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我正在尝试使用下面的网络服务

http://62.253.195.179/disaster/webservices/login.php?message=[{"email":"ugo.amanoh@transputec.com","password":"welcome"}]

这将返回一个 JSON 数组

[{"companyuserId":"2","name":"ben stein","superiorname":"Leon","departmentId":"26","departmentname":"Development","companyId":"23","UDID":"12345","isActive":"1","devicetoken":"12345","email":"ugo.amah@transputec.com","phone":"5456465465654","userrole":"1","chngpwdStatus":"1"}]

我的代码如下

try{

    String weblink = URLEncoder.encode("http://62.253.195.179/disaster/webservices/login.php?message=[{\"email\":\"ugo.amanoh@transputec.com\",\"password\":\"welcome\"}]");
    HttpParams httpParameters = new BasicHttpParams();
    int timeoutConnection = 7500;
    HttpConnectionParams.setConnectionTimeout(httpParameters,
            timeoutConnection);
    int timeoutSocket = 7500;
    HttpConnectionParams.setSoTimeout(httpParameters, timeoutSocket);
    HttpClient client = new DefaultHttpClient(httpParameters);
    HttpGet request = new HttpGet();
    URI link = new URI(weblink);
    request.setURI(link);
    HttpResponse response = client.execute(request);

    BufferedReader rd = new BufferedReader(new InputStreamReader(
            response.getEntity().getContent()));
    result = rd.readLine();
    JSONObject myData = new JSONObject(result);
    JSONArray jArray = myData.getJSONArray("");
    JSONObject steps = jArray.getJSONObject(0);
    String name = steps.getString("name");

} catch (JSONException e) {
    e.printStackTrace();
}

但它不起作用，我不能 100% 确定这是最好的方法。

11-10 10:49:55.489: E/AndroidRuntime(392): java.lang.IllegalStateException: Target host must not be null, or set in parameters.
4

1 回答 1

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你不需要HttpClient：

String weblink = URLEncoder.encode("http://62.253.195.179/disaster/webservices/login.php?message=[{\"email\":\"ugo.amanoh@transputec.com\",\"password\":\"welcome\"}]");
URI link = new URI(weblink);
URLConnection linkConnection = link.openConnection();
BufferedReader rd = new BufferedReader(new InputStreamReader(linkConnection.getInputStream());
....

我认为您犯的错误是您使用HttpGet request = new HttpGet();而不是HttpPost post = new HttpPost(webLink);.

于 2012-11-10T11:22:50.623 回答