Question

我有一个如下所示的集合

var flights = [{
    id: 1,
    legs:[{
        carrierName:'Pegasus' 
    }]
}, {
    id: 2,
    legs:[{
        carrierName: 'SunExpress'
    },{
        carrierName: 'SunExpress'
    }]
}, {
    id: 3,
    legs:[{
        carrierName: 'Pegasus'
    },{
        carrierName: 'SunExpress'
    }]
}]

我想过滤它，例如carrierName ='Pegasus'，然后我的结果就像这样

[{
    id: 1,
    legs:[{
        carrierName:'Pegasus' 
    }]
}, {
    id: 3,
    legs:[{
        carrierName: 'Pegasus'
    },{
        carrierName: 'SunExpress'
    }]
}]

Answer 1

您可以使用Array.prototype.filter()并使用Array.prototype.some()检查每个子数组，看看它是否包含搜索字词：

var carrierNames = ['Pegasus', 'SunExpress'];

flights.filter(function(item) {
  var predicate = this; // the carrierNames dictionary

  return item.legs.some(function(leg) {
    return predicate[leg.carrierName]; // see if carrierName exists in the dictionary
  });
}, carrierNames.reduce(function(obj, term) { // create a dictionary of carrierNames
  obj[term] = true;
  return obj;
}, Object.create(null)));

var flights = [{
    id: 1,
    legs:[{
        carrierName:'Pegasus' 
    }]
}, {
    id: 2,
    legs:[{
        carrierName: 'SunExpress'
    },{
        carrierName: 'SunExpress'
    }]
}, {
    id: 3,
    legs:[{
        carrierName: 'Pegasus'
    },{
        carrierName: 'SunExpress'
    }]
}];

var carrierNames = ['Pegasus', 'SunExpress'];

var result = flights.filter(function(item) {
  var predicate = this;
  
  return item.legs.some(function(leg) {
    return predicate[leg.carrierName];
  });
}, carrierNames.reduce(function(obj, term) {
  obj[term] = true;
  return obj;
}, Object.create(null)));

console.log(result);

使用ES6箭头函数和参数destructuring的相同逻辑：

const result = flights.filter(function({ legs }) {
  const predicate = this;
  return legs.some(({ carrierName }) => predicate.has(carrierName)); // keep if carrierName is in the Set
}, new Set(carrierNames)); // create a Set from the carrierNames

const flights = [{
  id: 1,
  legs: [{
    carrierName: 'Pegasus'
  }]
}, {
  id: 2,
  legs: [{
    carrierName: 'SunExpress'
  }, {
    carrierName: 'SunExpress'
  }]
}, {
  id: 3,
  legs: [{
    carrierName: 'Pegasus'
  }, {
    carrierName: 'SunExpress'
  }]
}];

const carrierNames = ['Pegasus', 'SunExpress'];

const result = flights.filter(function({ legs }) {
  const predicate = this;
  return legs.some(({ carrierName }) => predicate.has(carrierName));
}, new Set(carrierNames));

console.log(result);

Answer 2

检查一些航班的行程是否包含carrierName

_.filter(flights, function(flight) {
    return _.chain(flight)
        .get('legs')
        .map('carrierName')
        .includes('Pegasus')
        .value()
});

用于检查值的值

_.filter(flights, function(flight) {
    return _.chain(flight)
        .get('legs')
        .map('carrierName')
        .thru(function(names) {
            return _.every(valuesArr, function(val) { // _.some for OR, _.every for AND
                return _.includes(names, val);
            });
        })
        .value()
});

Answer 3

使用ES6，您可以使用一行进行过滤，使用Array#filter和Array#some。

＆＃13;

var flights = [{ id: 1, legs: [{ carrierName: 'Pegasus' }] }, { id: 2, legs: [{ carrierName: 'SunExpress' }, { carrierName: 'SunExpress' }] }, { id: 3, legs: [{ carrierName: 'Pegasus' }, { carrierName: 'SunExpress' }] }],
    search = new Set(['Pegasus', 'SunExpress']);
    result = flights.filter(a => a.legs.some(b => search.has(b.carrierName)));

console.log(result);

＆＃13;

.as-console-wrapper { max-height: 100% !important; top: 0; }

＆＃13;

Answer 4

谢谢大家。我解决了它，如下所示


    var airlines = ["Pegasus",SunExpress"];
    var result = _.filter(flights, function(item) {
                        return _.some(item.legs, function(leg) {
                            return _.includes(airlines, leg.carrierName);
                        });
                    });

带有lodash的嵌套集合过滤器

4 个答案: