blocks|key|1606011|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1606012|var+ar+=+[{"value":"14","label":"7"},{"value":"14","label":"7"},{"value":"18","label":"7"}];

for(var+i=0;+i+<+ar.length;+i%2B%2B)+{
+++if(ar[i].value+==+"14"+&&+ar[i].label+==+"7")
+++{
++++++ar.splice(i,1);
+++}
}|code-block|syntax|javascript|1606013|演示|offset|length|1606014|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/eorozco/95vs783e/3/^0|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Try:

<pre><code>var ar = [{"value":"14","label":"7"},{"value":"14","label":"7"},{"value":"18","label":"7"}];

for(var i=0; i &lt; ar.length; i++) {
 if(ar[i].value == "14" &amp;&amp; ar[i].label == "7")
 {
 ar.splice(i,1);
 }
}
</code></pre>

<a href="http://jsfiddle.net/eorozco/95vs783e/3/" rel="noreferrer">demo</a>

blocks|key|562997|text|我会创造一个新的数组。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|562998|var+original+=+[{"value":"14","label":"7"},{"value":"14","label":"7"},{"value":"18","label":"7"}]

var+result+=+[];

for+(var+i+=+0,+l+=+original.length;+i+<+l;+i%2B%2B)+{+//+Traverse+the+whole+array
++var+current+=+original[i];
++if+(!+(current.value+==+14+&&+current.label+==+7)+)+{
++++//+It+doesn't+match+the+criteria,+so+add+it+to+result
++++result.push(+current+);
++}
}|code-block|syntax|javascript|562999|编辑:我又读了一遍你的问题。您只想删除第一个元素？然后使用slice只获取数组的一部分。|offset|length|style|CODE|563000|var+result+=+original.slice(1,+original.length)|563001|或splice|563002|original.splice(0,1);+//+Remove+one+element+from+index+0.|563003|entityMap^0|0|0|T|5|0|0|1|6|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

I would create a new array...

<pre><code>var original = [{"value":"14","label":"7"},{"value":"14","label":"7"},{"value":"18","label":"7"}]

var result = [];

for (var i = 0, l = original.length; i &lt; l; i++) { // Traverse the whole array
 var current = original[i];
 if (! (current.value == 14 &amp;&amp; current.label == 7) ) {
 // It doesn't match the criteria, so add it to result
 result.push( current );
 }
}
</code></pre>

Edit: I've read your question once more. You only want to remove the first element? Then use <code>slice</code> to get only a part of the array.

<pre><code>var result = original.slice(1, original.length)
</code></pre>

or <code>splice</code>

<pre><code>original.splice(0,1); // Remove one element from index 0.
</code></pre>

blocks|key|2905268|text|您需要保留重复的对象吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2905269|用underscoreJS|2905270|var+oldarray+=+[{"value":"14","label":"7"},{"value":"14","label":"7"},{"value":"18","label":"7"}]
var+newarray+=+_.reject(oldarray,+function(o)+{+
++return+o+==+{"value":"14","label":"7"};+
});|code-block|syntax|javascript|2905271|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Do you need to keep the duplicate object?

With underscoreJS

<pre><code>var oldarray = [{"value":"14","label":"7"},{"value":"14","label":"7"},{"value":"18","label":"7"}]
var newarray = _.reject(oldarray, function(o) { 
 return o == {"value":"14","label":"7"}; 
});
</code></pre>

blocks|key|1606170|text|在ES6+(或使用es6-shim)中，您可以使用Array.prototype.findIndex和Array.prototype.splice|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1606171|arr.splice(arr.findIndex(matchesEl),+1);

function+matchesEl(el)+{
++++return+el.value+===+'14'+&&+el.label+===+'7';
}|code-block|syntax|javascript|1606172|或者，如果数组的副本是可以的(并且可以从ES5中获得)，那么Array.prototype.filter是最好的选择：|1606173|var+withoutEl+=+arr.filter(function+(el)+{+return+!matchesEl(el);+});|1606174|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/paulmillr/es6-shim/|1|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/findIndex|2|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/splice|3|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter^0|9|8|0|P|P|1|1F|M|2|0|0|U|M|3|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]|$A|14|B|15|1|16]|$A|17|B|18|1|19]]|C|$]]|$1|D|3|E|5|F|7|1A|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1B|8|@]|9|@$A|1C|B|1D|1|1E]]|C|$]]|$1|K|3|L|5|F|7|1F|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|1G|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]|W|$5|P|Q|R|C|$S|X]]|Y|$5|P|Q|R|C|$S|Z]]]]

In ES6 (or using <a href="https://github.com/paulmillr/es6-shim/">es6-shim</a>) you can use <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/findIndex">Array.prototype.findIndex</a> along with <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/splice">Array.prototype.splice</a>:

<pre><code>arr.splice(arr.findIndex(matchesEl), 1);

function matchesEl(el) {
 return el.value === '14' &amp;&amp; el.label === '7';
}
</code></pre>

Or if a copy of the array is ok (and available since ES5), <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter">Array.prototype.filter</a>'s the way to go:

<pre><code>var withoutEl = arr.filter(function (el) { return !matchesEl(el); });
</code></pre>

blocks|key|2905395|text|同一性解|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2905396|如果您具有对象标识，而不仅仅是对象相等(也就是说，您试图从数组中删除特定对象，而不仅仅是包含与现有对象相同的数据的对象)，那么您可以非常简单地使用剪接和indexOf来完成这个任务：|2905397|a+=+{x:1}
b+=+{x:2}
arr+=+[a,b]|code-block|syntax|javascript|2905398|假设你想移除b：|2905399|arr.splice(
++arr.indexOf(b),+1
);|2905400|平等解|2905401|这个问题使我们不清楚我们是在处理身份问题还是平等问题。如果您有相等(包含相同内容的两个对象，它们彼此看起来是相同的，但实际上不是同一个对象)，那么您将需要一个小的比较函数，我们可以将它传递给findIndex。|2905402|a+=+{x:+1};
b+=+{x:+2};
arr+=+[a,+b];|2905403|现在，我们要删除c，它等于，但不等于：|2905404|c+=+{x:+1}

index+=+arr.findIndex(
++(i)+=>+i.x+===+c.x
)

if+(index+!==+-1)+{
++arr.splice(
++++index,+1
++);
}++++|2905405|注意，我在这里传递一个ES6样式lambda函数作为findIndex的第一个参数。|2905406|除非您拥有只为常绿浏览器编写的特权，否则您可能希望使用像Babel这样的预告器来使用此解决方案。|2905407|不变性解|2905408|如果您关心不可变性，并且希望返回一个新数组，则可以使用筛选器：|2905409|a+=+{x:+1};
b+=+{x:+2};
arr+=+[a,+b];|2905410|让我们保留所有不是：|2905411|newArr+=+arr.filter((x)+=>+x+!=+a)|2905412|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1D|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1E|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|1F|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|1G|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|1H|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1I|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|1J|8|@]|9|@]|A|$G|H]]|$1|S|3|T|5|6|7|1K|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|1L|8|@]|9|@]|A|$G|H]]|$1|W|3|X|5|6|7|1M|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1N|8|@]|9|@]|A|$]]|$1|10|3|11|5|6|7|1O|8|@]|9|@]|A|$]]|$1|12|3|13|5|6|7|1P|8|@]|9|@]|A|$]]|$1|14|3|15|5|F|7|1Q|8|@]|9|@]|A|$G|H]]|$1|16|3|17|5|6|7|1R|8|@]|9|@]|A|$]]|$1|18|3|19|5|F|7|1S|8|@]|9|@]|A|$G|H]]|$1|1A|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]]|1B|$]]

<h2>Solution with Identity</h2>

If you have object identity not just object equality (i.e. you're trying to delete a specific object from the array, not just an object that contains the same data as an existing object) you can do this very simply with splice and indexOf:

<pre><code>a = {x:1}
b = {x:2}
arr = [a,b]
</code></pre>

Say you want to remove b:

<pre><code>arr.splice(
 arr.indexOf(b), 1
);
</code></pre>

<h2>Solution with Equality</h2>

The question makes it unclear whether we are dealing with identity or equality. If you have equality (two objects containing the same stuff, which look the same as each other, but are not actually the same object), you'll need a little comparison function, which we can pass to findIndex.

<pre><code>a = {x: 1};
b = {x: 2};
arr = [a, b];
</code></pre>

Now we want to remove c, which is equal to, but not the same as a:

<pre><code>c = {x: 1}

index = arr.findIndex(
 (i) =&gt; i.x === c.x
)

if (index !== -1) {
 arr.splice(
 index, 1
 );
} 
</code></pre>

Note that I'm passing an ES6 style lambda function here as the first parameter to findIndex. 

Unless you have the luxury of only writing for evergreen browsers, you'll probably want to be using a transpiler like Babel to use this solution.

<h2>Solution with Immutability</h2>

If you care about immutability, and you want to return a new array, you can do it with filter:

<pre><code>a = {x: 1};
b = {x: 2};
arr = [a, b];
</code></pre>

Let's keep everything that isn't a:

<pre><code>newArr = arr.filter((x) =&gt; x != a)
</code></pre>

blocks|key|1606248|text|使用ES6方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1606249|const+objIndex+=+objs.findIndex(o+=>+o.value+===+'14'+&&+o.label+===+'7')

if(objIndex+>+-1)+{
++objs.slice(objIndex,+1)
}|code-block|syntax|javascript|1606250|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Using ES6 approach:

<pre><code>const objIndex = objs.findIndex(o =&gt; o.value === '14' &amp;&amp; o.label === '7')

if(objIndex &gt; -1) {
 objs.slice(objIndex, 1)
}
</code></pre>

blocks|key|563165|text|function+removeDuplicateObject+(arr)+{
++let+temp+=+arr.map((item)+=>+{
++++return+JSON.stringify(item);
++});
++temp+=+Array.from(new+Set(temp));
++return+temp.map((item)+=>+{
++++return+JSON.parse(item);
++});
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|563166|然后是removeDuplicateObject(your+array)|unstyled|offset|length|style|CODE|563167|entityMap^0|0|3|X|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>function removeDuplicateObject (arr) {
 let temp = arr.map((item) =&gt; {
 return JSON.stringify(item);
 });
 temp = Array.from(new Set(temp));
 return temp.map((item) =&gt; {
 return JSON.parse(item);
 });
}
</code></pre>

Then <code>removeDuplicateObject(your array)</code>

blocks|key|563201|text|如果您不想更改数组，只需使用filter方法，Array.prototype.filter()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|563202|array.filter((i)=>+i+!=+"b");|563203|entityMap^0|N|O|0|0|T|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|I|$]]

If you dont want to mutate your array simply use filter method , <code>Array.prototype.filter()</code>

<code>array.filter((i)=&gt; i != "b");</code>

I have an array of objects:

<pre><code>[{"value":"14","label":"7"},{"value":"14","label":"7"},{"value":"18","label":"7"}]
</code></pre>

How I can delete this item <code>{"value":"14","label":"7"}</code> resulting in the new array:

<pre><code> [{"value":"14","label":"7"},{"value":"18","label":"7"}]
</code></pre>

?

Remove object from array of objects

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一组对象：[{"value":"14","label":"7"},{"value":"14","label":"7"},{"value":"18","label":"7"}]如何删除该项{"value":"14","label":"7"}，从而生成新的数组： [{"value":"14","label":"7"},...

问从对象数组中删除对象
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问从对象数组中删除对象EN