Filter array by finding multiple conditions in nested array of objects

Question

I have an array with a nested array of objects, I want to filter the data where the object of the nested arrays meets multiple conditions.

Here's the sample data.

const providerList = [
  {
    id: "bac4ef8d",
    provider_name: 'Paa Ra'
    provider_gender: "Male",
    provider_item: [
      {
        itemID: "5911319b"
        is_approved: true,
        is_active: true,
      },
      {
        itemID: "937a56d7"
        is_approved: true,
        is_active: true,
      },
    ],
  },
  {
    id: "9df373d5",
    provider_name: "Che Ta",
    provider_gender: "Female",
    provider_item: [
      {
        itemID: "5911319b"
        is_approved: true,
        is_active: true,
      }
    ],
  }
]

These are the filters, note that the itemID can have any number of elements.

const itemFilter = {
  itemID: ["5911319b", "937a56d7"],
  is_approved: [true],
  is_active: [true],
};

Here's my code, however the output does not return as desired.

const filterProviders = providerList.filter(provider =>
  provider.provider_item.every(item =>
    Object.entries(itemFilter).every(([k, v]) => v.includes(item[k]))),
);

I require to filter the providerList and returning providers where the provier_item matches all values in itemFilter. The expected output for the above itemFilter would be:

filterProviders = [
  {
    id: "bac4ef8d",
    provider_name: 'Paa Ra'
    provider_gender: "Male",
    provider_item: [
      {
        itemID: "5911319b"
        is_approved: true,
        is_active: true,
      },
      {
        itemID: "937a56d7"
        is_approved: true,
        is_active: true,
      },
    ],
  }
]

do you want to have all itemID of nested provider_item or just some? — Nina Scholz
– Nina Scholz, Commented Dec 21, 2021 at 19:16

Timur · Accepted Answer · 2021-12-23 16:37:09Z

1

Is this what you are after:

const providerList = [{
    id: "bac4ef8d",
    provider_name: 'Paa Ra',
    provider_gender: "Male",
    provider_item: [{
        itemID: "5911319b",
        is_approved: true,
        is_active: true,
      },
      {
        itemID: "937a56d7",
        is_approved: true,
        is_active: true,
      }
    ]
  },
  {
    id: "9df373d5",
    provider_name: "Che Ta",
    provider_gender: "Female",
    provider_item: [{
      itemID: "5911319b",
      is_approved: true,
      is_active: true,
    }],
  }
]

const itemFilter = {
  itemID: ["937a56d7", "5911319b"],
  is_approved: [true],
  is_active: [true],
}

const filterProviders = providerList.reduce((acc, provider) => {
  provider.provider_item = provider.provider_item.filter(item => (
    itemFilter.itemID.every(ai => provider.provider_item.map(i => i.itemID).includes(ai)) &&
    itemFilter.is_approved.includes(item.is_approved) &&
    itemFilter.is_active.includes(item.is_active)
  ))

  let providerCount = provider.provider_item.length

  if (providerCount > 0 && providerCount === itemFilter.itemID.length) {
    acc.push(provider)
  }

  return acc
}, [])

console.log(filterProviders)

edited Dec 23, 2021 at 16:37

answered Dec 21, 2021 at 19:56

Timur

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4 Comments

pr4d0g Over a year ago

Thanks, however, if I add another itemID to the itemFilter like itemID: ["937a56d7", "5911319b", the I get both results, however, I should get only provide with name 'Paa Ra'. The other user with name 'Che Ta' does not have both itemID so I should not return it

Timur Over a year ago

This important piece of information was missing at time of my posting. I have now updated my answer.

pr4d0g Over a year ago

something's not adding up, for Provider 'Paa Ra', let's say we add a third provider_item where is_active and is_approved equates to true, and in the itemFilter the itemID is not chosen, then I still get the result for 'Paa Ra' which should not be the case, it should have returned null, because the itemFilter.itemID should match what's shown in the provider array

Timur Over a year ago

In that case, you can additionaly check to see if filtered providerList has the same number of providers as the filter. I have updated my answer again.

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