blocks|key|1401396|text|我不认为有一个很好的"jQuery“方法可以做到这一点，但是如果你需要效率，可以通过某个键(唯一的对象字段之一)映射其中一个数组，然后通过循环遍历另一个数组并与您刚刚构建的映射或关联数组进行比较来进行比较。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1401397|如果效率不是问题，只需将A中的每个对象与B中的每个对象进行比较。只要%7CA%7C和%7CB%7C很小，就应该没问题。|1401398|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

I don't think there's a good "jQuery " way to do this, but if you need efficiency, map one of the arrays by a certain key (one of the unique object fields), and then do comparison by looping through the other array and comparing against the map, or associative array, you just built.

If efficiency is not an issue, just compare every object in A to every object in B. As long as |A| and |B| are small, you should be okay.

blocks|key|1401442|text|如果只想比较数组的内容，有一个有用的jQuery函数$.inArray()|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1401443|var+arr+=+[11,+"String+#1",+14,+"String+#2"];
var+arr_true+=+["String+#1",+14,+"String+#2",+11];+//+contents+are+the+same+as+arr
var+arr_false+=+["String+#1",+14,+"String+#2",+16];+//+contents+differ

function+test(arr_1,+arr_2)+{
++++var+equal+=+arr_1.length+==+arr_2.length;+//+if+array+sizes+mismatches,+then+we+assume,+that+they+are+not+equal
++++if+(equal)+{
++++++++$.each(arr_1,+function+(foo,+val)+{
++++++++++++if+(!equal)+return+false;
++++++++++++if+($.inArray(val,+arr_2)+==+-1)+{
++++++++++++++++equal+=+false;
++++++++++++}+else+{
++++++++++++++++equal+=+true;
++++++++++++}
++++++++});
++++}
++++return+equal;
}

alert('Array+contents+are+the+same?+'+%2B+test(arr,+arr_true));+//-+returns+true
alert('Array+contents+are+the+same?+'+%2B+test(arr,+arr_false));+//-+returns+false|code-block|syntax|javascript|1401444|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.jquery.com/Utilities/jQuery.inArray^0|Q|B|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]]]]

Well, if you want to compare only the contents of arrays, there's a useful jQuery function <a href="http://docs.jquery.com/Utilities/jQuery.inArray" rel="nofollow noreferrer">$.inArray()</a>

<pre><code>var arr = [11, "String #1", 14, "String #2"];
var arr_true = ["String #1", 14, "String #2", 11]; // contents are the same as arr
var arr_false = ["String #1", 14, "String #2", 16]; // contents differ

function test(arr_1, arr_2) {
 var equal = arr_1.length == arr_2.length; // if array sizes mismatches, then we assume, that they are not equal
 if (equal) {
 $.each(arr_1, function (foo, val) {
 if (!equal) return false;
 if ($.inArray(val, arr_2) == -1) {
 equal = false;
 } else {
 equal = true;
 }
 });
 }
 return equal;
}

alert('Array contents are the same? ' + test(arr, arr_true)); //- returns true
alert('Array contents are the same? ' + test(arr, arr_false)); //- returns false
</code></pre>

blocks|key|1401515|text|我找到这个讨论是因为我需要一种深入比较数组和对象的方法。使用这里的例子，我想出了以下几个方法(为清楚起见，分为3个方法)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1401516|jQuery.extend({
++++compare+:+function+(a,b)+{
++++++++var+obj_str+=+'[object+Object]',
++++++++++++arr_str+=+'[object+Array]',
++++++++++++a_type++=+Object.prototype.toString.apply(a),
++++++++++++b_type++=+Object.prototype.toString.apply(b);

++++++++++++if+(+a_type+!==+b_type)+{+return+false;+}
++++++++++++else+if+(a_type+===+obj_str)+{
++++++++++++++++return+$.compareObject(a,b);
++++++++++++}
++++++++++++else+if+(a_type+===+arr_str)+{
++++++++++++++++return+$.compareArray(a,b);
++++++++++++}
++++++++++++return+(a+===+b);
++++++++}
});

jQuery.extend({
++++compareArray:+function+(arrayA,+arrayB)+{
++++++++var+a,b,i,a_type,b_type;
++++++++//+References+to+each+other?
++++++++if+(arrayA+===+arrayB)+{+return+true;}

++++++++if+(arrayA.length+!=+arrayB.length)+{+return+false;+}
++++++++//+sort+modifies+original+array
++++++++//+(which+are+passed+by+reference+to+our+method!)
++++++++//+so+clone+the+arrays+before+sorting
++++++++a+=+jQuery.extend(true,+[],+arrayA);
++++++++b+=+jQuery.extend(true,+[],+arrayB);
++++++++a.sort();+
++++++++b.sort();
++++++++for+(i+=+0,+l+=+a.length;+i+<+l;+i%2B=1)+{
++++++++++++a_type+=+Object.prototype.toString.apply(a[i]);
++++++++++++b_type+=+Object.prototype.toString.apply(b[i]);

++++++++++++if+(a_type+!==+b_type)+{
++++++++++++++++return+false;
++++++++++++}

++++++++++++if+($.compare(a[i],b[i])+===+false)+{
++++++++++++++++return+false;
++++++++++++}
++++++++}
++++++++return+true;
++++}
});

jQuery.extend({
++++compareObject+:+function(objA,objB)+{

++++++++var+i,a_type,b_type;

++++++++//+Compare+if+they+are+references+to+each+other+
++++++++if+(objA+===+objB)+{+return+true;}

++++++++if+(Object.keys(objA).length+!==+Object.keys(objB).length)+{+return+false;}
++++++++for+(i+in+objA)+{
++++++++++++if+(objA.hasOwnProperty(i))+{
++++++++++++++++if+(typeof+objB[i]+===+'undefined')+{
++++++++++++++++++++return+false;
++++++++++++++++}
++++++++++++++++else+{
++++++++++++++++++++a_type+=+Object.prototype.toString.apply(objA[i]);
++++++++++++++++++++b_type+=+Object.prototype.toString.apply(objB[i]);

++++++++++++++++++++if+(a_type+!==+b_type)+{
++++++++++++++++++++++++return+false;+
++++++++++++++++++++}
++++++++++++++++}
++++++++++++}
++++++++++++if+($.compare(objA[i],objB[i])+===+false){
++++++++++++++++return+false;
++++++++++++}
++++++++}
++++++++return+true;
++++}
});|code-block|syntax|javascript|1401517|测试|1401518|var+a={a+:+{a+:+1,+b:+2}},
++++b={a+:+{a+:+1,+b:+2}},
++++c={a+:+{a+:+1,+b:+3}},
++++d=[1,2,3],
++++e=[2,1,3];

console.debug('a+and+b+=+'+%2B+$.compare(a,b));+//+a+and+b+=+true
console.debug('b+and+b+=+'+%2B+$.compare(b,b));+//+b+and+b+=+true
console.debug('b+and+c+=+'+%2B+$.compare(b,c));+//+b+and+c+=+false
console.debug('c+and+d+=+'+%2B+$.compare(c,d));+//+c+and+d+=+false
console.debug('d+and+e+=+'+%2B+$.compare(d,e));+//+d+and+e+=+true|1401519|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I found this discussion because I needed a way to deep compare arrays and objects. Using the examples here, I came up with the following (broken up into 3 methods for clarity):

<pre><code>jQuery.extend({
 compare : function (a,b) {
 var obj_str = '[object Object]',
 arr_str = '[object Array]',
 a_type = Object.prototype.toString.apply(a),
 b_type = Object.prototype.toString.apply(b);

 if ( a_type !== b_type) { return false; }
 else if (a_type === obj_str) {
 return $.compareObject(a,b);
 }
 else if (a_type === arr_str) {
 return $.compareArray(a,b);
 }
 return (a === b);
 }
});

jQuery.extend({
 compareArray: function (arrayA, arrayB) {
 var a,b,i,a_type,b_type;
 // References to each other?
 if (arrayA === arrayB) { return true;}

 if (arrayA.length != arrayB.length) { return false; }
 // sort modifies original array
 // (which are passed by reference to our method!)
 // so clone the arrays before sorting
 a = jQuery.extend(true, [], arrayA);
 b = jQuery.extend(true, [], arrayB);
 a.sort(); 
 b.sort();
 for (i = 0, l = a.length; i &lt; l; i+=1) {
 a_type = Object.prototype.toString.apply(a[i]);
 b_type = Object.prototype.toString.apply(b[i]);

 if (a_type !== b_type) {
 return false;
 }

 if ($.compare(a[i],b[i]) === false) {
 return false;
 }
 }
 return true;
 }
});

jQuery.extend({
 compareObject : function(objA,objB) {

 var i,a_type,b_type;

 // Compare if they are references to each other 
 if (objA === objB) { return true;}

 if (Object.keys(objA).length !== Object.keys(objB).length) { return false;}
 for (i in objA) {
 if (objA.hasOwnProperty(i)) {
 if (typeof objB[i] === 'undefined') {
 return false;
 }
 else {
 a_type = Object.prototype.toString.apply(objA[i]);
 b_type = Object.prototype.toString.apply(objB[i]);

 if (a_type !== b_type) {
 return false; 
 }
 }
 }
 if ($.compare(objA[i],objB[i]) === false){
 return false;
 }
 }
 return true;
 }
});
</code></pre>

Testing

<pre><code>var a={a : {a : 1, b: 2}},
 b={a : {a : 1, b: 2}},
 c={a : {a : 1, b: 3}},
 d=[1,2,3],
 e=[2,1,3];

console.debug('a and b = ' + $.compare(a,b)); // a and b = true
console.debug('b and b = ' + $.compare(b,b)); // b and b = true
console.debug('b and c = ' + $.compare(b,c)); // b and c = false
console.debug('c and d = ' + $.compare(c,d)); // c and d = false
console.debug('d and e = ' + $.compare(d,e)); // d and e = true
</code></pre>

blocks|key|1186168|text|有一个简单的方法..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1186169|$(arr1).not(arr2).length+===+0+&&+$(arr2).not(arr1).length+===+0|code-block|syntax|javascript|1186170|如果以上返回true，则即使元素的顺序不同，两个数组也是相同的。|1186171|注意:当使用JSON对象时，这只适用于低于3.0.0版的jquery|1186172|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

There is an easy way...

<pre><code>$(arr1).not(arr2).length === 0 &amp;&amp; $(arr2).not(arr1).length === 0
</code></pre>

If the above returns true, both the arrays are same even if the elements are in different order. 

NOTE: This works only for jquery versions &lt; 3.0.0 when using JSON objects

blocks|key|1186195|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1186196|function+check(c,d){
++var+a+=+c,+b+=+d,flg+=+0;
++if(a.length+==+b.length)+
++{+
+++++for(var+i=0;i<a.length;i%2B%2B)+
+++++++++++a[i]+!=+b[i]+?+flg%2B%2B+:+0;+
++}+
++else++
++{+
+++++flg+=+1;+
++}+
++return+flg+=+0;
}|code-block|syntax|javascript|1186197|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 

<pre><code>function check(c,d){
 var a = c, b = d,flg = 0;
 if(a.length == b.length) 
 { 
 for(var i=0;i&lt;a.length;i++) 
 a[i] != b[i] ? flg++ : 0; 
 } 
 else 
 { 
 flg = 1; 
 } 
 return flg = 0;
}
</code></pre>

blocks|key|1401616|text|作为jQuery全局方法的Sudhakar+R中的一个很好的线条。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1401617|/**
+*+Compare+two+arrays+if+they+are+equal+even+if+they+have+different+order.
+*
+*+@link+https://stackoverflow.com/a/7726509
+*/
jQuery.extend({
++/**
+++*+@param+{array}+a
+++*+++First+array+to+compare.
+++*+@param+{array}+b
+++*+++Second+array+to+compare.
+++*+@return+{boolean}
+++*+++True+if+both+arrays+are+equal,+otherwise+false.
+++*/
++arrayCompare:+function+(a,+b)+{
++++return+$(a).not(b).get().length+===+0+&&+$(b).not(a).get().length+===+0;
++}
});|code-block|syntax|javascript|1401618|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/7726509/1251219^0|D|A|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]]]]

The nice one liner from <a href="https://stackoverflow.com/a/7726509/1251219">Sudhakar R</a> as jQuery global method.

<pre><code>/**
 * Compare two arrays if they are equal even if they have different order.
 *
 * @link https://stackoverflow.com/a/7726509
 */
jQuery.extend({
 /**
 * @param {array} a
 * First array to compare.
 * @param {array} b
 * Second array to compare.
 * @return {boolean}
 * True if both arrays are equal, otherwise false.
 */
 arrayCompare: function (a, b) {
 return $(a).not(b).get().length === 0 &amp;&amp; $(b).not(a).get().length === 0;
 }
});
</code></pre>

blocks|key|1401691|text|我的方法完全不同--我使用JSON.stringify将两个集合拉平，并使用普通的字符串比较来检查是否相等。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1401692|也就是说。|1401693|var+arr1+=+[
+++++++++++++{Col:+'a',+Val:+1},+
+++++++++++++{Col:+'b',+Val:+2},+
+++++++++++++{Col:+'c',+Val:+3}
+++++++++++];

var+arr2+=+[
+++++++++++++{Col:+'x',+Val:+24},+
+++++++++++++{Col:+'y',+Val:+25},+
+++++++++++++{Col:+'z',+Val:+26}
+++++++++++];

if(JSON.stringify(arr1)+==+JSON.stringify(arr2)){
++++alert('Collections+are+equal');
}else{
++++alert('Collections+are+not+equal');
}|code-block|syntax|javascript|1401694|注意:请注意，他的方法假设两个集合都以类似的方式排序，如果不是这样，它将给你一个错误的结果！|1401695|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|$]]

My approach was quite different - I flattened out both collections using JSON.stringify and used a normal string compare to check for equality.

I.e.

<pre><code>var arr1 = [
 {Col: 'a', Val: 1}, 
 {Col: 'b', Val: 2}, 
 {Col: 'c', Val: 3}
 ];

var arr2 = [
 {Col: 'x', Val: 24}, 
 {Col: 'y', Val: 25}, 
 {Col: 'z', Val: 26}
 ];

if(JSON.stringify(arr1) == JSON.stringify(arr2)){
 alert('Collections are equal');
}else{
 alert('Collections are not equal');
}
</code></pre>

NB: Please note that his method assumes that both Collections are sorted in a similar fashion, if not, it would give you a false result!

blocks|key|1186372|text|在使用jQuery进行数组比较时，我也发现了这一点。在我的例子中，我知道字符串是数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1186373|var+needle+=+'apple+orange';
var+haystack+=+'kiwi+orange+banana+apple+plum';|code-block|syntax|javascript|1186374|但我关心它是完全匹配还是只是部分匹配，所以我根据Sudhakar+R的答案使用了类似以下内容：|1186375|function+compareStrings(+needle,+haystack+){
++var+needleArr+=+needle.split("+"),
++++haystackArr+=+haystack.split("+"),
++++compare+=+$(haystackArr).not(needleArr).get().length;

++if(+compare+==+0+){
++++return+'all';
++}+else+if+(+compare+==+haystackArr.length++)+{
++++return+'none';
++}+else+{
++++return+'partial';
++}
}|1186376|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I also found this when looking to do some array comparisons with jQuery. In my case I had strings which I knew to be arrays:

<pre><code>var needle = 'apple orange';
var haystack = 'kiwi orange banana apple plum';
</code></pre>

But I cared if it was a complete match or only a partial match, so I used something like the following, based off of Sudhakar R's answer:

<pre><code>function compareStrings( needle, haystack ){
 var needleArr = needle.split(" "),
 haystackArr = haystack.split(" "),
 compare = $(haystackArr).not(needleArr).get().length;

 if( compare == 0 ){
 return 'all';
 } else if ( compare == haystackArr.length ) {
 return 'none';
 } else {
 return 'partial';
 }
}
</code></pre>

blocks|key|1186416|text|将数组更改为字符串并进行比较|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1186417|var+arr+=+[1,2,3],+
arr2+=+[1,2,3];+
console.log(arr.toString()+===+arr2.toString());|code-block|syntax|javascript|1186418|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|$]]

Change array to string and compare

<pre><code>var arr = [1,2,3], 
arr2 = [1,2,3]; 
console.log(arr.toString() === arr2.toString());
</code></pre>

blocks|key|1401837|text|将两个数组都转换为字符串并进行比较|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1401838|if+(JSON.stringify(array1)+==+JSON.stringify(array2))
{
++++//+your+code+here
}|code-block|syntax|javascript|1401839|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|$]]

Convert both array to string and compare

<pre><code>if (JSON.stringify(array1) == JSON.stringify(array2))
{
 // your code here
}
</code></pre>

blocks|key|1401905|text|如果重复项很重要，那么[1,+1,+2]不应该等于[2,+1]，而应该等于[1,+2,+1]，下面是一个引用计数解决方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1401906|++const+arrayContentsEqual+=+(arrayA,+arrayB)+=>+{
++++if+(arrayA.length+!==+arrayB.length)+{
++++++return+false}

++++const+refCount+=+(function()+{
++++++const+refCountMap+=+{};
++++++const+refCountFn+=+(elt,+count)+=>+{
++++++++++refCountMap[elt]+=+(refCountMap[elt]+%7C%7C+0)+%2B+count}
++++++refCountFn.isZero+=+()+=>+{
++++++++for+(let+elt+in+refCountMap)+{
++++++++++if+(refCountMap[elt]+!==+0)+{
++++++++++++return+false}}
++++++++return+true}
++++++return+refCountFn})()

++++arrayB.map(eltB+=>+refCount(eltB,+1));
++++arrayA.map(eltA+=>+refCount(eltA,+-1));
++++return+refCount.isZero()}|code-block|syntax|javascript|1401907|Here+is+the+fiddle+to+play+with。|1401908|entityMap|0|LINK|mutability|MUTABLE|url|https://jsfiddle.net/whippy/9t30zutw/^0|B|9|P|6|11|9|0|0|0|V|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|D|@$9|13|A|14|1|15]]|E|$]]|$1|M|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

If duplicates matter such that <code>[1, 1, 2]</code> should not be equal to <code>[2, 1]</code> but should equal <code>[1, 2, 1]</code>, here is a reference counting solution:

<pre><code> const arrayContentsEqual = (arrayA, arrayB) =&gt; {
 if (arrayA.length !== arrayB.length) {
 return false}

 const refCount = (function() {
 const refCountMap = {};
 const refCountFn = (elt, count) =&gt; {
 refCountMap[elt] = (refCountMap[elt] || 0) + count}
 refCountFn.isZero = () =&gt; {
 for (let elt in refCountMap) {
 if (refCountMap[elt] !== 0) {
 return false}}
 return true}
 return refCountFn})()

 arrayB.map(eltB =&gt; refCount(eltB, 1));
 arrayA.map(eltA =&gt; refCount(eltA, -1));
 return refCount.isZero()}
</code></pre>

<a href="https://jsfiddle.net/whippy/9t30zutw/" rel="nofollow noreferrer">Here is the fiddle to play with</a>.

blocks|key|1186563|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1186564|var+arr1+=+[
+++++++++++++{name:+'a',+Val:+1},+
+++++++++++++{name:+'b',+Val:+2},+
+++++++++++++{name:+'c',+Val:+3}
+++++++++++];

var+arr2+=+[
+++++++++++++{name:+'c',+Val:+3},
+++++++++++++{name:+'x',+Val:+4},+
+++++++++++++{name:+'y',+Val:+5},+
+++++++++++++{name:+'z',+Val:+6}
+++++++++++];
var+_isEqual+=+_.intersectionWith(arr1,+arr2,+_.isEqual);//+common+in+both+array
var+_difference1+=+_.differenceWith(arr1,+arr2,+_.isEqual);//difference+from+array1+
var+_difference2+=+_.differenceWith(arr2,+arr1,+_.isEqual);//difference+from+array2+
console.log(_isEqual);//+common+in+both+array
console.log(_difference1);//difference+from+array1+
console.log(_difference2);//difference+from+array2+|code-block|syntax|javascript|1186565|<script+src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.js"></script>|1186566|1186567|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|M|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|I|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

<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 arr1 = [
 {name: 'a', Val: 1}, 
 {name: 'b', Val: 2}, 
 {name: 'c', Val: 3}
 ];

var arr2 = [
 {name: 'c', Val: 3},
 {name: 'x', Val: 4}, 
 {name: 'y', Val: 5}, 
 {name: 'z', Val: 6}
 ];
var _isEqual = _.intersectionWith(arr1, arr2, _.isEqual);// common in both array
var _difference1 = _.differenceWith(arr1, arr2, _.isEqual);//difference from array1 
var _difference2 = _.differenceWith(arr2, arr1, _.isEqual);//difference from array2 
console.log(_isEqual);// common in both array
console.log(_difference1);//difference from array1 
console.log(_difference2);//difference from array2 </code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.js"&gt;&lt;/script&gt;</code></pre>
</div>
</div>

I have two arrays of JavaScript Objects that I'd like to compare to see if they are the same. The objects may not (and most likely will not) be in the same order in each array. Each array shouldn't have any more than 10 objects. I thought jQuery might have an elegant solution to this problem, but I wasn't able to find much online.

I know that a brute nested <code>$.each(array, function(){})</code> solution could work, but is there any built in function that I'm not aware of?

Thanks.

Using jQuery to compare two arrays of Javascript objects

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我有两个JavaScript对象数组，我想对它们进行比较，看看它们是否相同。对象在每个数组中的顺序可能不同(并且很可能不会)。每个数组不应该有超过10个对象。我以为jQuery可能有一个很好的解决方案来解决这个问题，但我在网上找不到太多。我知道粗暴的嵌套$.each(array, function(){})解决方案可以...

问使用jQuery比较两个Javascript对象数组
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