underscoreJS- 形成一个包含 array1 中但不在 array2 中的项目的新数组

Question

假设我有 array1 和 array2，需要找到一种方法来找到 array1 而不是 array2 中的项目，

array1":[{"name":"tom","lastname":"jacob","emailId":"tomjacob@gmail.com","$$key":"object:4"},{"name":"will","lastname":"rogers","emailId":"willrogers@gmail.com","$$key":"object:24"},{"name":"matt","lastname":"jacob","emailId":"mattjacob@gmail.com","$$key":"object:84"}]
array2":[{"name":"tom","lastname":"jacob","emailId":"tomjacob@gmail.com","$$key":"object:4"},{"name":"matt","lastname":"jacob","emailId":"mattjacob@gmail.com","$$key":"object:84"}]
**i need to get a new array** 
**result:**[{"name":"will","lastname":"rogers","emailId":"willrogers@gmail.com","$$key":"object:24"}]

Answer 1

您可以使用 reject and some:

var result = _.reject(data.array1, item => _.some(data.array2, { $$key : item.$$key}));

或没有箭头功能：

var result = _.reject(data.array1, function(item) {
    return _.some(data.array2, { $$key : item.$$key}); 
});

var data = {
 "array1":[
  {"name":"tom","lastname":"jacob","emailId":"tomjacob@gmail.com","$$key":"object:4"},
  {"name":"will","lastname":"rogers","emailId":"willrogers@gmail.com","$$key":"object:24"},
  {"name":"matt","lastname":"jacob","emailId":"mattjacob@gmail.com","$$key":"object:84"}],

 "array2":[
  {"name":"tom","lastname":"jacob","emailId":"tomjacob@gmail.com","$$key":"object:4"},
  {"name":"matt","lastname":"jacob","emailId":"mattjacob@gmail.com","$$key":"object:84"}]
}

var result = _.reject(data.array1, item => _.some(data.array2, { $$key : item.$$key}));

document.getElementById('result').textContent = JSON.stringify(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.2/underscore.js"></script>

<p>
  <pre id="result"></pre>
</p>

underscoreJS- 形成一个包含 array1 中但不在 array2 中的项目的新数组

underscoreJS- form a new array with items in array1 but not in array2

underscore.js