javascript - For 循环无法正确访问 JSON 对象？

Question

由于某种原因，当我完成测验时，没有访问正确的 JSON 对象属性。一旦字符串被识别，它就应该访问该属性。如果我硬编码它。 [0]、[1] 等有效。我怎样才能解决这个问题？另外，如何使这段代码更简洁？我觉得除了使用这么多 if 语句之外，可能还有另一种方法。谢谢。

有问题的对象

var personTypes = [{ 
        type : "INTJ",
        typeInfo: "Imaginative and strategic thinkers, with a plan for everything.",

        },
        {type : "INTP",
        typeInfo: "Innovative inventors with an unquenchable thirst for knowledge.",

        },
        {type : "ENTJ",
        typeInfo: "Bold, imaginative and strong-willed leaders, always finding a way – or making one.",

        },
        {type : "ENTP",
        typeInfo: "Smart and curious thinkers who cannot resist an intellectual challenge.",

        },
        {type : "INFJ",
        typeInfo: "Quiet and mystical, yet very inspiring and tireless idealists",

        },
        {type : "INFP",
        typeInfo: "Poetic, kind and altruistic people, always eager to help a good cause.",

        },
        {type : "ENFJ",
        typeInfo: "Charismatic and inspiring leaders, able to mesmerize their listeners.",

        },
        {type : "ENFP",
        typeInfo: "Enthusiastic, creative and sociable free spirits, who can always find a reason to smile.",

        },
        {type : "ISTJ",
        typeInfo: "Practical and fact-minded individuals, whose reliability cannot be doubted.",

        },
        {type : "ISFJ",
        typeInfo: "Very dedicated and warm protectors, always ready to defend their loved ones.",

        },
        {type : "ESTJ",
        typeInfo: "Excellent administrators, unsurpassed at managing things – or people.",

        },
        {type : "ESFJ",
        typeInfo: "Extraordinarily caring, social and popular people, always eager to help.",

        },
        {type : "ISTP",
        typeInfo: "Bold and practical experimenters, masters of all kinds of tools.",

        },
        {type : "ISFP",
        typeInfo: "Flexible and charming artists, always ready to explore and experience something new.",

        },
        {type : "ESTP",
        typeInfo: "Smart, energetic and very perceptive people, who truly enjoy living on the edge",

        },
        {type : "ESFP",
        typeInfo: "Spontaneous, energetic and enthusiastic people – life is never boring around them.",

        },
];

访问对象

 // once done with quiz
   if (questionNum === 3) {

        //concat radio inputs
        var typeConcat = this.ei + this.sn+ this.tf+ this.pj;
        //output concat to screen
        $("p").show();
        $("h2").text("Your type is " + typeConcat);



            //use this data inside the <p>
        for (i = 0; i < personTypes.length; i++) {    
            if (typeConcat=="INTJ") {
                $("p").text(personTypes[i].typeInfo);
            }
            if (typeConcat=="INTP") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="ENTJ") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="ENTP") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="INFJ") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="INFP") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="ENFJ") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="ENFP") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="ISTJ") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="ISFJ") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="ESTJ") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="ESFJ") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="ISTP") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="ISFP") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="ESTP") {
                $("p").text(personTypes[i].typeInfo);
            }
             if (typeConcat=="ESFP") {
                $("p").text(personTypes[i].typeInfo);
            }
        }
    }

Answer 1

我认为当您指定了类型时，您不需要所有的 if 语句，对吧？这会起到同样的作用。

编辑:不要忘记在 i = 0 之前包含 var。

for (var i = 0; i < personTypes.length; i++) {    
    if (typeConcat==personTypes[i].type) {
        $("p").text(personTypes[i].typeInfo);
    }
}