Javascript:如何迭代对象[Key:value]，但跳过一些键([最快的方法])

如果给定一个脚本，如：

 <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>

我以前也这样做过，但不知怎么的，我的大脑现在一片空白。。。。任何建议都将不胜感激。（请使用jQuery运行）

作为JQuery文档状态，

返回非false与for循环中的continue语句相同；它将立即跳到下一次迭代。

因此，按照您的示例，您可以使用以下方法跳过具有不需要属性的对象：

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

（不是jquery-vanilla）迭代属性并跳过特定的键：（没有测试语法，但应该可以）

    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

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