javascript - 返回json中数组的所有值

Question

我有一个包含所有帖子的 json 我的博客，我想返回“标签”数组的所有值。因此，请参阅下面的 json 示例：

"posts":[
  {
     "id":"89319077059",
     "title":"Post Title 01",
     "tags":[
        "Politcs",
        "Science",
     ]
  },
  {
     "id":"89318918989",
     "title":"Post Title 02",
     "tags":[
        "Football",
        "Soccer",
     ]
  },
]

所以，我需要在循环中获取唯一的标签值，例如：

for (var i = 0; i < posts.length; i++) {
    console.info("Here [i = 0] should be show the Politcs and Science and after [i = 1] show Football and Soccer");
}

我尝试创建另一个循环来搜索标签并使用标签 [1]、标签 [2] 等，但不起作用。

var tags = [];

for (var tag in posts[i].tags) {
     tags.push(tag);
}

任何想法？

Answer 1

If I understand it well, you can use

• Loop:

  var tags = [];
  for (var i = 0; i < posts.length; ++i) {
       tags.push(posts[i].tags);
  }
  

• ES5 map:

  var tags = posts.map(function(post){
      return post.tags;
  });
  

• ES5 map + ES6 arrow functions:

  var tags = posts.map(post => post.tag);
  

Answer 2

Here is code :

var posts  = [
  {
     "id":"89319077059",
     "title":"Post Title 01",
     "tags":[
        "Politcs",
        "Science",
     ]
  },
  {
     "id":"89318918989",
     "title":"Post Title 02",
     "tags":[
        "Football",
        "Soccer",
     ]
  }
];

var tags = [];

for (var index in posts) {
    var tagsArray = posts[index].tags;
     tags.push(tagsArray);
}
console.log(tags);

Jsbin

Answer 3

In your for (var tag in posts[i].tags) loop, tag contains the keys, not the values.

So a quick fix would be:

for (var tag in posts[i].tags) {
    tags.push(posts[i].tags[tag]);
}

But as Oriol pointed out, for..in loops are meant to iterate through object properties, not arrays.

Just use a for loop like your first one.

Answer 4

尝试这个，

var posts = {"posts":[
  {
     "id":"89319077059",
     "title":"Post Title 01",
     "tags":[
        "Politcs",
        "Science",
     ]
  },
  {
     "id":"89318918989",
     "title":"Post Title 02",
     "tags":[
        "Football",
        "Soccer",
     ]
  },
]};


var tags = [];

for(var i=0;i<posts.posts.length;i++) {
    for(var j=0;j<posts.posts[i].tags.length;j++) {
        tags.push(posts.posts[i].tags[j]);
    }
}

console.log(tags);

Answer 5

您可以尝试使用 for 循环：

for (var i = 0; i < posts.length; i++) {
    for(var j = 0; j < posts[i].tags.length; j++) {
        console.log(posts[i].tags[j]);
    }
}

干杯。