android和php api的响应是正确的，而通过android App删除数据时没有删除数据？

Question

在 android 应用程序上按下删除按钮时，产品记录没有从数据库中删除....而它在 android 中的响应与 Api 中的 ..Api 响应相同，通过 POSTMAN 软件检查是正确的，并且正在从数据库中删除记录。

Android 代码：

case R.id.delete:
    StringRequest request = new StringRequest(Request.Method.GET, getResources().getString(R.string.Delete_URL) + "?action=deleteProduct&product_id=" + menuItem.getItemId() + "", new Response.Listener < String > () {
        @Override
        public void onResponse(String response) {
            try {
                JSONObject jsonObject = new JSONObject(response);
                boolean error = jsonObject.getBoolean("error");
                if (error) {
                    Toast.makeText(getActivity(), "Error=>" + jsonObject.getString("error_message"), Toast.LENGTH_SHORT).show();

                } else {

                    Toast.makeText(getActivity(), jsonObject.getString("success_msg"), Toast.LENGTH_SHORT).show();

                }
            } catch (JSONException e) {
                e.printStackTrace();
                Toast.makeText(getActivity(), "JSON error : " + e.getMessage(), Toast.LENGTH_SHORT).show();
            }
        }
    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {
            Toast.makeText(getActivity(), error.getMessage() + "<==Error Connecting", Toast.LENGTH_SHORT).show();

        }
    });

    RequestQueue queue = Volley.newRequestQueue(getActivity());
    queue.add(request);


    break;

    }
    return false;
}
});
}
});
mProductListview.setAdapter(adapter);

php 代码

case 'deleteProduct':
    if(isset($_GET['product_id'])){
        $ID=$_GET['product_id'];
        $query="DELETE FROM `products` WHERE `product_id`='$ID'";
        if(mysqli_query($con,$query)){
            $response['error']=false;
            $response['success_msg']='Product Deleted successfully';

        }else{
            $response['error']=true;
            $response['error_message']='Product not exist';
        }
        }else{
            echo 'some worng';
        }
    break;

Answer 1

在 php 端试试这个代码：

case 'deleteProduct':
    if(isset($_GET['product_id'])){
        $ID=$_GET['product_id'];
        $query="DELETE FROM products WHERE product_id='".$ID."'";
        if(mysqli_query($con,$query)){
            $response['error']=false;
            $response['success_msg']='Product Deleted successfully';
        }
        else{
            $response['error']=true;
            $response['error_message']='Product not exist';
        }
    }else{
        echo 'some worng';
    }
break;