英文:
Symfony Assert\Expression compare entity property with current users property
问题
你能否在Symfony的Assert\Expression中获取当前已验证的用户对象？
我想要使用类似以下的方式：
#[Assert\Expression(
this.getType() != user.getType(),
message: 'Invalid type'
)]
我尝试了但遇到错误：
Variable "user" is not valid around position 60 for expression
英文:
Is it possible to get current authenticated user object in symfony Assert\Expression ?
I want to use something like
#[Assert\Expression(
this.getType() != user.getType()",
message: 'Invalid type'
)]
I tried but getting an error
Variable \"user\" is not valid around position 60 for expression
答案1
得分: 1
我不认为直接从你的实体中获取当前经过身份验证的用户是可能的。
在这种情况下，你应该按照这里解释的方式创建一个自定义约束和验证器：https://symfony.com/doc/current/validation/custom_constraint.html#creating-the-constraint-class。
你可以创建一个HasValidType和HasValidTypeValidator，在其中你可以注入Symfony的Security组件并执行$this->security->getUser()以获取当前经过身份验证的用户。
英文:
I don't think it is possible to get the current authenticated user directly from your entity.
In this case, you should create a custom constraint & validator as explained here : https://symfony.com/doc/current/validation/custom_constraint.html#creating-the-constraint-class.
You could create a HasValidType and HasValidTypeValidator in which you can inject the Security component of Symfony and perform a $this->security->getUser() to get the current authenticated user.
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