Android 截取json响应
我有以下json响应Android 截取json响应,android,json,android-volley,Android,Json,Android Volley,我有以下json响应 {"multicast_id":8XXXD,"success":1,"failure":0,"canonical_ids":0,"results":[{"message_id":"0:14XX"}]} 我想检查java中是否存在failure=0。下面是我的截击代码 StringRequest stringRequest = new StringRequest(Request.Method.POST, CONN_URL.send_single_push,
{"multicast_id":8XXXD,"success":1,"failure":0,"canonical_ids":0,"results":[{"message_id":"0:14XX"}]}
我想检查java中是否存在failure=0。下面是我的截击代码
StringRequest stringRequest = new StringRequest(Request.Method.POST, CONN_URL.send_single_push,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
Toast.makeText(getActivity(), response, Toast.LENGTH_LONG).show();
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
}
}) {
@Override
protected Map<String, String> getParams() throws AuthFailureError {
Map<String, String> params = new HashMap<>();
params.put("title", title);
params.put("message", message);
if (!TextUtils.isEmpty(image))
params.put("image", image);
params.put("course", course);
return params;
}
};
MySingleton.getInstance(getActivity()).addToRequestQueue(stringRequest);
StringRequest StringRequest=新的StringRequest(Request.Method.POST、CONN\u URL.send\u single\u push、,
新的Response.Listener(){
@凌驾
公共void onResponse(字符串响应){
Toast.makeText(getActivity(),response,Toast.LENGTH_LONG).show();
}
},
新的Response.ErrorListener(){
@凌驾
公共无效onErrorResponse(截击错误){
}
}) {
@凌驾
受保护的映射getParams()引发AuthFailureError{
Map params=新的HashMap();
参数put(“标题”,标题);
参数put(“消息”,消息);
如果(!TextUtils.isEmpty(图像))
参数put(“图像”,图像);
参数put(“航向”,航向);
返回参数;
}
};
getInstance(getActivity()).addToRequestQueue(stringRequest);
我是android新手。请帮我解决这个问题。你可以这样做 首先,您需要将字符串响应转换为JSON响应
StringRequest stringRequest = new StringRequest(Request.Method.POST, CONN_URL.send_single_push,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
JSONObject obj = new JSONObject(response);
int failure = obj.optInt("failure");
if (failure == 0){
}else{
}
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
}
}) {
@Override
protected Map<String, String> getParams() throws AuthFailureError {
Map<String, String> params = new HashMap<>();
params.put("title", title);
params.put("message", message);
if (!TextUtils.isEmpty(image))
params.put("image", image);
params.put("course", course);
return params;
}
};
StringRequest StringRequest=新的StringRequest(Request.Method.POST、CONN\u URL.send\u single\u push、,
新的Response.Listener(){
@凌驾
公共void onResponse(字符串响应){
JSONObject obj=新的JSONObject(响应);
内部故障=对象选项(“故障”);
如果(失败==0){
}否则{
}
}
},
新的Response.ErrorListener(){
@凌驾
公共无效onErrorResponse(截击错误){
}
}) {
@凌驾
受保护的映射getParams()引发AuthFailureError{
Map params=新的HashMap();
参数put(“标题”,标题);
参数put(“消息”,消息);
如果(!TextUtils.isEmpty(图像))
参数put(“图像”,图像);
参数put(“航向”,航向);
返回参数;
}
};
你可以这样做
首先,您需要将字符串响应转换为JSON响应
StringRequest stringRequest = new StringRequest(Request.Method.POST, CONN_URL.send_single_push,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
JSONObject obj = new JSONObject(response);
int failure = obj.optInt("failure");
if (failure == 0){
}else{
}
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
}
}) {
@Override
protected Map<String, String> getParams() throws AuthFailureError {
Map<String, String> params = new HashMap<>();
params.put("title", title);
params.put("message", message);
if (!TextUtils.isEmpty(image))
params.put("image", image);
params.put("course", course);
return params;
}
};
StringRequest StringRequest=新的StringRequest(Request.Method.POST、CONN\u URL.send\u single\u push、,
新的Response.Listener(){
@凌驾
公共void onResponse(字符串响应){
JSONObject obj=新的JSONObject(响应);
内部故障=对象选项(“故障”);
如果(失败==0){
}否则{
}
}
},
新的Response.ErrorListener(){
@凌驾
公共无效onErrorResponse(截击错误){
}
}) {
@凌驾
受保护的映射getParams()引发AuthFailureError{
Map params=新的HashMap();
参数put(“标题”,标题);
参数put(“消息”,消息);
如果(!TextUtils.isEmpty(图像))
参数put(“图像”,图像);
参数put(“航向”,航向);
返回参数;
}
};
1)json无效。2) 您可以通过json.getInt(“failure”)获得它;1) json无效。2) 您可以通过json.getInt(“failure”)获得它;