我有一个家庭作业,我必须从文件中加载代码行并将它们输出到log cat中。这是我的代码:
private void readFromURL (String requestedURL){
try {
URL myurl = new URL (requestedURL);
InputStream mystream = myurl.openStream();
Scanner myscan = new Scanner (mystream);
while (myscan.hasNextLine()) {
String aLine = myscan.nextLine();
Log.d ("works", aLine);
}
}
catch (MalformedURLException oops) {
Log.d ("ERROR" , "Are you sure the URL is correct?" + oops);
}
catch (IOException oops_again)
{
Log.d ("ERROR", "Can't access the remote resource: " + oops_again);
}
}
通过单击按钮调用readFromURL方法。每当我运行应用程序时,我都会收到IOException并显示“无法访问远程资源:”。在日志猫。
我试图从中获取信息的网址是:http://www.cis.gvsu.edu/~dulimarh/CS163H/courses.cgi
我已将清单中的权限添加到互联网中。
感谢您的帮助!
答案 0 :(得分:0)
使用以下内容获取InputStream
数据以执行您所需的操作:
private InputStream downloadUrl(String url) {
HttpURLConnection con = null;
URL url;
InputStream is=null;
try {
url = new URL(url);
con = (HttpURLConnection) url.openConnection();
con.setRequestMethod("GET");
con.setDoInput(true);
// Start the query
con.connect();
is = con.getInputStream();
}catch (IOException e) {
//handle the exception !
e.printStackTrace();
}
return is;
}
使用此功能,您应该能够使用InputStream
从Log.d
回显 - 您应该能够找出该部分;)