小编Asf*_*fab的帖子

是否有可能从config中的子目录导入services.yml文件？

我有这样的结构:

[Acme/MyBundle/Resources]

--[config]

----[routing]

----[services]

------[User]

--------services.yml

----[validation]

----routing.yml

----services.yml

----validation.yml

现在我想导入文件Acme/MyBundle/Resources/config/services/User/services.yml 到Acme/MyBundle/Resources/config/services.yml,这是在装DependencyInjection/AcmeMyExtension.代码Acme/MyBundle/Resources/config/services.yml是:

imports:
    - { resource: "@AcmeMyBundle/Resources/config/services/User/services.yml" }

services:
    //other services

我收到错误:

Fatal error: Uncaught exception 'InvalidArgumentException' with message 'The file "@AcmeMyBundle/Resources/config/services/User/services.yml" does not exist (in: ., C:\wamp\www\dir\src\Acme\MyBundle\DependencyInjection/../Resources/config).' in C:\wamp\www\dir\app\cache\dev\classes.php:1518 
Stack trace: #0 C:\wamp\www\dir\vendor\symfony\symfony\src\Symfony\Component\Config\Loader\FileLoader.php(70): Symfony\Component\Config\FileLocator->locate('@AcmeMyBundle/...', '.', false) #1 C:\wamp\www\dir\vendor\symfony\symfony\src\Symfony\Component\DependencyInjection\Loader\YamlFileLoader.php(97): Symfony\Component\Config\Loader\FileLoader->import('@AcmeMyBundle/...', NULL, false, 'services.yml') #2 C:\wamp\www\dir\vendor\symfony\symfony\src\Symfony\Component\DependencyInjection\Loader\YamlFileLoader.php(54): Symfony\Component\DependencyInjection\Loader\YamlFileLoader->parseImports(Array, 'services.yml') #3 C:\wamp\www\dir\src\Acme\MyBundle\DependencyInjection\AcmeMyExtension.php(26): Symfony\Co in C:\wamp\www\dir\vendor\symfony\symfony\src\Symfony\Component\Config\Loader\FileLoader.php on line 100

Acme/MyBundle/DependencyInjection/AcmeMyExtension.php的代码:

<?php

namespace Core\MpgBundle\DependencyInjection;

use Symfony\Component\DependencyInjection\ContainerBuilder; …

我只是为用户注册创建一个表单代码在这里

forms.py

from django import forms
from django.conf import settings
from django.contrib.auth import get_user_model



def lowercase_email(email):
        """
        Normalize the address by lowercasing the domain part of the email
        address.
        """
        email = email or ''
        try:
            email_name, domain_part = email.strip().rsplit('@', 1)
        except ValueError:
            pass
        else:
            email = '@'.join([email_name.lower(), domain_part.lower()])
        return email

class SignupForm (forms.ModelForm):

    username =forms.CharField( 
        label='username',required=True,max_length=20,min_length=3)
    email = forms.EmailField( 
        label='email',required=True)
    password =forms.CharField( 
        label='pssword',required=True,max_length=20,min_length=6,widget = forms.PasswordInput)
    confirm_password= forms.CharField(
        label='confirm_email',required=True,max_length=20,min_length=6,widget = forms.PasswordInput)

    class Meta:
        model = get_user_model()
        fields = ("username","email","password","confirm_password",)




    def …