英文:
undefined local variable or method `asistencia_params' for #<AsistenciaController:0x00000000083450>
问题
class AsistenciaController < ApplicationControllerdef asistencia@username = params[:user_username]@entry = params[:user_entry]if @asistencia.update(asistencia_params)redirect_to asistencia_path, notice: "Asistencia Guardada"elserender :edit, status: :unprocessable_entityendendend未定义局部变量或方法 `asistencia_params` 用于 #<AsistenciaController:0x00000000083450>我期望能够通过名称来更新该条目我想强调在其他控制器中，所有的用户名和条目都已经定义。
英文:
class AsistenciaController < ApplicationControllerdef asistencia@username = params[:user_username]@entry = params[:user_entry]if @asistencia.update(asistencia_params)redirect_to asistencia_path, notice: "Asistencia Guardada"elserender :edit, status: :unprocessable_entityendendendundefined local variable or method `asistencia_params' for #<AsistenciaController:0x00000000083450>
I expected to be able to update the entry with the use of the name
I want to emphasize that in other controllers all the usernames, entries are already defined.
答案1
得分: 1
你需要在你的控制器中定义方法 asistencia_params，像这样：
def asistencia_params# 在这里你可以允许或进行其他你需要的操作# (例如，params 赋值或只允许处理所需的参数)params.require(:asistencia).permit(:user_name, :user_entry, 等等..)end
英文:
You have to define the method asistencia_params like this in your controller.
def asistencia_params#here you can permit or do whatever you want(eg params assignment or permiting only required params for processing)params.require(:asistencia).permit(:user_name, :user_entry, etc..)end
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