Javascript-计算重复的JSON值并将计数与关联键一起排序？

可能重复：

我希望获得JSON中某些值的最佳结果。通过示例更容易解释：

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”或其他许可证的出现次数为X，然后我希望能够对要插入DOM的key:val对进行排序，作为“前X个最流行的许可证是：

• 许可证ABC-100
• XYZ-70许可证
• 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 );
}();

---

但我不知道该怎么做。我只需要对数字右边的列进行排序，并让相关的键保持不变。想法？谢谢！

你不能根据对象的值对其进行排序。你可以做的是将其转换为一个对象数组，然后进行排序。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; });
}

---

/代码未测试

谢谢fencliff！插入它，效果很好。仍然需要一些时间来理解逻辑，但正是我所需要的。谢谢！@Matt我在代码中添加了一些注释来解释逻辑。如果解决方案有效，不要忘了将答案标记为已接受！哇！非常感谢fencliff的注释。解决方案有效从一开始就很好，但是，理解你正在插入的代码总是更好！这很有帮助。再次感谢！@Matt，不客气。我不会使用在互联网上找到的任何代码，除非我理解：）

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; });
}

var sorted = rank(jsonData.bom, "License");
var first = sorted[0].value;