blocks|key|1339444|text|应该很简单。假设arr1是带有价格的数组，而arr2是具有名称的数组：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1339445|var+newarr+=+[],
++++len+=+(number+of+items);

for(var+i=0;i<len;i%2B%2B){
++++newarr.push({
++++++++Price:arr1[i];
++++++++Name:arr2[i];
++++});
}|code-block|syntax|javascript|1339446|entityMap^0|8|4|M|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Should be pretty straightforward. Assuming <code>arr1</code> is your array with prices, and <code>arr2</code> is your array with names:

<pre><code>var newarr = [],
 len = (number of items);

for(var i=0;i&lt;len;i++){
 newarr.push({
 Price:arr1[i];
 Name:arr2[i];
 });
}
</code></pre>

blocks|key|1339448|text|在underscore.js中使用zip函数的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1339449|var+prices+=+["EUR+1,99",+"EUR+0,99",+"EUR+4,99",+"EUR+2,29",+"EUR+1,43",+"EUR+1,60",+"EUR+1,79",+"EUR+1,79",+"EUR+1,39",+"EUR+6,30",+"EUR+1,43",+"EUR+1,78",+"EUR+1,90",+"EUR+1,24",+"EUR+1,41"];

var+products+=+["Popken",+"Lucky+Animal",+"mein-terrarium",+"zooup",+"zoofair",+"XL-Hundeshop",+"tiierisch-de",+"Zoo+Galaxie",+"Petshop",+"Danto+GmbH",+"Heimtierbedarf-Mazinke",+"TIERKOSMOS",+"Gazoma",+"Zooheld",+"dasko24"];

var+result+=+_.map(_.zip(products,+prices),+function(arr)+{
++++return+{
++++++++Price:+arr[1],
++++++++Name:+arr[0]
++++};
});


console.log(result);|code-block|syntax|javascript|1339450|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|$]]

Solution using the zip-function in underscore.js

<pre><code>var prices = ["EUR 1,99", "EUR 0,99", "EUR 4,99", "EUR 2,29", "EUR 1,43", "EUR 1,60", "EUR 1,79", "EUR 1,79", "EUR 1,39", "EUR 6,30", "EUR 1,43", "EUR 1,78", "EUR 1,90", "EUR 1,24", "EUR 1,41"];

var products = ["Popken", "Lucky Animal", "mein-terrarium", "zooup", "zoofair", "XL-Hundeshop", "tiierisch-de", "Zoo Galaxie", "Petshop", "Danto GmbH", "Heimtierbedarf-Mazinke", "TIERKOSMOS", "Gazoma", "Zooheld", "dasko24"];

var result = _.map(_.zip(products, prices), function(arr) {
 return {
 Price: arr[1],
 Name: arr[0]
 };
});


console.log(result);
</code></pre>

blocks|key|1959766|text|这是一个通用的解决方案--它将“折叠”一个对象，这样您就可以得到一个对象列表，而不是一个列表对象。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1959767|function+refold_list_from_object(data)+{
++++"use+strict"++++
++++var+keys+=+Object.keys(data),
++++++++descriptor+=+{configurable:+true,+enumerable:+true,+writable:+true,+value:+null},
++++++++properties+=+keys.reduce(function(p,+k){
++++++++++++p[k]+=+descriptor
++++++++++++return+p
++++++++},+{}),
++++++++nrecords+=+Math.max.apply(null,+keys.map(function(k){return+data[k].length})),
++++++++records++=+[],+vi
++++function+makerecord()+{
++++++++return+Object.create(null,+properties)
++++}
++++function+record_from_value_index(vi)+{
++++++++return+keys.reduce(function(obj,+k)+{
++++++++++++obj[k]+=+data[k][vi]+%7C%7C+null
++++++++++++return+obj
++++++++},+makerecord())
++++}
++++for+(vi+=+0;+vi+<+nrecords;+vi%2B%2B)+{
++++++++records.push(record_from_value_index(vi))
++++}
++++return+records
}|code-block|syntax|javascript|1959768|data应该是一个具有按字段名索引的列表的对象：|offset|length|style|CODE|1959769|var+data+=+{
++++Price:+["EUR+1,99",+"EUR+0,99",+"EUR+4,99",+"EUR+2,29",+"EUR+1,43",+"EUR+1,60",+"EUR+1,79",+"EUR+1,79",+"EUR+1,39",+"EUR+6,30",+"EUR+1,43",+"EUR+1,78",+"EUR+1,90",+"EUR+1,24",+"EUR+1,41"],
++++Name:+["Popken",+"Lucky+Animal",+"mein-terrarium",+"zooup",+"zoofair",+"XL-Hundeshop",+"tiierisch-de",+"Zoo+Galaxie",+"Petshop",+"Danto+GmbH",+"Heimtierbedarf-Mazinke",+"TIERKOSMOS",+"Gazoma",+"Zooheld",+"dasko24"],
++++//+An+extra+field+to+demonstrate+padding:
++++Description:+['Foo',+'Bar',+'Baz']
}

var+recordlist+=+refold_list_from_object(data)|1959770|recordlist将如下所示：|1959771|[
++{"Price":"EUR+1,99",+"Name":"Popken",+++++++++"Description":"Foo"},
++{"Price":"EUR+0,99",+"Name":"Lucky+Animal",+++"Description":"Bar"},
++{"Price":"EUR+4,99",+"Name":"mein-terrarium",+"Description":"Baz"},
++{"Price":"EUR+2,29",+"Name":"zooup",++++++++++"Description":null},
++{"Price":"EUR+1,43",+"Name":"zoofair",++++++++"Description":null},
++{"Price":"EUR+1,60",+"Name":"XL-Hundeshop",+++"Description":null},
++{"Price":"EUR+1,79",+"Name":"tiierisch-de",+++"Description":null},
++{"Price":"EUR+1,79",+"Name":"Zoo+Galaxie",++++"Description":null},
++{"Price":"EUR+1,39",+"Name":"Petshop",++++++++"Description":null},
++{"Price":"EUR+6,30",+"Name":"Danto+GmbH",+++++"Description":null},
++{"Price":"EUR+1,43",+"Name":"Heimtierbedarf-Mazinke",+"Description":null},
++{"Price":"EUR+1,78",+"Name":"TIERKOSMOS",+++++"Description":null},
++{"Price":"EUR+1,90",+"Name":"Gazoma",+++++++++"Description":null},
++{"Price":"EUR+1,24",+"Name":"Zooheld",++++++++"Description":null},
++{"Price":"EUR+1,41",+"Name":"dasko24",++++++++"Description":null}
]|1959772|entityMap^0|0|0|0|4|0|0|0|A|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|$]]

Here is a generic solution--it will "fold" an object so that you get a list of objects instead of an object of lists.



<pre><code>function refold_list_from_object(data) {
 "use strict" 
 var keys = Object.keys(data),
 descriptor = {configurable: true, enumerable: true, writable: true, value: null},
 properties = keys.reduce(function(p, k){
 p[k] = descriptor
 return p
 }, {}),
 nrecords = Math.max.apply(null, keys.map(function(k){return data[k].length})),
 records = [], vi
 function makerecord() {
 return Object.create(null, properties)
 }
 function record_from_value_index(vi) {
 return keys.reduce(function(obj, k) {
 obj[k] = data[k][vi] || null
 return obj
 }, makerecord())
 }
 for (vi = 0; vi &lt; nrecords; vi++) {
 records.push(record_from_value_index(vi))
 }
 return records
}
</code></pre>

<code>data</code> should be an object with lists indexed by their field names:

<pre><code>var data = {
 Price: ["EUR 1,99", "EUR 0,99", "EUR 4,99", "EUR 2,29", "EUR 1,43", "EUR 1,60", "EUR 1,79", "EUR 1,79", "EUR 1,39", "EUR 6,30", "EUR 1,43", "EUR 1,78", "EUR 1,90", "EUR 1,24", "EUR 1,41"],
 Name: ["Popken", "Lucky Animal", "mein-terrarium", "zooup", "zoofair", "XL-Hundeshop", "tiierisch-de", "Zoo Galaxie", "Petshop", "Danto GmbH", "Heimtierbedarf-Mazinke", "TIERKOSMOS", "Gazoma", "Zooheld", "dasko24"],
 // An extra field to demonstrate padding:
 Description: ['Foo', 'Bar', 'Baz']
}

var recordlist = refold_list_from_object(data)
</code></pre>

<code>recordlist</code> will look like this:



<pre><code>[
 {"Price":"EUR 1,99", "Name":"Popken", "Description":"Foo"},
 {"Price":"EUR 0,99", "Name":"Lucky Animal", "Description":"Bar"},
 {"Price":"EUR 4,99", "Name":"mein-terrarium", "Description":"Baz"},
 {"Price":"EUR 2,29", "Name":"zooup", "Description":null},
 {"Price":"EUR 1,43", "Name":"zoofair", "Description":null},
 {"Price":"EUR 1,60", "Name":"XL-Hundeshop", "Description":null},
 {"Price":"EUR 1,79", "Name":"tiierisch-de", "Description":null},
 {"Price":"EUR 1,79", "Name":"Zoo Galaxie", "Description":null},
 {"Price":"EUR 1,39", "Name":"Petshop", "Description":null},
 {"Price":"EUR 6,30", "Name":"Danto GmbH", "Description":null},
 {"Price":"EUR 1,43", "Name":"Heimtierbedarf-Mazinke", "Description":null},
 {"Price":"EUR 1,78", "Name":"TIERKOSMOS", "Description":null},
 {"Price":"EUR 1,90", "Name":"Gazoma", "Description":null},
 {"Price":"EUR 1,24", "Name":"Zooheld", "Description":null},
 {"Price":"EUR 1,41", "Name":"dasko24", "Description":null}
]
</code></pre>

What is the best way to merge these two objects into one and add keys as well. I tried a lot already but I am lost with how javascript works on objects and arrays... Who can guide me into the right direction? Either with plain JS or jquery.

Object 1

<pre><code>["EUR 1,99", "EUR 0,99", "EUR 4,99", "EUR 2,29", "EUR 1,43", "EUR 1,60", "EUR 1,79", "EUR 1,79", "EUR 1,39", "EUR 6,30", "EUR 1,43", "EUR 1,78", "EUR 1,90", "EUR 1,24", "EUR 1,41"]
</code></pre>

Object 2

<pre><code>["Popken", "Lucky Animal", "mein-terrarium", "zooup", "zoofair", "XL-Hundeshop", "tiierisch-de", "Zoo Galaxie", "Petshop", "Danto GmbH", "Heimtierbedarf-Mazinke", "TIERKOSMOS", "Gazoma", "Zooheld", "dasko24"]
</code></pre>

Desired object

<pre><code>[[Price="EUR 1,99",Name="Popken"],[Price="EUR 0,99",Name="Lucky Animal"],[Price="EUR 4,99",Name="mein-terrarium"], etc....] 
</code></pre>

merge two javascript objects to new key : value object

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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将这两个对象合并为一个并添加键的最佳方法是什么。我已经尝试了很多，但是我对javascript在对象和数组上的工作方式感到困惑.谁能指引我走向正确的方向？使用普通的JS或jquery。对象1["EUR 1,99", "EUR 0,99", "EUR 4,99", "EUR 2,29", "EUR 1,43", "EUR...

问将两个javascript对象合并到新的key : value对象
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问将两个javascript对象合并到新的key : value对象EN