blocks|key|4151490|text|使用Array.filter，我们可以返回符合条件的项。在本例中，条件是“不在第二个列表中”，我们可以用Array.some来表示。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4151491|function+getItemsOnlyInFirst(first,+second)+{
++return+first.filter(
++++item+=>
++++++!second.some(
++++++++compareItem+=>
++++++++++item.name+===+compareItem.name+&&+item.group+===+compareItem.group
++++++)
++);
}

let+first+=+[{+name:+"Sam",+group:+"test"+},+{+name:+"John",+group:+"it"+}];
let+second+=+[{+name:+"John",+group:+"it"+},+{+name:+"Tim",+group:+"hr"+}];

getItemsOnlyInFirst(first,+second);+//+{name:+"Sam",+group:+"test"}|code-block|syntax|javascript|4151492|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter|1|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/some^0|2|C|0|1G|A|1|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|R]]]]

Using <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter" rel="nofollow noreferrer">Array.filter</a>, we can return items that pass a condition. In this case, the condition is "not in the second list", which we can express with <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/some" rel="nofollow noreferrer">Array.some</a>.

<pre><code>function getItemsOnlyInFirst(first, second) {
 return first.filter(
 item =&gt;
 !second.some(
 compareItem =&gt;
 item.name === compareItem.name &amp;&amp; item.group === compareItem.group
 )
 );
}

let first = [{ name: "Sam", group: "test" }, { name: "John", group: "it" }];
let second = [{ name: "John", group: "it" }, { name: "Tim", group: "hr" }];

getItemsOnlyInFirst(first, second); // {name: "Sam", group: "test"}
</code></pre>

blocks|key|4151910|text|问题是不能用==运算符比较两个对象：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4151911|var+obj1+=+{test:+1}
var+obj2+=+{test:+1}

console.log(obj1+==+obj2)+//+false|code-block|syntax|javascript|4151912|一种简单的解决方案是将对象转换为JSON格式。|4151913|这之所以有效，是因为您可以比较字符串：|4151914|var+obj1+=+JSON.stringify({test:+1})
var+obj2+=+JSON.stringify({test:+1})

console.log(obj1+==+obj2)+//+true|4151915|使用JSON.stringify的解决方案：|4151916|var+first++=+[{name:+"Sam",+group:+"test"},+{name:+"John",+group:+"it"}];
var+second+=+[{name:+"John",+group:+"it"}+,+{name:+"Tim",+group:+"hr"}];

//+Convert+the+objects+to+JSON
var+second_str+=+second.map(JSON.stringify)

var+filtered+=+first.filter(function(elem)+{
++++return+second_str.indexOf(JSON.stringify(elem))+==+-1
})

console.log(filtered)|4151917|entityMap^0|6|2|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|13|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|V|$]]

The problem is you cannot compare two objects with the <code>==</code> operator:

<pre><code>var obj1 = {test: 1}
var obj2 = {test: 1}

console.log(obj1 == obj2) // false
</code></pre>

An easy solution is to convert the objects into the JSON format.

This works because you can compare strings:

<pre><code>var obj1 = JSON.stringify({test: 1})
var obj2 = JSON.stringify({test: 1})

console.log(obj1 == obj2) // true
</code></pre>

Solution using JSON.stringify:

<pre><code>var first = [{name: "Sam", group: "test"}, {name: "John", group: "it"}];
var second = [{name: "John", group: "it"} , {name: "Tim", group: "hr"}];

// Convert the objects to JSON
var second_str = second.map(JSON.stringify)

var filtered = first.filter(function(elem) {
 return second_str.indexOf(JSON.stringify(elem)) == -1
})

console.log(filtered)
</code></pre>

blocks|key|4152016|text|我知道它们都有相同的界面？对于复杂的对象，你可以使用lodash和例如：_.differenceBy函数，但这里的情况很简单(仍然，方法是非常通用的，不依赖于对象内部的键，但它们需要在平面结构中没有递归)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4152017|let+isOnlyInFirst+=+first.filter(el+=>+{
++++return+!second.some(z+=>+{
++++++++const+el_keys+=+Object.keys(el)
++++++++const+z__keys+=+Object.keys(z)
++++++++let+flag+=+true
++++++++for+(const+key+of+el_keys)+{
++++++++++++if+(!z.hasOwnProperty(key)+%7C%7C+z[key]+!=+el[key])+flag+=+false
++++++++}
++++++++if+(el_keys.length+===+z__keys.length+&&+flag)
++++++++++++return+true
++++++++return+false
++++})
})|code-block|syntax|javascript|4152018|entityMap^0|10|E|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

I understand they both share same interface? For complex objects you could use lodash with eg: <code>_.differenceBy</code> function, but here the case is simple (still, approach is quite generic, does not depends on keys inside objects, but they need to be in flat structure [no recursion])

<pre><code>let isOnlyInFirst = first.filter(el =&gt; {
 return !second.some(z =&gt; {
 const el_keys = Object.keys(el)
 const z__keys = Object.keys(z)
 let flag = true
 for (const key of el_keys) {
 if (!z.hasOwnProperty(key) || z[key] != el[key]) flag = false
 }
 if (el_keys.length === z__keys.length &amp;&amp; flag)
 return true
 return false
 })
})
</code></pre>

I have two JS objects:

<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var first = [{name: "Sam", group: "test"}, {name: "John", group: "it"}];
var second = [{name: "John", group: "it"}, {name: "Tim", group: "hr"}];
 
for (var k = 0; k &lt; first.length; k++) {
 if (first.indexOf(second[k]) == -1) {
 console.log('found');
 }
}</code></pre>
</div>
</div>


I am trying to filter out a value that is present at <code>first</code>, but missing at <code>second</code>. How can I do that?

Ideal return value: <code>{name: "Sam", group: "test"}</code>

Tried many things, nothing works at the moment. For example:

Two JS objects, get a value from first object that it missing in the second

JSON

我有两个JS对象：​var first = [{name: "Sam", group: "test"}, {name: "John", group: "it"}];var second = [{name: "John", group: "it"}, {name: "Tim", group: "hr"}];    for...

问两个JS对象，从第一个对象中获取在第二个对象中缺失的值
EN

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