Volley不为我的自定义请求调用getParams？
Volley不为我的自定义请求调用getParams？
提问于 2013-08-28 17:52:35
请问Volley会自动将我的GET参数添加到URL中吗？对我来说，它不能工作，所以在查看源代码时，我就是找不到任何对getParams方法的调用。那么，我应该自己构建URL吗？这完全没问题，我只是想，当有像getParams这样的方法时，它可以为我做到这一点：)
更新:下面是我的代码..
public class BundleRequest extends com.android.volley.Request<Bundle>{
private String token;
private OnAuthTokenValidatorResponseListener mListener;
private final Map<String, String> mParams = new HashMap<String, String>();;
public BundleRequest(int method, String url, Response.ErrorListener listener) {
super(method, url, listener);
}
public BundleRequest(int method, String url,OnAuthTokenValidatorResponseListener providedListener, Response.ErrorListener listener, String token) {
super(method, url, listener);
this.token = token;
mListener = providedListener;
mParams.put(AuthenticatorConfig.TOKEN_VALIDATION_PARAMNAME, token);
}
@Override
public Map<String, String> getParams() throws AuthFailureError {
return mParams;
}
@Override
protected Response<Bundle> parseNetworkResponse(NetworkResponse httpResponse) {
switch (httpResponse.statusCode) {
case AuthTokenValidator.TOKEN_VALID_RESPONSE_CODE:
//token is ok
JSONObject response;
try {
response = new JSONObject(new String(httpResponse.data, HttpHeaderParser.parseCharset(httpResponse.headers)));
Bundle userDataResponse = new Bundle();
userDataResponse.putInt("responseCode", httpResponse.statusCode);
userDataResponse.putString("username", response.getString("user_id"));
userDataResponse.putString("email", response.getString("user_email"));
userDataResponse.putString("expiresIn", response.getString("expires_in"));
userDataResponse.putString("scope", response.getJSONArray("scope").getString(0));
userDataResponse.putString("token", token);
return Response.success(userDataResponse, HttpHeaderParser.parseCacheHeaders(httpResponse));
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
return Response.error(new VolleyError("Unsupported encoding"));
} catch (JSONException e) {
e.printStackTrace();
return Response.error(new VolleyError("Problem while parsing JSON"));
}
case AuthTokenValidator.TOKEN_INVALID_RESPONSE_CODE:
//token is not valid
mListener.onValidatorResponse(httpResponse.statusCode);
try {
mListener.onValidatorResponse(parseOnErrorResponse(new String(httpResponse.data, HttpHeaderParser.parseCharset(httpResponse.headers))));
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}
default:
return Response.error(new VolleyError("Error status code:" + httpResponse.statusCode));
}
}
protected int parseOnErrorResponse(String responseBody) {
try {
JSONObject response = new JSONObject(responseBody);
String moreInfo = response.getString("more_info");
if (moreInfo.equals("Token was not recognised")) {
return AuthTokenValidator.TOKEN_WAS_NOT_RECOGNISED;
} else if (moreInfo.equals("Token has expired")) {
return AuthTokenValidator.TOKEN_HAS_EXPIRED;
} else if (moreInfo.equals("Client doesn't exist anymore")) {
return AuthTokenValidator.CLIENT_DOES_NOT_EXIST_ANYMORE;
} else if (moreInfo.equals("Client is locked")) {
return AuthTokenValidator.CLIENT_IS_LOCKED;
} else {
return AuthTokenValidator.UNKNOWN_ERROR;
}
} catch (JSONException e) {
e.printStackTrace();
return AuthTokenValidator.UNKNOWN_ERROR;
}
}
@Override
protected void deliverResponse(Bundle response) {
mListener.onGetUserDataResponse(response);
}
}实际上，params参数现在是多余的
回答 3
Stack Overflow用户
回答已采纳
发布于 2013-08-28 22:24:06
GET方法不会调用getParams()，因此在发送请求之前，您似乎必须将其添加到URL中。
查看JavaDoc：
返回用于POST或PUT请求的参数映射。可以抛出{@link AuthFailureError}，因为可能需要身份验证才能提供这些值。
请注意，您可以直接覆盖自定义数据的{@link #getBody()}。
@在身份验证失败时抛出AuthFailureError
Stack Overflow用户
发布于 2015-09-09 16:17:56
对于Itai Hanski answer，这是一个实现该功能的示例：
for(String key: params.keySet()) {
url += "&"+key+"="+params.get(key);
}Stack Overflow用户
发布于 2013-08-28 17:56:33
尝尝这个,
public class LoginRequest extends Request<String> {
// ... other methods go here
private Map<String, String> mParams;
public LoginRequest(String param1, String param2, Listener<String> listener, ErrorListener errorListener) {
super(Method.POST, "http://test.url", errorListener);
mListener = listener;
mParams.put("paramOne", param1);
mParams.put("paramTwo", param2);
}
@Override
public Map<String, String> getParams() {
return mParams;
}
}另请参阅此示例。
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原文链接：
https://stackoverflow.com/questions/18484647
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