2011-09-17 42 views
0
And some times i'm getting java.net.MalFormedURLException what's the reason behind this and how can i resolve this.. 

My code is as follows.. 


SAXParserFactory spf = SAXParserFactory.newInstance(); 
     SAXParser sp = spf.newSAXParser(); 
     XMLReader xr = sp.getXMLReader(); 

     /** Send URL to parse XML Tags */ 
     URL sourceUrl = new URL(
       "http://w3devadv.liveproj.com /api/apiRequest.php?Method=getdealdetails&DealId=2&SessionId=EA3JQ0RZJT4e66223143fc5"); 

     /** 
     * Create handler to handle XML Tags (extends DefaultHandler) 
     */ 
     DealsHandler myXMLHandler = new DealsHandler(); 
     xr.setContentHandler(myXMLHandler); 
     xr.parse(new InputSource(sourceUrl.openStream())); 


In handler i'm writing the following code in startelement 

public void startElement(String uri, String localName, String qName, 
     Attributes attributes) throws SAXException { 
    currentElement = true; 
    if (localName.equalsIgnoreCase("data")) { 
     dealsdata = new DealsData(); 
    } else if (localName.equalsIgnoreCase("dealdetails")) { 
     deals = new Deals(); 
    } else if (localName.equalsIgnoreCase("title")) { 
     deals.title = attributes.getLocalName(0); 
    } 

i'm getting the above said exception how can i resolve this. 

回答

2

您收到此錯誤支持,因爲.....

"This exception is thrown when a program attempts to create an URL from an 
incorrect specification." 
0
"http://w3devadv.liveproj.com /api/apiRequest.php?Method=getdealdetails&DealId=2&SessionId=EA3JQ0RZJT4e66223143fc5" 

這可能是由於URL中的空間。

1

傳遞字符串URL 才能將字符串到標準的URL格式

這樣..

String url = new String(str.trim().replace(" ", "%20").replace("&", "%26") 
.replace(",", "%2c").replace("(", "%28").replace(")", "%29") 
.replace("!", "%21").replace("=", "%3D").replace("<", "%3C") 
.replace(">", "%3E").replace("#", "%23").replace("$", "%24") 
.replace("'", "%27").replace("*", "%2A").replace("-", "%2D") 
.replace(".", "%2E").replace("/", "%2F").replace(":", "%3A") 
.replace(";", "%3B").replace("?", "%3F").replace("@", "%40") 
.replace("[", "%5B").replace("\\", "%5C").replace("]", "%5D") 
.replace("_", "%5F").replace("`", "%60").replace("{", "%7B") 
.replace("|", "%7C").replace("}", "%7D")); 

您可以使用此功能,方便您隨時隨地獲取URL,並用它,所以你可以克服這個錯誤並讓你的字符串工作。

我知道你接受了答案,但這也可以幫助其他人也我認爲 謝謝你。

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