HttpGet無法識別url

所以我使用下面的代碼從一個不同的舊帖子，但有一個部分的問題，行：HttpGet request = new HttpGet(url);不起作用。在url中，我放置了類似www.stackoverflow.com的東西，但是其中一部分不會讓代碼編譯。我基本上試圖從HTML網站拉文字。完整的代碼：HttpGet無法識別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(); 
}

來源

2011-04-22 Clozecall

for HttpGet request = new HttpGet（www.stackoverflow.com）;我改成它（「www.stackoverflow.com」）;正如許多人所建議的那樣，但其他行如「HttpResponse response = client.execute（request）;」返回編譯錯誤，所有這些：未處理的異常類型IOException，eclipse中的大多數快速修復提示「Surroned with try/catch」 – Clozecall 2011-04-22 19:08:16

試試這個：HttpGet request = new HttpGet("www.stackoverflow.com");

來源

2011-04-22 18:52:14 Zepplock

我試着改變它，所以url有「」，但是這會導致如下行：「HttpResponse response = client.execute（request）;「，」InputStream in = response.getEntity（）。getContent（）;「，還有一些錯誤。 – Clozecall 2011-04-22 19:01:26

「出錯」是指編譯錯誤還是運行時異常？如果是例外情況，請發佈。 – djg 2011-04-22 19:03:25

編譯錯誤，所有這些：未處理的異常類型IOException，eclipse中的大多數快速修復提示「與try/catch一起使用Surroned」 – Clozecall 2011-04-22 19:06:09

HTTPGet預計的URL或字符串，所以要儘量改變您的要求爲：

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

來源

2011-04-22 18:54:21

使用形式的字符串：

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

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

來源

2011-04-22 18:54:43 djg

添加上面給出的答案用try和catch語句圍繞您的代碼來捕獲異常。

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 
}

來源

2011-04-22 19:19:23 primpap

HttpGet無法識別url

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