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|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|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|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|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|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|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|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|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>

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>

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...

问获取数组中符合条件的对象的索引
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