2 u013866421 u013866421 于 2014.09.10 16:55 提问

HttpGet抛出异常,不明白!求指导~

想用数据库做一个登录验证的功能,服务器端响应正常,json数据包 {"ret":1,"num":123}
ret是登录正常时的状态值,num是号码:123
url是192.168.1.3:8080/web/getActInfo
params:num=123
通过logcat发现在httpGet中抛出了一个异常,但是搞不明白!!痛苦了一下午求点拨!

public String httpGet(String url, String params) throws Exception
{
String response = null;
if (null!=params&&!params.equals(""))
{

        url += "?" + params;
    }

    int timeoutConnection = 8000;  
    int timeoutSocket = 10000;  
    HttpParams httpParameters = new BasicHttpParams();// Set the timeout in milliseconds until a connection is established.  
    HttpConnectionParams.setConnectionTimeout(httpParameters, timeoutConnection);// Set the default socket timeout (SO_TIMEOUT) // in milliseconds which is the timeout for waiting for data.  
    HttpConnectionParams.setSoTimeout(httpParameters, timeoutSocket);  

    HttpClient httpClient = new DefaultHttpClient(httpParameters);  
    HttpGet httpGet = new HttpGet(url);
    try
    {
        HttpResponse httpResponse = httpClient.execute(httpGet);
        int statusCode = httpResponse.getStatusLine().getStatusCode();
        if (statusCode == HttpStatus.SC_OK) //SC_OK = 200
        {
            response = EntityUtils.toString(httpResponse.getEntity());
        }
        else
        {
            response = "状态码"+statusCode;
        }
    } catch (Exception e)
    {
        throw new Exception(e);         
    } 

1个回答

u013866421
u013866421   2014.12.17 11:03
已采纳

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