可能重复:
按属性值对JavaScript对象进行排序
我想获得JSON中某个值的最高结果。更容易解释的例子:
var jsonData = {
bom: [
{
"Component":"Some Thing",
"Version":"Version ABC",
"License":"License ABC",
},
{
"Component":"Another Thing",
"Version":"Version XYZ",
"License":"License ABC",
},
etc ....
]
}
因此,我的目标是确定"许可证ABC"或其他许可证有X次出现,然后我希望能够将这些密钥:val对排序为"最受欢迎的X个许可证是:
- 许可证ABC-100
- 许可XYZ-70
- 许可证123-25
现在我有这个:
var countResults = function() {
var fileLicenses = [];
for ( var i = 0, arrLen = jsonData.files.length; i < arrLen; ++i ) {
fileLicenses.push(jsonData.files[i]["License"]);
}
keyCount = {};
for(i = 0; i < fileLicenses.length; ++i) {
if(!keyCount[fileLicenses[i]]) {
keyCount[fileLicenses[i]] = 0;
}
++keyCount[fileLicenses[i]];
}
console.log( keyCount );
}();
哪个得到了我想要的东西,一个带有关键字的对象:values
{
thisKey : 78,
thatKey :125,
another key : 200,
another key : 272,
another key : 45,
etc ...
}
但我不知道该怎么办。我只需要对右边的数字列进行排序,并让相关的键在旅途中保持不变。想法?非常感谢。
不能根据对象的值对其进行排序。您可以做的是将其转换为一个对象数组,然后对其进行排序。类似于:
var rank = function(items, prop) {
//declare a key->count table
var results = {}
//loop through all the items we were given to rank
for(var i=0;len=items.length;i<len;i++) {
//get the requested property value (example: License)
var value = items[i][prop];
//increment counter for this value (starting at 1)
var count = (results[value] || 0) + 1;
results[value] = count;
}
var ranked = []
//loop through all the keys in the results object
for(var key in results) {
//here we check that the results object *actually* has
//the key. because of prototypal inheritance in javascript there's
//a chance that someone has modified the Object class prototype
//with some extra properties. We don't want to include them in the
//ranking, so we check the object has it's *own* property.
if(results.hasOwnProperty(key)) {
//add an object that looks like {value:"License ABC", count: 2}
//to the output array
ranked.push({value:key, count:results[key]});
}
}
//sort by count descending
return ranked.sort(function(a, b) { return b.count - a.count; });
}
用法:
var sorted = rank(jsonData.bom, "License");
var first = sorted[0].value;
/代码未测试