过滤数据数组多个值 javascript

Question

I know this question has been asked before, but existing solutions do not seem to work for my case. I am trying to filter out data on the basis of multiple properties/values in a varied array of objects.

My sample data looks like this:

const products = [
          { name: 'A', color: 'Blue', size: 50, locations: 'USA' },
          { name: 'B', color: 'Blue', size: 60, locations: 'Europe' },
          { name: 'C', color: 'Blue', size: 30, locations: 'Europe' },
          { name: 'D', color: 'Black', size: 70, locations: 'Japan' },
          { name: 'E', color: 'Green', size: 50, locations: 'Europe' },
          { name: 'F', color: 'Blue', size: 50, locations: 'Brazil' },
          { name: 'G', color: 'Black', size: 40, locations: 'Australia' },
        ];

This is what I require the filters result to be:

const filters_one = ['Blue'];

const requiredResult_One = [
      { name: 'A', color: 'Blue', size: 50, locations: 'USA' }, 
      { name: 'B', color: 'Blue', size: 60, locations: 'Europe' },
      { name: 'C', color: 'Blue', size: 30, locations: 'Europe' },
      { name: 'E', color: 'Blue', size: 50, locations: 'Brazil' }, 
]


const filters_two = ['Blue', 'Europe'];

const requiredResult_Two = [
      { name: 'B', color: 'Blue', size: 60, locations: 'Europe' },
      { name: 'C', color: 'Blue', size: 30, locations: 'Europe' }, 
]


----------OR-------------------------
const filters_three = ['Black'];

const requiredResult_three = [
      { name: 'D', color: 'Black', size: 70, locations: 'Japan' },
      { name: 'G', color: 'Black', size: 40, locations: 'Australia' }, 
]

const filters_four = ['Black', 'Japan'];

const requiredResult_Four = [
      { name: 'D', color: 'Black', size: 70, locations: 'Japan' },
]

this is what I have achieved so far:

const filterdata = (products, filter) => {
      // where, filter can be equal to *filters_one*, *filters_two*, *filters_three*, 
         or *filters_four* anyone of it. //

      const keysExact = ['color', 'locations'];
      const valuesExact = filter.map(col => col.toLowerCase());

      const resultExact = products.filter((prod) =>
          keysExact.every((key) => valuesExact.includes(prod[key].toLowerCase()))
      );

      console.log(resultExact);
};

This seems to work partially or not looks like a good approach to me. If anyone can help me with a way better solution on this, will be really helpful.

Thanks in advance!

Answer 1

You could get the values from the object and check if all values of the query is in the values array.

 const filter = (data, query) => data.filter(o => { const values = Object.values(o); return query.every(q => values.includes(q)); }), products = [{ name: 'A', color: 'Blue', size: 50, locations: 'USA' }, { name: 'B', color: 'Blue', size: 60, locations: 'Europe' }, { name: 'C', color: 'Blue', size: 30, locations: 'Europe' }, { name: 'D', color: 'Black', size: 70, locations: 'Japan' }, { name: 'E', color: 'Green', size: 50, locations: 'Europe' }, { name: 'F', color: 'Blue', size: 50, locations: 'Brazil' }, { name: 'G', color: 'Black', size: 40, locations: 'Australia' }]; console.log(filter(products, ['Blue'])); console.log(filter(products, ['Blue', 'Europe'])); console.log(filter(products, ['Black'])); console.log(filter(products, ['Black', 'Japan']));

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

Answer 2

const filterArray = (filter, arr) => {
   return arr.filter(val => (filter[1] 
                       ? val.color === filter[0] && val.locations === filter[1] 
                       : val.color === filter[0]));
};

Answer 3

You can create a filter based on color and location

 const products = [ { name: 'A', color: 'Blue', size: 50, locations: 'USA' }, { name: 'B', color: 'Blue', size: 60, locations: 'Europe' }, { name: 'C', color: 'Blue', size: 30, locations: 'Europe' }, { name: 'D', color: 'Black', size: 70, locations: 'Japan' }, { name: 'E', color: 'Green', size: 50, locations: 'Europe' }, { name: 'F', color: 'Blue', size: 50, locations: 'Brazil' }, { name: 'G', color: 'Black', size: 40, locations: 'Australia' }, ]; function filterProducts(filterParams) { const [color, country] = filterParams; const output = products.filter((product) => product.color === color && (country? product.locations === country: true)) return output; } const filters_one = ['Blue']; const requiredResult_One = filterProducts(filters_one); console.log(requiredResult_One); const filters_two = ['Blue', 'Europe']; const requiredResult_Two = filterProducts(filters_two); console.log(requiredResult_Two); const filters_three = ['Black']; const requiredResult_three = filterProducts(filters_three); console.log(requiredResult_three); const filters_four = ['Black', 'Japan']; const requiredResult_Four = filterProducts(filters_four); console.log(requiredResult_Four);

Answer 4

You can use an object to filter list:

 let products = [{ name: 'A', color: 'Blue', size: 50, locations: 'USA' }, { name: 'B', color: 'Blue', size: 60, locations: 'Europe' }, { name: 'C', color: 'Blue', size: 30, locations: 'Europe' }, { name: 'D', color: 'Black', size: 70, locations: 'Japan' }, { name: 'E', color: 'Green', size: 50, locations: 'Europe' }, { name: 'F', color: 'Blue', size: 50, locations: 'Brazil' }, { name: 'G', color: 'Black', size: 40, locations: 'Australia' },]; function* filter(array, list = {}) { loop: for (let item of array) { for (let [k, v] of Object.entries(list)) if (item[k];= v) continue loop; yield item. } } console:log("size,50". [..,filter(products: { size. 50 })]) console:log("color,Blue". [..,filter(products: { color. 'Blue' })]) console:log("color,Blue: size,50". [..,filter(products: { color, 'Blue': size. 50 })]) console:log("color,Blue: locations,Europe". [..,filter(products: { color, 'Blue': locations: 'Europe' })])

This example is very fast, flexible and lazy!