Question

此代码应该将我的JSON值发送到服务器。当我点击按钮时，出现错误 - Unexpected reponse code 404.

有人可以解释问题是什么以及如何解决它？

这是服务器端错误吗？

这是我的代码......

b2.setOnClickListener(new View.OnClickListener(){
    @Override 
    public void onClick(View v){
      getResults();
      StringRequest request=new StringRequest(Request.Method.POST,server_url,new Response.Listener<String>() {

          @Override 
          public void onResponse(String response){

          }
      },new Response.ErrorListener(){
          @Override
          public void onErrorResponse(VolleyError error) {

          }
        })

    {
    protected Map<String,String>getParams() throws AuthFailureError{
    Map<String,String> parameters = new HashMap<String, String>();
    getResults().put(parameters);

return parameters;
}
            };
            requestQueue.add(request);
        }
    });
}


private JSONArray getResults() {

    String myPath = "/data/data/com.example.sebastian.patientdetails/databases/" + "MyDBName.db";

    String myTable = "patients";


    SQLiteDatabase myDataBase = SQLiteDatabase.openDatabase(myPath, null, SQLiteDatabase.OPEN_READONLY);

    String searchQuery = "SELECT  * FROM " + myTable;
    Cursor cursor = myDataBase.rawQuery(searchQuery, null);

    JSONArray resultSet = new JSONArray();

    cursor.moveToFirst();
    JSONArray jsonArray = null;
    while (!cursor.isAfterLast()) {

        int totalColumn = cursor.getColumnCount();
        JSONObject rowObject = new JSONObject();

        jsonArray = new JSONArray();
        for (int i = 0; i < totalColumn; i++) {
            if (cursor.getColumnName(i) != null) {

                JSONObject object = new JSONObject();
                try {
                    if (cursor.getString(i) != null) {
                        Log.d("TAG_NAME", cursor.getString(i));
                        object.put(cursor.getColumnName(i), cursor.getString(i));

                    } else {
                        object.put(cursor.getColumnName(i), "");
                    }

                    jsonArray.put(object);
                } 
                catch (Exception e) {
                    Log.d("TAG_NAME", e.getMessage());
                }
            }
        } 
        jsonArray.put(rowObject);
        resultSet.put(rowObject);
        cursor.moveToNext();
    }
    return resultSet;
}
}

Answer 1

404表示未找到，server_url指向不存在的网址，或者您请求的网页将您重定向到404。

如果您打开浏览器并将AJAX中的POST请求发送到相同的URL和相同的参数，那么您应该会遇到相同的行为。

将数据发送到服务器时出错404

1 个答案: