Question

如何获取嵌套在JSON值中的数组的值以及嵌套在数组中的JSON？

var contacts = [
        {
            "firstName": "Akira",
            "lastName": "Laine",
            "number": "0543236543",
            "likes": ["Pizza", "Coding", "Brownie Points"]
        },
        {
            "firstName": "Harry",
            "lastName": "Potter",
            "number": "0994372684",
            "likes": ["Hogwarts", "Magic", "Hagrid"]
        },
        {
            "firstName": "Sherlock",
            "lastName": "Holmes",
            "number": "0487345643",
            "likes": ["Intriguing Cases", "Violin"]
        },
        {
            "firstName": "Kristian",
            "lastName": "Vos",
            "number": "unknown",
            "likes": ["Javascript", "Gaming", "Foxes"]
        }
    ];

/ ** /

如果我想获得喜欢＆amp;的价值“firstName”：“Akira”，我该怎么办？有人可以帮忙吗？

       var firstName = "Akira", prop = "likes";
       for (var i = 0; i < contacts.length; i++) {
            var a;
            console.log("contacts[i].firstName = "+contacts[i].firstName);
            if(contacts[i].firstName == firstName){
                a = true;
            }else{
                a = false;
            }
         //get the cales of likes?
         console.log("contacts[i].prop[i].value() = "+contacts[i].prop[i].value() );
         //contacts[i].prop[i].value();  ?
        }

1

Answer 1

contacts数组中的对象没有名称为prop的属性。幸运的是，您可以通过名称访问对象的属性，类似于从数组中获取值的方式，因此如果使用contacts[i][prop]，它将获得属性的值，该属性的名称是变量的值prop，在本例中为"likes"。由于likes也是一个数组，因此您可以通过索引与contacts数组索引相同的方式从中获取值。

var firstName = "Akira", prop = "likes";
for (var i = 0; i < contacts.length; i++) {
    var a;
    console.log("contacts[" + i + "].firstName = " + contacts[i].firstName);
    if (contacts[i].firstName == firstName){
        a = true;
    } else {
        a = false;
    }

    // get the likes array and loop through to print the values
    // you might want to put this where a = true; is if you only want to get 
    // the likes where the firstName matches
    var likes = contacts[i][prop];
    for (var j = 0; j < likes.length; j++) {
        console.log("contacts[" + i + "]." + prop + "[" + j + "] = " + likes[j]);
    }
}

Answer 2

Yoo可能如下所示;

＆＃13;

var contacts = [
        {
            "firstName": "Akira",
            "lastName": "Laine",
            "number": "0543236543",
            "likes": ["Pizza", "Coding", "Brownie Points"]
        },
        {
            "firstName": "Harry",
            "lastName": "Potter",
            "number": "0994372684",
            "likes": ["Hogwarts", "Magic", "Hagrid"]
        },
        {
            "firstName": "Sherlock",
            "lastName": "Holmes",
            "number": "0487345643",
            "likes": ["Intriguing Cases", "Violin"]
        },
        {
            "firstName": "Kristian",
            "lastName": "Vos",
            "number": "unknown",
            "likes": ["Javascript", "Gaming", "Foxes"]
        }
    ],
  getMeObj = (d,p,v,t) => d.filter(o => o[p] === v)[0][t];
  console.log(getMeObj(contacts,"firstName", "Akira", "likes"));

＆＃13;

Answer 3

以下代码可以使用

    var firstName = "Akira", prop = "likes";
    for(var i in contacts)
    {
      if(contacts[i].firstName == firstName){
      var likes = contacts[i][prop];
      console.log(likes);
    }

Answer 4

我修复了你的代码。如果你需要更多细节问我，但我认为代码很简单

var contacts = [{
  "firstName": "Akira",
  "lastName": "Laine",
  "number": "0543236543",
  "likes": ["Pizza", "Coding", "Brownie Points"]
}, {
  "firstName": "Harry",
  "lastName": "Potter",
  "number": "0994372684",
  "likes": ["Hogwarts", "Magic", "Hagrid"]
}, {
  "firstName": "Sherlock",
  "lastName": "Holmes",
  "number": "0487345643",
  "likes": ["Intriguing Cases", "Violin"]
}, {
  "firstName": "Kristian",
  "lastName": "Vos",
  "number": "unknown",
  "likes": ["Javascript", "Gaming", "Foxes"]
}];
var firstName = "Akira"
  , prop = "likes";
for (var i = 0; i < contacts.length; i++) {
  var a;
  console.log("contacts[" + i + "].firstName = " + contacts[i].firstName);
  if (contacts[i].firstName == firstName) {
    a = true;
  } else {
    a = false;
  }
  for (var n = 0; n < contacts[i].likes.length; n++) {
    console.log("contacts[" + i + "].likes[" + n + "] = " + contacts[i].likes[n]);
  }
}

作为旁注：答案应与提问者处于同一级别以帮助学习：）

如何获取嵌套在JSON值中的Array的值以及嵌套在数组中的JSON？

4 个答案: