Question

鉴于此：

var results = [
  {
        "Title": "Battle of Baekgang",
        "Space": ["South Korea"]
    },
    {
        "Title": "Victory Tests",
        "Space": ["United Kingdom"]
    },
    {
        "Title": "Everett massacre",
        "Space": ["United States"]
    },
    {
        "Title": "Bologna massacre",
        "Space": ["Italy"]
    },
    {
        "Title": "Milano massacre",
        "Space": ["Italy"]
    }
];

我如何循环才能说：

If Space value匹配myNation {do this}？

设置var countries = ["Italy", United Kingdom", "South Korea"];

我有var regex = new RegExp(countries.join("|"), "i");

我知道我可以做if(location.match(regex)) {

但我需要先存储location我猜是Space value

的名称

还是其他/更好的方式？

更新

我有一个带有多边形的map，每个多边形都有类似的类：

<path class="italy france germany">

我可以为每个国家/地区设置N个类名称，json我会得到一个名为Space的对象，其中包含一个或多个countries，所以我' d需要检查json个国家/地区是否匹配polygons paths上任意map中的任何类，如果是，则路径应添加class active。

Answer 1

如果我理解你的问题，这应该有效：

var results = [{
        "Title": "Battle of Baekgang",
        "Space": ["South Korea"]
    }, {
        "Title": "Victory Tests",
        "Space": ["United Kingdom"]
    }, {
        "Title": "Everett massacre",
        "Space": ["United States"]
    }, {
        "Title": "Bologna massacre",
        "Space": ["Italy"]
    }, {
        "Title": "Milano massacre",
        "Space": ["Italy"]
    }
];

var countries = ["Italy", "United Kingdom", "South Korea"];
for (var i = 0; i < results.length; i++) {
    for (var j = 0; j < results[i]["Space"].length; j++) {
        if(countries.includes(results[i]["Space"][j])) {
            var matchedCountry = results[i]["Space"][j];
            var currentTitle = results[i]["Title"];
            //Do other stuff
        }
    }
}

Answer 2

 results.filter(
   ({Space})=>countries.includes(Space[0])
 ).forEach(el=>console.log(el))

只需过滤，然后循环过滤：/

更新：您可以只创建一个空间集：

var spaces = new Set(
  results.reduce((acc,el)=>[...acc,...el.Space])
 );

然后你可以回顾你的元素：

 var els = document.querySelectorAll("path");
 els.forEach(function(el){
  if(el.className.split(" ").some(clas=>spaces.has(clas))){
   //do sth
  }
 });

Answer 3

这就是我要去的方式：

＆＃13;

var results = [
  {
    "Title": "Battle of Baekgang",
    "Space": ["South Korea"]
  },
  {
    "Title": "Victory Tests",
    "Space": ["United Kingdom"]
  },
  {
    "Title": "Everett massacre",
    "Space": ["United States"]
  },
  {
    "Title": "Bologna massacre",
    "Space": ["Italy"]
  },
  {
    "Title": "Milano massacre",
    "Space": ["Italy"]
  }
];

var countries = ["Italy"];

for(var i = 0; i < results.length; i++) {
  if(countries.includes(results[i]["Space"][0])) { // the [0] is because of Space being an Array
    console.log(results[i]["Title"]);
  }
}

＆＃13;

Answer 4

在ES6解决方案下找到Space数组中countries的所有条目

＆＃13;

var countries = ["Italy", "United Kingdom", "South Korea"];
var regex = new RegExp(countries.join("|"), "i");
var results = [
  {
        "Title": "Battle of Baekgang",
        "Space": ["South Korea"]
    },
    {
        "Title": "Victory Tests",
        "Space": ["United Kingdom"]
    },
    {
        "Title": "Everett massacre",
        "Space": ["United States"]
    },
    {
        "Title": "Bologna massacre",
        "Space": ["Italy"]
    },
    {
        "Title": "Milano massacre",
        "Space": ["Italy"]
    }
];

let matches = results.reduce((acc,{Space}) => regex.test(...Space)?acc.concat(...Space):acc, []);

matches.forEach((i) => {
jQuery(`path.${i.toLowerCase()}`).addClass('active')
})

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<path class="italy france germany">
</path>
<path class="should not be updated">
</path>

＆＃13;

如何遍历对象数组

4 个答案: