我打算将一个参数发送到一个php页面,并且在以json格式执行特定处理后,我得到了相同的页面。我写了以下代码:
//Showing the progress dialog
final ProgressDialog loading = ProgressDialog.show(this,"در حال ارتباط با سرور","لطفا صبر کنید",false,false);
final JsonArrayRequest jsonArrReq = new JsonArrayRequest(GetFavTickets_url,
new Response.Listener<JSONArray>() {
@Override
public void onResponse(JSONArray jsonArray) {
// Log.e("LOG", jsonArray.toString());
for (int i = 0 ; i < jsonArray.length() ; i++)
{
try {
JSONObject jsonObject = jsonArray.getJSONObject(i);
Log.e("LOG", "object : " + jsonObject.getString("name"));
name.add(i,jsonObject.getString("name"));
family.add(i,jsonObject.getString("family"));
} catch (JSONException e) {
e.printStackTrace();
}
}
loading.dismiss();
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError volleyError) {
//Dismissing the progress dialog
loading.dismiss();
//Showing toast
Toast.makeText(Favorite.this, volleyError.getMessage().toString(), Toast.LENGTH_LONG).show();
Log.e("Volley",volleyError.getMessage().toString());
}
}){
@Override
protected Map<String, String> getParams() throws AuthFailureError {
//Converting Bitmap to String
//Creating parameters
Map<String,String> params = new Hashtable<String, String>();
//Adding parameters
params.put("id", id);
//returning parameters
return params;
}
};
//Creating a Request Queue
RequestQueue requestQueue = Volley.newRequestQueue(this);
//Adding request to the queue
requestQueue.add(jsonArrReq);
}
Volley给出了以下信息:
org.json.JSONException:值类型为java.lang.String的错误不能 转换为
JSONArray
答案 0 :(得分:0)
我的PHP代码:
<?php
if($_SERVER['REQUEST_METHOD']=='POST'){
@$national_code = $_POST['national_code'];
$Tickets = array(null);
$result = mysqli_query($con,"SELECT * FROM favorites WHERE user_national_code='$national_code'");
$count=mysqli_num_rows($result);
while($row = mysqli_fetch_array($result))
{
$Tickets[] = $row['ticket_id'];
}
$length = count($Tickets)-1;
for ($i=1; $i<=$length; $i++){
$result = mysqli_query($con,"SELECT * FROM `ticket` WHERE id='$Tickets[i]'");
while($row = mysqli_fetch_array($result)){
$output[]=$row;
}
}
}else{
echo "Error";
}
?>
答案 1 :(得分:0)
尝试替换此
final JsonArrayRequest jsonArrReq = new JsonArrayRequest(GetFavTickets_url, new Response.Listener()
与
StringRequest stringRequest = new StringRequest(Request.Method.GET, url,
new Response.Listener<String>() ...