url.substring(url.indexOf("?") + 1).trim();
        System.out.println(city);
        // Check the weather of the city: 14#Cloudy
        // Remove city name
        // Remove last #
        if (weather.toLowerCase().contains(city.toLowerCase())) {
            // Get condition: 
            String condition = weather.substring(weather.indexOf("#") + 1,
                    weather.length() - 2);
            System.out.println(condition);
            // Split with # sign and you have a list of conditions
            String[] information = condition.split("#");
            for (int i = 0; i < information.length; i++) {
                System.out.println(information[i]);
            }
        }

            @Override

            protected void onPause() {
                // TODO Auto-generated method stub
                super.onPause();
            }

如何解决无效布局错误?

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1 回答 1

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更改字符串“ `url =http://myexample.info/?cities `

字符串 url = "http://myexample.info/?cities";

由于 //.

于 2012-12-07T11:21:15.107 回答