我正在做一个名为“天气预报”的项目,信息来自这个静态URL。我要弄到一张列有他们的条件的城市名单。例如,当我输入东京时,它应该显示如下:
Tokyo, Japan#14#Cloudy ##我还创建了包含城市列表的ArrayList。我已经完成了布局xml文件。
输出格式是纯文本:
(\n)结束。(##)的一行结束。它给我带来了错误:
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这是我的密码:
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();
}如何解决无效的布局错误?
发布于 2012-12-07 11:21:15
更改字符串“` url =http://myexample.info/?cities‘
至
字符串url = "http://myexample.info/?cities";
"url“前面的‘是搞乱了您的字符串,并在http: a注释之后进行了所有操作,这是因为//。
https://stackoverflow.com/questions/13761940
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