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JS過濾器中的數組

[英]JS filter array by array within

 原文  2018-11-13 05:03:55  9   5    javascript/ arrays/ array-filter/ arrayobject

問題描述

我有如下數組 

  [{
    "id": 68,
    "proffesional_photo": "",
    "top_image": "https://sampleimage.jpg",
    "ratings": "1",
    "price": 690,
    "description": null,
    "type": true,
    "promo": 0,
    "status": true,
    "item": {
      "Item_name": "Dark Chocolate Latte"
    },
    "restaurant_dish_menus": [
      {
        "id": 1,
        "res_dish_id": 1318,
        "menu_id": 4,
        "createdAt": "2018-11-13T04:28:17.000Z",
        "updatedAt": "2018-11-13T04:28:17.000Z"
      },
      {
        "id": 1,
        "res_dish_id": 1318,
        "menu_id": 3,
        "createdAt": "2018-11-13T04:28:17.000Z",
        "updatedAt": "2018-11-13T04:28:17.000Z"
      }
     ]
  },
  {
    "id": 69,
    "proffesional_photo": "",
    "top_image": "https://sampleimage2.jpg",
    "ratings": "1",
    "price": 700,
    "description": null,
    "type": true,
    "promo": 0,
    "status": true,
    "item": {
      "Item_name": "Latte"
    },
    "restaurant_dish_menus": [
      {
        "id": 1,
        "res_dish_id": 1318,
        "menu_id": 3,
        "createdAt": "2018-11-13T04:28:17.000Z",
        "updatedAt": "2018-11-13T04:28:17.000Z"
      }
     ]
  }
  ],

當用戶選擇某個菜單時，需要對其進行過濾，

每個餐具對象可能具有多個menu_id，

我嘗試使用array.filter但是我在弄清楚如何從Dish array通過子數組進行過濾時遇到了麻煩。

我嘗試的代碼（ filterBy = 4 ） 

let result = data.filter(function(row) {
  row.restaurant_dish_menus.filter(function(i) {
    return i.menu_id == filterBy;
  });
});

console.log(result)給了我一個空數組。

如果filterBy is = 4則預期輸出為 

{
    "id": 68,
    "proffesional_photo": "",
    "top_image": "https://sampleimage.jpg",
    "ratings": "1",
    "price": 690,
    "description": null,
    "type": true,
    "promo": 0,
    "status": true,
    "item": {
      "Item_name": "Dark Chocolate Latte"
    },
    "restaurant_dish_menus": [
      {
        "id": 1,
        "res_dish_id": 1318,
        "menu_id": 4,
        "createdAt": "2018-11-13T04:28:17.000Z",
        "updatedAt": "2018-11-13T04:28:17.000Z"
      },
      {
        "id": 1,
        "res_dish_id": 1318,
        "menu_id": 3,
        "createdAt": "2018-11-13T04:28:17.000Z",
        "updatedAt": "2018-11-13T04:28:17.000Z"
      }
     ]
  }

如果filterBy為3，則兩個對象都應該是輸出

5 個解決方案

解決方案1
 3 已采納  2018-11-13 05:21:43

這個怎么樣 

var data =   [{
    "id": 68,
    "proffesional_photo": "",
    "top_image": "https://sampleimage.jpg",
    "ratings": "1",
    "price": 690,
    "description": null,
    "type": true,
    "promo": 0,
    "status": true,
    "item": {
      "Item_name": "Dark Chocolate Latte"
    },
    "restaurant_dish_menus": [
      {
        "id": 1,
        "res_dish_id": 1318,
        "menu_id": 4,
        "createdAt": "2018-11-13T04:28:17.000Z",
        "updatedAt": "2018-11-13T04:28:17.000Z"
      },
      {
        "id": 1,
        "res_dish_id": 1318,
        "menu_id": 3,
        "createdAt": "2018-11-13T04:28:17.000Z",
        "updatedAt": "2018-11-13T04:28:17.000Z"
      }
     ]
  }];

  var result = data.filter(function(m) {
    return m.restaurant_dish_menus.some(function(d) {
      return d.menu_id === 4;
    });
  })

