Question

我需要遍历相同的对象并根据用户所食用的n餐数进行打印。我有以下数组

[id, name, food1, food2, food3],
[id, name, food1, food2, ''],
[id, name, food1, '', '']

在数据库中应该是这样的：

1 | Ronald| chicken rice     | NoFood1
2 | Ronald| Tortillas azadas | NoFood2
3 | Ronald| stewed chicken   | NoFood3
4 | Paul  | Arepas           | NoFood1
5 | Paul  | Torillas         | NoFood2
6 | Raul  | noodles          | NoFood1

因为记录是基于每个用户的食物量，但是通过使用代码，我为我带来了每个对象的第一个食物，所以我看到有些示例包含链表和.push，但是如果有的话我可以使用另一个代码，但是如果您可以或多或少地向我展示使用我的代码，或者如果有另一个实现，请多多指教。

My function

ConditionDataToSend(lines, values: Array<any>){
while (lines !== null) {
const info: FoodDto = {food: '', name: '', id:0, noFood:''}

if (values.length >= 4 && values[4] !== '') {

    info.name = values[1]
    info.food = values[2]
    info.noFood = 'food1'
    info.id = id

    return info
}

if (values.length >= 5 && values[5] !== '') {
   
    info.name = values[1]
    info.food = values[2]
    info.nofood = 'food2'
    info.id = id

    return info
}

....

Answer 1

我对这个问题（特别是noFood属性）感到有些困惑，但这能满足您的要求吗？

const data = [
  [1, 'Jim', 'raspberry', 'blackberry', 'lingonberry'],
  [2, 'Tina', 'boysenberry', 'strawberry', ''],
  [3, 'Sarah', 'blueberry', '', '']
];

function infoTransform(values) {
  return values.map(value => {
    [id, name, ...foods] = value;

    return {
      id,
      name,
      foods: foods.filter(food => food !== '')
    };
  });
}

console.log(infoTransform(data));

/* Log:
[
  {
    id: 1,
    name: 'Jim',
    foods: [ 'raspberry', 'blackberry', 'lingonberry' ]
  },
  { id: 2, name: 'Tina', foods: [ 'boysenberry', 'strawberry' ] },
  { id: 3, name: 'Sarah', foods: [ 'blueberry' ] }
]
*/

编辑：使用noFood更新了代码：

const data = [
  [1, 'Jim', 'raspberry', 'blackberry', 'lingonberry'],
  [2, 'Tina', 'boysenberry', 'strawberry', ''],
  [3, 'Sarah', 'blueberry', '', '']
];

function infoTransform(values) {
  return values.map(value => {
    // Destructure
    let [id, name, ...foods] = value;

    // Filter out empty strings
    foods = foods.filter(food => food !== '');

    // Set food and noFood properties
    let noFood = null;
    let food = null;

    if (foods.length === 1) {
      noFood = foods[0];
      food = foods[0];
    } else if (foods.length > 1) {
      noFood = foods[foods.length - 1];
      food = foods[0];
    }

    return {
      id,
      name,
      food,
      noFood
    };
  });
}

console.log(infoTransform(data));

/* Log:
[
  { id: 1, name: 'Jim', food: 'raspberry', noFood: 'lingonberry' },
  { id: 2, name: 'Tina', food: 'boysenberry', noFood: 'strawberry' },
  { id: 3, name: 'Sarah', food: 'blueberry', noFood: 'blueberry' }
]
*/

如何遍历同一数组对象n次

1 个答案: