blocks|key|337810|text|正如TJ·克劳德所说，每种方式都会有某种隐藏的迭代，有了lodash，这就变成了：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|337811|var+index+=+_.findIndex(array,+{prop2:+'yutu'})|code-block|syntax|javascript|337812|entityMap|0|LINK|mutability|MUTABLE|url|https://lodash.com/docs#findIndex^0|S|6|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

What TJ Crowder said, everyway will have some kind of hidden iteration, with <a href="https://lodash.com/docs#findIndex">lodash</a> this becomes:

<pre><code>var index = _.findIndex(array, {prop2: 'yutu'})
</code></pre>

blocks|key|120371|text|var+CarId+=+23;

//x.VehicleId+property+to+match+in+the+object+array
var+carIndex+=+CarsList.map(function+(x)+{+return+x.VehicleId;+}).indexOf(CarId);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|120372|对于基本数组编号，您也可以这样做：|unstyled|120373|var+numberList+=+[100,200,300,400,500];
var+index+=+numberList.indexOf(200);+//+1|120374|如果在数组中找不到值，您将得到-1。|120375|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>var CarId = 23;

//x.VehicleId property to match in the object array
var carIndex = CarsList.map(function (x) { return x.VehicleId; }).indexOf(CarId);
</code></pre>

And for basic array numbers you can also do this:

<pre><code>var numberList = [100,200,300,400,500];
var index = numberList.indexOf(200); // 1
</code></pre>

You will get -1 if it cannot find a value in the array.

blocks|key|337867|text|var+index;
yourArray.some(function+(elem,+i)+{
++++return+elem.prop2+===+'yutu'+?+(index+=+i,+true)+:+false;
});|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|337868|迭代数组的所有元素。如果条件不匹配，则返回索引和true或false。|unstyled|337869|重要的是显式返回值true+(或布尔结果为true的值)。单次赋值是不够的，因为可能会有索引为0+(Boolean(0)+===+false)，这不会导致错误，但会禁用迭代的中断。|337870|编辑|337871|上面的一个更简短的版本：|337872|yourArray.some(function+(elem,+i)+{
++++return+elem.prop2+===+'yutu'+&&+~(index+=+i);
});|337873|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$B|C]]|$1|O|3|-4|5|F|7|W|8|@]|9|@]|A|$]]]|P|$]]

<pre><code>var index;
yourArray.some(function (elem, i) {
 return elem.prop2 === 'yutu' ? (index = i, true) : false;
});
</code></pre>

Iterate over all elements of array.
It returns either the index and true or false if the condition does not match.

Important is the explicit return value of true (or a value which boolean result is true). The single assignment is not sufficient, because of a possible index with 0 (Boolean(0) === false), which would not result an error but disables the break of the iteration.

Edit

An even shorter version of the above: 

<pre><code>yourArray.some(function (elem, i) {
 return elem.prop2 === 'yutu' &amp;&amp; ~(index = i);
});
</code></pre>

blocks|key|337982|text|使用Array.map()和Array.indexOf(string)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|337983|​|337984|const+arr+=+[{
++prop1:+"abc",
++prop2:+"qwe"
},+{
++prop1:+"bnmb",
++prop2:+"yutu"
},+{
++prop1:+"zxvz",
++prop2:+"qwrq"
}]

const+index+=+arr.map(i+=>+i.prop2).indexOf("yutu");

console.log(index);|code-block|syntax|javascript|337985|337986|entityMap^0|2|B|E|L|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|V|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|W|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|O|$]]

Using <code>Array.map()</code> and <code>Array.indexOf(string)</code>
<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>const arr = [{
 prop1: "abc",
 prop2: "qwe"
}, {
 prop1: "bnmb",
 prop2: "yutu"
}, {
 prop1: "zxvz",
 prop2: "qwrq"
}]

const index = arr.map(i =&gt; i.prop2).indexOf("yutu");

console.log(index);</code></pre>
</div>
</div>

blocks|key|120241|text|您可以通过以下方式使用Array.prototype.some()+(如其他答案中所述)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|120242|https://jsfiddle.net/h1d69exj/2/|120243|function+findIndexInData(data,+property,+value)+{
++++var+result+=+-1;
++++data.some(function+(item,+i)+{
++++++++if+(item[property]+===+value)+{
++++++++++++result+=+i;
++++++++++++return+true;
++++++++}
++++});
++++return+result;
}
var+data+=+[{prop1:"abc",prop2:"qwe"},{prop1:"bnmb",prop2:"yutu"},{prop1:"zxvz",prop2:"qwrq"}]



alert(findIndexInData(data,+'prop2',+"yutu"));+//+shows+index+of+1|code-block|syntax|javascript|120244|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/some|1^0|B|M|0|0|0|W|1|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|D|3|E|5|6|7|X|8|@]|9|@$A|Y|B|Z|1|10]]|C|$]]|$1|F|3|G|5|H|7|11|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|12|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|E]]]]

You can use the <a href="https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/some" rel="nofollow">Array.prototype.some()</a> in the following way (as mentioned in the other answers):

<a href="https://jsfiddle.net/h1d69exj/2/" rel="nofollow">https://jsfiddle.net/h1d69exj/2/</a>

<pre><code>function findIndexInData(data, property, value) {
 var result = -1;
 data.some(function (item, i) {
 if (item[property] === value) {
 result = i;
 return true;
 }
 });
 return result;
}
var data = [{prop1:"abc",prop2:"qwe"},{prop1:"bnmb",prop2:"yutu"},{prop1:"zxvz",prop2:"qwrq"}]



alert(findIndexInData(data, 'prop2', "yutu")); // shows index of 1
</code></pre>

blocks|key|120334|text|function+findIndexByKeyValue(_array,+key,+value)+{
++++for+(var+i+=+0;+i+<+_array.length;+i%2B%2B)+{+
++++++++if+(_array[i][key]+==+value)+{
++++++++++++return+i;
++++++++}
++++}
++++return+-1;
}
var+a+=+[
++++{prop1:"abc",prop2:"qwe"},
++++{prop1:"bnmb",prop2:"yutu"},
++++{prop1:"zxvz",prop2:"qwrq"}];
var+index+=+findIndexByKeyValue(a,+'prop2',+'yutu');
console.log(index);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|120335|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>function findIndexByKeyValue(_array, key, value) {
 for (var i = 0; i &lt; _array.length; i++) { 
 if (_array[i][key] == value) {
 return i;
 }
 }
 return -1;
}
var a = [
 {prop1:"abc",prop2:"qwe"},
 {prop1:"bnmb",prop2:"yutu"},
 {prop1:"zxvz",prop2:"qwrq"}];
var index = findIndexByKeyValue(a, 'prop2', 'yutu');
console.log(index);
</code></pre>

blocks|key|337920|text|格奥尔格已经提到过，ES6已经为此提供了Array.findIndex。其他一些答案是使用Array.some方法的ES5的变通方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|337921|一种更优雅的方法可以是|337922|var+index;
for(index+=+yourArray.length;+index--+>+0+&&+yourArray[index].prop2+!==+"yutu";);|code-block|syntax|javascript|337923|同时，我想强调的是，Array.some可以使用二进制或其他有效的搜索技术来实现。因此，在某些浏览器中，它可能比for循环执行得更好。|337924|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Georg have already mentioned ES6 have Array.findIndex for this.
And some other answers are workaround for ES5 using Array.some method.

One more elegant approach can be

<pre><code>var index;
for(index = yourArray.length; index-- &gt; 0 &amp;&amp; yourArray[index].prop2 !== "yutu";);
</code></pre>

At the same time I will like to emphasize, Array.some may be implemented with binary or other efficient searching technique. So, it might perform better over for loop in some browser.

blocks|key|120387|text|试试这段代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|120388|var+x+=+[{prop1:"abc",prop2:"qwe"},{prop1:"bnmb",prop2:"yutu"},{prop1:"zxvz",prop2:"qwrq"}]
let+index+=+x.findIndex(x+=>+x.prop1+===+'zxvz')|code-block|syntax|javascript|120389|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|$]]

Try this code

<pre><code>var x = [{prop1:"abc",prop2:"qwe"},{prop1:"bnmb",prop2:"yutu"},{prop1:"zxvz",prop2:"qwrq"}]
let index = x.findIndex(x =&gt; x.prop1 === 'zxvz')
</code></pre>

blocks|key|120140|text|为什么你不想精确地迭代呢？新的Array.prototype.forEach在这方面做得很好！|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|120141|如果需要，您可以使用二进制搜索树通过单个方法调用来查找。这是JS+-+https://github.com/vadimg/js_bintrees中BTree和红黑搜索树的巧妙实现--但我不确定您是否能同时找到索引。|120142|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/Array/forEach|1|https://github.com/vadimg/js_bintrees^0|F|N|0|0|Z|11|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|F|3|-4|5|6|7|X|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]|N|$5|I|J|K|C|$L|O]]]]

Why do you not want to iterate exactly ? The new <a href="https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/Array/forEach" rel="nofollow">Array.prototype.forEach</a> are great for this purpose!

You can use a Binary Search Tree to find via a single method call if you want. This is a neat implementation of BTree and Red black Search tree in JS - <a href="https://github.com/vadimg/js_bintrees" rel="nofollow">https://github.com/vadimg/js_bintrees</a> - but I'm not sure whether you can find the index at the same time.

blocks|key|120316|text|只需一步即可使用Array.reduce()+-无需jQuery|type|unstyled|depth|inlineStyleRanges|entityRanges|data|120317|var+items+=+[{id:+331},+{id:+220},+{id:+872}];

var+searchIndexForId+=+220;
var+index+=+items.reduce(function(searchIndex,+item,+index){
++if(item.id+===+searchIndexForId)+{+
++++console.log('found!');
++++searchIndex+=+index;
++}
++return+searchIndex;
},+null);|code-block|syntax|javascript|120318|如果未找到索引，将返回null。|offset|length|style|CODE|120319|entityMap^0|0|0|B|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

One step using Array.reduce() - no jQuery

<pre><code>var items = [{id: 331}, {id: 220}, {id: 872}];

var searchIndexForId = 220;
var index = items.reduce(function(searchIndex, item, index){
 if(item.id === searchIndexForId) { 
 console.log('found!');
 searchIndex = index;
 }
 return searchIndex;
}, null);
</code></pre>

will return <code>null</code> if index was not found.

blocks|key|120331|text|var+list+=++[
++++++++++++++++{prop1:"abc",prop2:"qwe"},
++++++++++++++++{prop1:"bnmb",prop2:"yutu"},
++++++++++++++++{prop1:"zxvz",prop2:"qwrq"}
++++++++++++];

var+findProp+=+p+=>+{
++++var+index+=+-1;
++++$.each(list,+(i,+o)+=>+{
++++++++if(o.prop2+==+p)+{
++++++++++++index+=+i;
++++++++++++return+false;+//+break
++++++++}
++++});
++++return+index;+//+-1+==+not+found,+else+==+index
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|120332|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>var list = [
 {prop1:"abc",prop2:"qwe"},
 {prop1:"bnmb",prop2:"yutu"},
 {prop1:"zxvz",prop2:"qwrq"}
 ];

var findProp = p =&gt; {
 var index = -1;
 $.each(list, (i, o) =&gt; {
 if(o.prop2 == p) {
 index = i;
 return false; // break
 }
 });
 return index; // -1 == not found, else == index
}
</code></pre>

blocks|key|120400|text|另一种简单的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|120401|+function+getIndex(items)+{
++++++++for+(const+[index,+item]+of+items.entries())+{
++++++++++++if+(item.prop2+===+'yutu')+{
++++++++++++++++return+index;
++++++++++++}
++++++++}
++++}

const+myIndex+=+getIndex(myArray);|code-block|syntax|javascript|120402|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|$]]

Another easy way is :
<pre><code> function getIndex(items) {
 for (const [index, item] of items.entries()) {
 if (item.prop2 === 'yutu') {
 return index;
 }
 }
 }

const myIndex = getIndex(myArray);
</code></pre>

I have an array like this:

<pre><code>[{prop1:"abc",prop2:"qwe"},{prop1:"bnmb",prop2:"yutu"},{prop1:"zxvz",prop2:"qwrq"},...]
</code></pre>

How can I get the index of the object that matches a condition, without iterating over the entire array?

For instance, given <code>prop2=="yutu"</code>, I want to get index <code>1</code>.

I saw <code>.indexOf()</code> but think it's used for simple arrays like <code>["a1","a2",...]</code>. I also checked <code>$.grep()</code> but this returns objects, not the index.

Get the index of the object inside an array, matching a condition

jQuery

我有一个这样的数组：[{prop1:"abc",prop2:"qwe"},{prop1:"bnmb",prop2:"yutu"},{prop1:"zxvz",prop2:"qwrq"},...]如何在不遍历整个数组的情况下获得匹配条件的对象的索引？例如，在给定prop2=="yutu"的情况下，我希望获得索引1。我看到过.indexOf()，但我认为它用于像["a1","a2",...]这样的简单

问获取数组中符合条件的对象的索引
EN

回答 12

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问获取数组中符合条件的对象的索引EN

回答 12

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问获取数组中符合条件的对象的索引
EN