6
也许该方法返回它应该如何,但我基本上只是做了一个测试方法,看起来像这样创建ASP.net WebService的返回而不是XML JSON
[WebMethod]
[ScriptMethod(ResponseFormat = ResponseFormat.Json)]
public string TestJSON()
{
var location = new Location[2];
location[0] = new Location();
location[0].Latitute = "19";
location[0].Longitude = "27";
location[1] = new Location();
location[1].Latitute = "-81.9";
location[1].Longitude = "28";
return new JavaScriptSerializer().Serialize(location);
}
当我打这个从我的android应用程序我得到这样
Value <?xml of type java.lang.String cannot be converted to JSONArray
我觉得这个方法将返回只是直JSON异常,但是这是Web服务方法返回
<?xml version="1.0" encoding="utf-8"?>
<string xmlns="http://tempuri.org/">[{"Latitute":"19","Longitude":"27"},{"Latitute":"-81.9","Longitude":"28"}]</string>
是不是应该这样?有没有办法删除JSON之外的XML内容?我不知道我有我的web服务做,使其返回数据
代码在Android上使用来调用web服务
public String readWebService(String method)
{
StringBuilder builder = new StringBuilder();
HttpClient client = new DefaultHttpClient();
HttpGet httpGet = new HttpGet("http://myserver.com/WebService.asmx/" + method);
Log.d(main.class.toString(), "Created HttpGet Request object");
try
{
HttpResponse response = client.execute(httpGet);
Log.d(main.class.toString(), "Created HTTPResponse object");
StatusLine statusLine = response.getStatusLine();
Log.d(main.class.toString(), "Got Status Line");
int statusCode = statusLine.getStatusCode();
if (statusCode == 200) {
HttpEntity entity = response.getEntity();
InputStream content = entity.getContent();
BufferedReader reader = new BufferedReader(new InputStreamReader(content));
String line;
while ((line = reader.readLine()) != null) {
builder.append(line);
}
return builder.toString();
} else {
Log.e(main.class.toString(), "Failed to contact Web Service: Status Code: " + statusCode);
}
}
catch (ClientProtocolException e) {
Log.e(main.class.toString(), "ClientProtocolException hit");
e.printStackTrace();
}
catch (IOException e) {
Log.e(main.class.toString(), "IOException hit");
e.printStackTrace();
}
catch (Exception e) {
Log.e(main.class.toString(), "General Exception hit");
}
return "WebService call failed";
}
那么正确的格式我的地方调用这个方法在代码如
try {
JSONArray jsonArray = new JSONArray(readWebService("TestJSON"));
Log.i(main.class.toString(), "Number of entries " + jsonArray.length());
....
}
...
你是怎么从android调用它的?您是否在该请求中指定了任何内容类型? –
我不是,但我只是尝试添加httpGet.setHeader(“Content-Type”,“application/json”);当我添加这个web服务返回一个500服务器错误状态代码。我会更新我的问题,包括我用来调用Web服务方法的Android代码 –
看起来像别人有类似的问题,我在我的广泛研究(5分钟谷歌搜索)中错过了http://stackoverflow.com/questions/2058454/asp-net-webservice-is-wrapping-my-json-reponse -with-xml-tags?rq = 1 ...显然,如果我使用POST而不是get的内容类型设置到application/json –