解決方案2
 2  2018-11-13 05:15:23

.filter期望傳遞的函數返回布爾值。 在您的情況下，該函數不返回任何內容（或undefined ），這總是錯誤的。

一種選擇是在嵌套過濾器中使用.find ，並根據結果是否為undefined返回一個布爾值。

這是一個片段。 

 let data = [{ "id": 68, "proffesional_photo": "", "top_image": "https://sampleimage.jpg", "ratings": "1", "price": 690, "description": null, "type": true, "promo": 0, "status": true, "item": { "Item_name": "Dark Chocolate Latte" }, "restaurant_dish_menus": [{ "id": 1, "res_dish_id": 1318, "menu_id": 4, "createdAt": "2018-11-13T04:28:17.000Z", "updatedAt": "2018-11-13T04:28:17.000Z" }, { "id": 1, "res_dish_id": 1318, "menu_id": 3, "createdAt": "2018-11-13T04:28:17.000Z", "updatedAt": "2018-11-13T04:28:17.000Z" } ] }, { "id": 69, "proffesional_photo": "", "top_image": "https://sampleimage.jpg", "ratings": "1", "price": 690, "description": null, "type": true, "promo": 0, "status": true, "item": { "Item_name": "Dark Chocolate Latte" }, "restaurant_dish_menus": [{ "id": 1, "res_dish_id": 1318, "menu_id": 6, "createdAt": "2018-11-13T04:28:17.000Z", "updatedAt": "2018-11-13T04:28:17.000Z" }, { "id": 1, "res_dish_id": 1318, "menu_id": 5, "createdAt": "2018-11-13T04:28:17.000Z", "updatedAt": "2018-11-13T04:28:17.000Z" } ] }, ] let filterBy = 4; let result = data.filter(function(row) { return row.restaurant_dish_menus.find(function(i) { return i.menu_id == filterBy; }) !== undefined; }); console.log(result);

解決方案3
 1  2018-11-13 05:28:10

您可以如下使用“過濾器” 

 var data = [{ "id": 68, "proffesional_photo": "", "top_image": "https://sampleimage.jpg", "ratings": "1", "price": 690, "description": null, "type": true, "promo": 0, "status": true, "item": { "Item_name": "Dark Chocolate Latte" }, "restaurant_dish_menus": [ { "id": 1, "res_dish_id": 1318, "menu_id": 4, "createdAt": "2018-11-13T04:28:17.000Z", "updatedAt": "2018-11-13T04:28:17.000Z" }, { "id": 1, "res_dish_id": 1318, "menu_id": 3, "createdAt": "2018-11-13T04:28:17.000Z", "updatedAt": "2018-11-13T04:28:17.000Z" } ] }, { "id": 69, "proffesional_photo": "", "top_image": "https://sampleimage2.jpg", "ratings": "1", "price": 700, "description": null, "type": true, "promo": 0, "status": true, "item": { "Item_name": "Latte" }, "restaurant_dish_menus": [ { "id": 1, "res_dish_id": 1318, "menu_id": 3, "createdAt": "2018-11-13T04:28:17.000Z", "updatedAt": "2018-11-13T04:28:17.000Z" } ] } ] function filterBy(f) { return data.filter(d => d.restaurant_dish_menus.some(({ menu_id }) => menu_id == f)) } console.log(filterBy(4)) console.log(filterBy(3))

解決方案4
 0  2018-11-13 05:12:33

您也可以使用grep函數 

var menus=  {
    "id": 68,
    "proffesional_photo": "",
    "top_image": "https://sampleimage.jpg",
    "ratings": "1",
    "price": 690,
    "description": null,
    "type": true,
    "promo": 0,
    "status": true,
    "item": {
      "Item_name": "Dark Chocolate Latte"
    },
    "restaurant_dish_menus": [
      {
        "id": 1,
        "res_dish_id": 1318,
        "menu_id": 4,
        "createdAt": "2018-11-13T04:28:17.000Z",
        "updatedAt": "2018-11-13T04:28:17.000Z"
      },
      {
        "id": 1,
        "res_dish_id": 1318,
        "menu_id": 3,
        "createdAt": "2018-11-13T04:28:17.000Z",
        "updatedAt": "2018-11-13T04:28:17.000Z"
      }
     ]
  };

var found_names = $.grep(menus.restaurant_dish_menus, function(v) {
    return v.menu_id ===4;
});

console.log(found_names);

http://jsfiddle.net/ejPV4/

解決方案5
 0  2018-11-13 05:15:45

您的問題尚不清楚最終目標，但如果您要過濾頂級對象，即，當且僅當它具有menu_id === filterBy菜時，才必須存在頂級對象，則： 

let result = data.filter(row => {
  return row.restaurant_dish_menus.some(({menu_id}) => menu_id === filterBy);
});

如果restaurant_dish_menus menu_id === filterBy項為menu_id === filterBy則以上內容僅過濾您的行。 但是這類對象的restaurant_dish_menus仍未過濾。

結果： 

[{
  "id": 68,
  // skipped
  "item": {
    "Item_name": "Dark Chocolate Latte"
  },
  "restaurant_dish_menus": [
    {
      "id": 1,
      "res_dish_id": 1318,
      "menu_id": 4,
      // skipped
    },
    {
      "id": 1,
      "res_dish_id": 1318,
      "menu_id": 3,
      // skipped
    }
  ]
}]

但是，如果您要過濾頂級商品並同時過濾restaurant_dish_menus ，即修改頂層對象，則： 

let result = data.filter(row => {
  return row.restaurant_dish_menus.some(({menu_id}) => menu_id === filterBy);
}).map(row => {
  return {...row, restaurant_dish_menus: row.restaurant_dish_menus.filter(i => i.menu_id === filterBy)};
});

這將首先過濾具有menu_id === filterBy的行對象，然后再過濾restaurant_dish_menus ，使其僅包含一次menu_id === filterBy ，從而有效地修改行對象，即map 。

結果： 

[{
  "id": 68,
  // skipped
  "item": {
    "Item_name": "Dark Chocolate Latte"
  },
  "restaurant_dish_menus": [
    {
      "id": 1,
      "res_dish_id": 1318,
      "menu_id": 4,
      // skipped
    }
  ]
}]

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