如何仅在不存在其他属性时才允许一个属性,并在JSON模式中允许其他属性 [英] How to allow one property only if other is not present and allow any of other propeties in json schema
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问题描述
我有六个属性:姓名,年龄,电话,deletePhone,地址,deleteAddress.
I have six properties: name, age, phone, deletePhone, address, deleteAddress.
我想创建允许上述任何属性的模式,但phone不应与deletePhone一起出现,而address不应与deleteAddress一起出现(反之亦然).
I want to create schema that allows anyOf above properties but phone should not be present with deletePhone and address should not be present with deleteAddress(and vice versa).
我尝试过以下模式:
{
"type": "object",
"properties": {
"name": {
"type": "string"
},
"age": {
"type": "number"
},
"phone": {
"type": "number"
},
"deletePhone": {
"type": "boolean"
},
"address": {
"type": "string"
},
"deleteAddress": {
"type": "boolean"
}
},
"allOf": [
{
"oneOf": [
{
"required": ["phone"]
},
{
"required": ["deletePhone"]
}
]
},
{
"oneOf": [
{
"required": ["address"]
},
{
"required": ["deleteAddress"]
}
]
}
]
}
它证实为真
{
"name": "my name",
"address": "some addr",
"phone": 34
}
并验证
{
"address": "some addr",
"phone": 34,
"deletePhone": true
}
这是正确的,但它也验证了错误的
which is correct but it also validates false for
{
"phone": 34
}
or
{
"name": "some name"
}
我想验证真,我知道我缺少anyOf,oneOf的某种组合,还是有更好的方法?
Which i want to validate true, i know i'm missing some combination of anyOf, oneOf, Or is there any better way?
推荐答案
我认为这是最简单的解决方案.
I think this is the simplest solution.
{
"type": "object",
"properties": {
"name": { "type": "string" },
"age": { "type": "number" },
"phone": { "type": "number" },
"deletePhone": { "type": "boolean" },
"address": { "type": "string" },
"deleteAddress": { "type": "boolean" }
},
"allOf": [
{ "not": { "required": ["phone", "deletePhone"] } },
{ "not": { "required": ["address", "deleteAddress"] } }
]
}
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