Javascript：如何迭代对象 [Key: value] 但跳过一些键（[可能的最快方法]）

Question

如果给出如下脚本：

 <html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Test DOC</title>

<script type="text/javascript" src="../nt_scr/jq_1.9.js"></script>

<script type="text/javascript">

 var mainObj = { 0: {   "fruitName":"Lemon",
                        "fruitColor":"green",    
                        "fruitTaste":"bad"},

                 1:  {  "fruitName":"Mango",
                        "fruitColor":"Yellow",    
                        "fruitTaste":"Yammy"},
                 2:   {


                        "fruitName":"Banana",
                        "fruitColor":"Yellow",    
                        "fruitTaste":"Yammy"},


                "skip_these":   

                     {
                        "team1":"Liverpool",
                        "team2":"Manchester United"

                        } 

}

var collect_data = $.parseJSON(JSON.stringify(mainObj)),
    getFruitNames=[],getFruitColors=[], getFruitTastes=[];

$.each( collect_data,function(index,val){

     //console.log(val); //Un-comment this console.log to see the contents of 'val'

    //----->Here is the Challenge:,---------\\
    if(/*How do you SKIP if val is something like */ "val.team1" || "val.team2"){

     getFruitNames.push(val.fruitName);
     getFruitColors.push(val.fruitColor);
     getFruitTastes.push(val.fruitTaste);

        //This works well so long as we have not yet reached the "skip_these":   
        //Once there, it reports an error because there is no "skip_these".fruitName or    "skip_these".fruitColor
        }

        console.log( getFruitNames[index])// Run a test out put :: NOTICE the "Undefined" in the Console. How to avoid that? 
        //To see well, Comment the previous console.log  <<the one on top>>
     })


</script>
</head>

<body>


</body>
</html>

我以前有时这样做过，但不知何故，我的大脑现在一片空白……非常感谢任何建议。（请运行加上你的jQuery）

Answer 1

（不是 jquery - 原版）迭代属性并跳过特定键：（没有测试语法但应该没问题）

    var skipped = ['keySkipped'];
    for (var someProperty in someObject) {
        if (someObject.hasOwnProperty(someProperty) && skipped.indexOf(someObject[someProperty] === -1)) {
            // do stuff
        }
    }

解释：迭代属性，如果属性确实包含在对象中但不包含在skiped中，则执行任何操作

Answer 2

作为JQuery docs状态，

Returning non-false is the same as a continue statement in a for loop; it will skip immediately to the next iteration.

因此，按照您的示例，您可以使用以下方法跳过具有您不需要的属性的对象：

JavaScript

$.each( collect_data,function(index,val){

     //console.log(val); //Un-comment this console.log to see the contents of 'val'

    //----->Here is the Challenge:,---------\\
    if(val.team1 || val.team2){
      return true;
    }

     getFruitNames.push(val.fruitName);
     getFruitColors.push(val.fruitColor);
     getFruitTastes.push(val.fruitTaste);

        //This works well so long as we have not yet reached the "skip_these":   
        //Once there, it reports an error because there is no "skip_these".fruitName or    "skip_these".fruitColor


        console.log( getFruitNames[index])//No "Undefined"
     })

https://plnkr.co/edit/9vOACIpnlWRtSWjmAm5x?p=preview

Javascript：如何迭代对象 [Key: value] 但跳过一些键（[可能的最快方法]）

Javascript: How to Iterate Object [Key: value] but skip some keys ([fastest way possible])

javascript

iteration

key

object