在子对象数组中搜索 - JavaScript答案

【问题标题】：search inside child Array of objects - JavaScript在子对象数组中搜索 - JavaScript
【发布时间】：2019-02-12 11:34:16
【问题描述】：

目前我有以下对象数组

obj = [
    {
        "id":28,
        cities: [
            {
                cityTypes: "AA",
                citySource: "sdsf"
            },
            {
                cityTypes: "BB",
                citySource: "sdsgf"
            },
            {
                cityTypes: "CC",
                citySource: "fgsdfgd"
            }
        ]
    }, 
    {
        "id":56,
        cities: [
            {
                cityTypes: "DD",
                citySource: "sdsf"
            },
            {
                cityTypes: "EE",
                citySource: "sdsgf"
            },
            {
                cityTypes: "FF",
                citySource: "fgsdfgd"
            }
        ]
    }, 
    {
        "id":89,
        cities: [
            {
                cityTypes: "GG",
                citySource: "sdsf"
            },
            {
                cityTypes: "HH",
                citySource: "sdsgf"
            },
            {
                cityTypes: "II",
                citySource: "fgsdfgd"
            }
        ]
    }
]

我需要在整个Object 中搜索特定值的cityTypes。

假设，我需要搜索cityTypes = BB

如果BB存在于整个对象中，则返回true

如果BB没有预设，返回false。

这是我尝试过的，但似乎不起作用。

for(let k=0; k<obj.length; k++){
    if(obj[k].cities){
        let cityObj = obj[k].cities;
        for(let city in cityObj){
            city.cityTypes !== "BB" ? "true" : "false"
        }
    }
}

实现这一目标的正确方法是什么？

【问题讨论】：

标签： javascript arrays loops object search

【解决方案1】：

您可以在另一个.some 中使用.some：

const obj=[{"id":28,cities:[{cityTypes:"AA",citySource:"sdsf"},{cityTypes:"BB",citySource:"sdsgf"},{cityTypes:"CC",citySource:"fgsdfgd"}]},{"id":56,cities:[{cityTypes:"DD",citySource:"sdsf"},{cityTypes:"EE",citySource:"sdsgf"},{cityTypes:"FF",citySource:"fgsdfgd"}]},{"id":89,cities:[{cityTypes:"GG",citySource:"sdsf"},{cityTypes:"HH",citySource:"sdsgf"},{cityTypes:"II",citySource:"fgsdfgd"}]}];

console.log(
  obj.some(({ cities }) => cities.some(({ cityTypes }) => cityTypes === 'BB'))
);
console.log(
  obj.some(({ cities }) => cities.some(({ cityTypes }) => cityTypes === 'foobar'))
);

【讨论】：

【解决方案2】：

由于以下几个原因，您现有的代码无法正常工作：

您需要执行let city of cityObj，因为您在内部循环中以city.cityTypes 的形式访问city 的属性，因此使用of 将为您提供city 变量中的每个对象。
您需要使用if 条件实际检查是否找到匹配项。如果找到与您期望的 cityTypes 匹配，则 break 内部循环。如果找到匹配项，还要break 外循环。

var obj = [{
    "id": 28,
    cities: [{
        cityTypes: "AA",
        citySource: "sdsf"
      },
      {
        cityTypes: "BB",
        citySource: "sdsgf"
      },
      {
        cityTypes: "CC",
        citySource: "fgsdfgd"
      }
    ]
  },
  {
    "id": 56,
    cities: [{
        cityTypes: "DD",
        citySource: "sdsf"
      },
      {
        cityTypes: "EE",
        citySource: "sdsgf"
      },
      {
        cityTypes: "FF",
        citySource: "fgsdfgd"
      }
    ]
  },
  {
    "id": 89,
    cities: [{
        cityTypes: "GG",
        citySource: "sdsf"
      },
      {
        cityTypes: "HH",
        citySource: "sdsgf"
      },
      {
        cityTypes: "II",
        citySource: "fgsdfgd"
      }
    ]
  }
]
var found = false;
for (let k = 0; k < obj.length; k++) {
  if (obj[k].cities) {
    let cityObj = obj[k].cities;
    for (let city of cityObj) {
      found = city.cityTypes === "BB";
      if (found) {
        break;
      }
    }
  }
  if (found) {
    break;
  }
}
console.log(found);

【讨论】：

【解决方案3】：

使用双重减少应该可以工作

var obj = [{
    "id": 28,
    cities: [{
        cityTypes: "AA",
        citySource: "sdsf"
      },
      {
        cityTypes: "BB",
        citySource: "sdsgf"
      },
      {
        cityTypes: "CC",
        citySource: "fgsdfgd"
      }
    ]
  },
  {
    "id": 56,
    cities: [{
        cityTypes: "DD",
        citySource: "sdsf"
      },
      {
        cityTypes: "EE",
        citySource: "sdsgf"
      },
      {
        cityTypes: "FF",
        citySource: "fgsdfgd"
      }
    ]
  },
  {
    "id": 89,
    cities: [{
        cityTypes: "GG",
        citySource: "sdsf"
      },
      {
        cityTypes: "HH",
        citySource: "sdsgf"
      },
      {
        cityTypes: "II",
        citySource: "fgsdfgd"
      }
    ]
  }
]
var cityTypes1 = 'BB';
console.log(obj.reduce((total, cur) => 
    total||cur.cities.reduce((total, cur) => 
        total||cur.cityTypes === cityTypes1, false), 
    false))

【讨论】：

【解决方案4】：

如果至少有一个城市具有所需的类型，则此简短方法返回 true：

function findMyCode(code) {
   return obj.some((d) => {
       return d.cities.some((city) => {
           return city.cityTypes === code;
       });
   });
}

或者你可以尝试使用lodash库。

所以你可以使用它：

var hasType = findMyCode('BB'); // returns true/false

【讨论】：

【解决方案5】：

试试下面的代码。

var result = false;
for(let k=0; k<obj.length; k++){
    if(obj[k].cities){
        let cityObj = obj[k].cities;
        for(let city in cityObj){
            if(cityObj[city].cityTypes == "BB"){
                result = true;
            }
        }
    }
}

返回结果或打印控制台日志。

【讨论】：

【解决方案6】：

var filtered = obj.filter((o) => {
    var found=false;
    for(let i=0;  i<o.cities.length; i++) {
        let city = o.cities[i];
        if(city.cityTypes === "BB"){
            found=true;
            break;
        }
    }
    return found;
})
console.log(filtered)

【讨论】：