2011-12-20 35 views
0

我无法用put方法将数据发送到特定的链接。请找到code.Let下面我知道,如果我不得不做出任何改变无法使用HTTP PUT方法将数据发送到一个URL

public class test { 
    private static String url = "http://semldvw0728.google.net/.svc/web/testrequests/"; 
    private static String urlconnection; 

    public static void main(String[] args) throws IOException { 

    StringBuffer xmlString = new StringBuffer(); 

    xmlString.append("<TestRequest>"); 
    xmlString.append("<DateRequested>2011-12-20</DateRequested>"); 
    xmlString.append("<DemCapID>893467</DemCapID>"); 
    xmlString.append("<DemCapVersion>1</DemCapVersion>"); 
    xmlString.append("<IBIS_ID>13530</IBIS_ID>"); 
    xmlString.append("<ProjectName>LTS</ProjectName>"); 
    xmlString.append("<RequestedBy>ktmq331</RequestedBy>"); 
    xmlString.append("<SampleNumber>SN1033645061</SampleNumber>"); 
    xmlString.append("<Status>Sample Ordered</Status>"); 
    xmlString.append("</TestRequest>"); 

    System.out.println("xmlString :" + xmlString.toString()); 
    url = url + 893467; 
    System.out.println("URL : " + url); 

    try { 
     System.out.println("URL : " + url); 
     HttpClient client = new HttpClient(); 
     PutMethod putMethod = new PutMethod(url); 
     client.setConnectionTimeout(8000); 
     putMethod.setRequestBody(xmlString.toString()); 
     System.out.println("statusLine>>>" + putMethod.getStatusLine()); 
     System.out.println("statusLine>>>"+   putMethod.getResponseBodyAsString()); 
     putMethod.releaseConnection(); 
    } catch (Exception e) { 
     e.printStackTrace(); 
    } 
} 
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更具体。添加更多信息。问题是什么?有错误吗? etc ... – 2011-12-20 20:36:41

回答

1

我想你需要拨打execute,或者你可以使用你创建的HttpClient对象call this

所以,也许加入这一行你设置请求体后您的putMethod ...

client.executMethod(putMethod); 
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你能否请我解释一下如何创建HttpClient – Winz 2011-12-20 20:45:38

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我添加了如何给你提供的代码打电话并将你链接到文档。这应该足够了。 – Dave 2011-12-20 20:46:13

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@ user1042983:'HttpClient client = new DefaultHttpClient();' – 2011-12-20 20:48:22