如何跳过特定约束的验证？

Question

假设我有以下Foo对象:

Foo foo = new Foo(foo: "foo", bar: "bar", baz: "baz")

我知道如何验证特定约束:

foo.validate(["foo", "bar"]) // validates the "foo" property and the "bar" property, but not the "baz" property

我也知道如何放弃验证:

foo.save(validate: false)

但我不知道如何告诉Grails验证除列表中的约束之外的所有约束.我可以创建一个满足我想要的方法,但我想确保没有Groovy方法可以先做.

更新

如果没有"更加时髦"的方式,我将如何做到这一点.

    // This method exists in my Util.groovy class in my src/groovy folder
    static def validateWithBlacklistAndSave(def obj, def blacklist = null) {
        def propertiesToValidate = obj.domainClass.constraints.keySet().collectMany{ !blacklist?.contains(it)?  [it] : [] }
        if(obj.validate(propertiesToValidate)) {
            obj.save(flush: true, validate: false)
        }
        obj
    }

Answer 1

考虑以下Foo域类

class Foo {
    String foo
    String bar
    String baz

    static constraints = {
        foo size: 4..7
        bar size: 4..7
        baz size: 4..7
    }
}

验证baz可以排除如下:

Foo foo = new Foo(foo: "fool", bar: "bars", baz: "baz")

//Gather all fields
def allFields = foo.class.declaredFields
                         .collectMany{!it.synthetic ? [it.name] : []}
//Gather excluded fields
def excludedFields = ['baz'] //Add other fields if necessary

//All but excluded fields
def allButExcluded = allFields - excludedFields

assert foo.validate(allButExcluded)
assert foo.save(validate: false) //without validate: false, validation kicks in
assert !foo.errors.allErrors

没有直接的方法来发送排除字段列表以进行验证.