从JSON模式中的特定属性中仅选择一个随机键值对

Question

如何从给定的JSON模式中获取新的JSON模式，只有一个键值对从＆＃34;属性＆＃34;中随机选择。属性。它还应该有＆＃34;标题＆＃34;和＆＃34;键入＆＃34;属性。

AuthenticationEvents::AUTHENTICATION_SUCCESS

Answer 1

首先，您可以获得0-4之间的随机数，可用于随机获取properties密钥，并使用该密钥将其添加到新JSON对象的properties中：

var jsonSchema = { "title": "animals object",
   "type": "object",
   "properties":
     {
     'cat': 'meow',
     'dog': 'woof',
     'cow': 'moo',
     'sheep': 'baaah',
     'bird': 'tweet'
     }
};
var newJSON = { "title": "animals object",
   "type": "object",
   "properties":{}
};

var randomNumber = Math.floor(Math.random() * 5);
var randomPropertyKey = Object.keys(jsonSchema.properties)[randomNumber];
newJSON.properties[randomPropertyKey] = jsonSchema.properties[randomPropertyKey];
console.log(newJSON);

Answer 2

这应该可以解决问题：

var originalJson = {
      "title": "animals object",
      "type": "object",
      "properties": {
        'cat': 'meow',
        'dog': 'woof',
        'cow': 'moo',
        'sheep': 'baaah',
        'bird': 'tweet'
      }
    };

    // deep copy of the json object
    var jsonCopy = JSON.parse(JSON.stringify(originalJson)); 
    //gets a random property of a collection
    var propToKeep = pickRandomProperty(jsonCopy.properties);
   
    //deletes all properties of the copied collection except the randomly chosen one
    for (var key in jsonCopy.properties) {
          if(key.toString() !== propToKeep) {
            delete jsonCopy.properties[key]
        }    
    }

    function pickRandomProperty(obj) {
      var result;
      var count = 0;
      
      for (var key in obj){
        //if there's only one left, take it.
        if (Math.random() < 1 / ++count){
          result = key;
        }
      }
      return result;
    }
    console.log("copy", jsonCopy);
    console.log("original", originalJson);