使用 Underscorejs 从 json 中提取值

Question

我有一个 json，我需要从中过滤 name 值与传递的输入匹配的对象，然后从相应的 links 中提取 href 值ref 值为 self 的对象。我附上了代码片段以及我的过滤方法。我正在使用 Underscore.js.

var jsonObj = [
{
  "links": [
    {
      "rel": "self",
      "href": "http://hostname:port/state/644444",
      "variables": [],
      "variableNames": [],
      "templated": false
    },
    {
      "rel": "city",
      "href": "http://hostname:port/city/6184",
      "variables": [],
      "variableNames": [],
      "templated": false
    }
  ],
  "stateId": 65986,
  "countyId": 6184,
  "name": "AAATest",
  "population": 20000,
  "location": "SFS CSR"
},
{
  "links": [
    {
      "rel": "self",
      "href": "http://hostname:port/state/65987",
      "variables": [],
      "variableNames": [],
      "templated": false
    },
    {
      "rel": "city",
      "href": "http://hostname:port/city/6184",
      "variables": [],
      "variableNames": [],
      "templated": false
    }
  ],
  "stateId": 65987,
  "countyId": 6184,
  "name": "2k8std511",
  "population": 20000,
  "location": "SFS CSR"
  }
]

var keywords='2k8std511';

var filtered = _.filter(jsonObj, function (item) {
                return (_.contains('2k8std511', item['name']));
            });
console.log(_.chain(jsonObj)
 .pluck("links")
 .flatten()
 .value());

请告诉我如何根据给定的关键字进行过滤：

示例输入：AAATest 预期输出：[href:'http://hostname:port/state/644444']

此致，普拉迪普

Answer 1

你可以用普通的 JavaScript:

var result = jsonObj.filter(function(obj) {
  return obj.name === 'AAATest'
}).reduce(function(acc, obj) {
  var links = obj.links.filter(function(link) {
    return link.rel === 'self'
  }).map(function(link) {
    return link.href
  })
  return acc.concat(links)
},[])

console.log(result) //=> ["http://hostname:port/state/644444"]

Answer 2

下划线解决方案：

var jsonObj = [{
  "links": [{
    "rel": "self",
    "href": "http://hostname:port/state/644444",
    "variables": [],
    "variableNames": [],
    "templated": false
  }, {
    "rel": "city",
    "href": "http://hostname:port/city/6184",
    "variables": [],
    "variableNames": [],
    "templated": false
  }],
  "stateId": 65986,
  "countyId": 6184,
  "name": "AAATest",
  "population": 20000,
  "location": "SFS CSR"
}, {
  "links": [{
    "rel": "self",
    "href": "http://hostname:port/state/65987",
    "variables": [],
    "variableNames": [],
    "templated": false
  }, {
    "rel": "city",
    "href": "http://hostname:port/city/6184",
    "variables": [],
    "variableNames": [],
    "templated": false
  }],
  "stateId": 65987,
  "countyId": 6184,
  "name": "2k8std511",
  "population": 20000,
  "location": "SFS CSR"
}]

var filtername = 'AAATest';
var answer = _.chain(jsonObj)
  .filter(function(obj) {
    return obj.name === filtername;
  })
  .map(function(obj) {
    return obj.links
  })
  .first()
  .filter(function(obj) {
    return obj.rel === 'self'
  })
  .map(function(obj) {
    return obj.href
  })
  .value();
console.log(answer);

<script src="http://underscorejs.org/underscore-min.js"></script>

Answer 3

另一个使用下划线的版本：

var result = _.chain(jsonObj)
    .where({ name: 'AAATest' })
    .pluck('links')
    .flatten()
    .where({ rel: 'self'})
    .pluck('href')
    .value();

使用 Underscorejs 从 json 中提取值

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