HttpGet无法识别url

所以我使用下面的代码从一个不同的旧帖子，但有一个部分的问题，行：HttpGet request = new HttpGet(url);不起作用。在url中，我放置了类似www.stackoverflow.com的东西，但是其中一部分不会让代码编译。我基本上试图从HTML网站拉文字。完整的代码：HttpGet无法识别url

public void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.main); 

    HttpClient client = new DefaultHttpClient(); 
    HttpGet request = new HttpGet(www.stackoverflow.com); 
    HttpResponse response = client.execute(request); 

    String html = "Toronto-GTA"; 
    InputStream in = response.getEntity().getContent(); 
    BufferedReader reader = new BufferedReader(new InputStreamReader(in)); 
    StringBuilder str = new StringBuilder(); 
    String line = null; 
    while((line = reader.readLine()) != null) 
    { 
     str.append(line); 
    } 
    in.close(); 
    html = str.toString(); 
}

来源

2011-04-22 Clozecall

for HttpGet request = new HttpGet（www.stackoverflow.com）;我改成它（“www.stackoverflow.com”）;正如许多人所建议的那样，但其他行如“HttpResponse response = client.execute（request）;”返回编译错误，所有这些：未处理的异常类型IOException，eclipse中的大多数快速修复提示“Surroned with try/catch” – Clozecall 2011-04-22 19:08:16

试试这个：HttpGet request = new HttpGet("www.stackoverflow.com");

来源

2011-04-22 18:52:14 Zepplock

我试着改变它，所以url有“”，但是这会导致如下行：“HttpResponse response = client.execute（request）;“，”InputStream in = response.getEntity（）。getContent（）;“，还有一些错误。 – Clozecall 2011-04-22 19:01:26

“出错”是指编译错误还是运行时异常？如果是例外情况，请发布。 – djg 2011-04-22 19:03:25

编译错误，所有这些：未处理的异常类型IOException，eclipse中的大多数快速修复提示“与try/catch一起使用Surroned” – Clozecall 2011-04-22 19:06:09

HTTPGet预计的URL或字符串，所以要尽量改变您的要求为：

HttpGet request = new HttpGet("http://www.stackoverflow.com/");

来源

2011-04-22 18:54:21

使用形式的字符串：

[scheme:][//authority][path][?query][#fragment]

ie "http://www.stackoverflow.com"

来源

2011-04-22 18:54:43 djg

添加上面给出的答案用try和catch语句围绕您的代码来捕获异常。

try{ 
HttpClient client = new DefaultHttpClient(); 
    HttpGet request = new HttpGet("www.stackoverflow.com"); 
    HttpResponse response = client.execute(request); 
    InputStream in = response.getEntity().getContent(); 
    BufferedReader reader = new BufferedReader(new InputStreamReader(in)); 
    StringBuilder str = new StringBuilder(); 
    String line, html = null; 
    while((line = reader.readLine()) != null) 
    { 
     str.append(line); 
    } 
    in.close(); 
    html = str.toString(); 
} 
catch(Exception e){ 
//Do something here like printing the stacktrace 
}

来源

2011-04-22 19:19:23 primpap

HttpGet无法识别url

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