Is there a way to obtain a non-matching string in JavaScript?

Question

Is there a way to obtain a non-matching string in JavaScript?

I have an array of bookings and need to search for a specific value inside the array using searchValue.

In this case, I need to check the booking id field. If the booking id matches the searchValue, then I need to push that object into the result array.

Here's an example of the result array:

Result array example:

let searchValue = "12,13,15"

Result:

[{ name:"user 3", bookingid:12, product: "ui" },
    { name:"user 4", bookingid:13, product: "ef" }]

Expected Output:

Since 12 and 13 are matched in the booking array, the remaining value in the searchValue is "15". Can someone please provide guidance on how to handle this?

let bookingArr = [
    { name:"user 1", bookingid:10, product: "ab" },
    { name:"user 1", bookingid:10, product: "cd" },

    { name:"user 2", bookingid:11, product: "ui" },
    { name:"user 1", bookingid:10, product: "ef" },

    { name:"user 3", bookingid:12, product: "ui" },
    { name:"user 4", bookingid:13, product: "ef" },
];
let searchValue = "12,13,15";
let set = new Set(searchValue.split(",").map(Number)); // for faster lookup
let res = bookingArr.filter(x => set.has(x.bookingid));
console.log(res);

// how can i get not matched searchValue 
// expected result notmatchedsearchValue ="15"

Check out the demo here

javascript arrays data-structures set reduce

Answer 1

Answer №1

The original poster was close to the solution. To remove each array item's bookingid number-value from the previously generated set, one must use the delete function. Afterwards, creating an array from the modified set and joining its items/values will result in the final string-value output.

The OP's code with slight modifications ...

let bookingArr = [
    { name:"user 1", bookingid:10, product: "ab" },
    { name:"user 1", bookingid:10, product: "cd" },

    { name:"user 2", bookingid:11, product: "ui" },
    { name:"user 1", bookingid:10, product: "ef" },

    { name:"user 3", bookingid:12, product: "ui" },
    { name:"user 4", bookingid:13, product: "ef" },
];
let searchValue = '12,13,15';

let set = new Set(searchValue.split(',').map(Number)); // for faster lookup

bookingArr.forEach(item => set.delete(item.bookingid));

let notmatchedSearchValue = Array.from(set).join(',');

console.log({ searchValue, notmatchedSearchValue });
// expected result notmatchedsearchValue = '15'

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

A more versatile approach would involve a reducer function that includes a filter-process independent of a specific array item property name. This reducer is capable of both filtering array items based on a custom property name (key) and providing the final string-value of missing (non-matching) values ... something along the lines of ...

function collectMatchingItemAndMissingValue(collector, item) {
  const { key, searchValues, missingValues, matchingItems } = collector;
  const value = item[key];

  if (searchValues.has(value)) {

    matchingItems.push(item);
    missingValues.delete(value);    
  }
  return collector;
}

function getMatchingItemsAndMissingValues(arr, key, searchValues, castValue) {
  searchValues = new Set(
    searchValues.split(',').map(castValue)
  );
  const {
    missingValues,
    matchingItems,
  } = arr.reduce(collectMatchingItemAndMissingValue, {
    key,
    searchValues,
    missingValues: new Set([...searchValues]),
    matchingItems: [],
  });
  return {
    matchingItems,
    missingValues: [...missingValues].join(', '),
  };
}

const bookingArr = [
  { name:"user 1", bookingid:10, product: "ab" },
  { name:"user 1", bookingid:10, product: "cd" },
  { name:"user 2", bookingid:11, product: "ui" },
  { name:"user 1", bookingid:10, product: "ef" },
  { name:"user 3", bookingid:12, product: "ui" },
  { name:"user 4", bookingid:13, product: "ef" },
];
const searchValues = '12, 13, 15';
const castValue = str => Number(str.trim());
const {

  matchingItems,
  missingValues,

} = getMatchingItemsAndMissingValues(bookingArr, 'bookingid', searchValues, castValue);

console.log({ matchingItems, missingValues, searchValues });

console.log({
  ...getMatchingItemsAndMissingValues(
    bookingArr,
    'product',
    'xy, ui, mn, qr, ef',
    str => str.trim(),
  ),
  searchValues: 'xy, ui, mn, qr, ef'
});

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

Answer 2

The original poster was close to the solution. To remove each array item's bookingid number-value from the previously generated set, one must use the delete function. Afterwards, creating an array from the modified set and joining its items/values will result in the final string-value output.

The OP's code with slight modifications ...

let bookingArr = [
    { name:"user 1", bookingid:10, product: "ab" },
    { name:"user 1", bookingid:10, product: "cd" },

    { name:"user 2", bookingid:11, product: "ui" },
    { name:"user 1", bookingid:10, product: "ef" },

    { name:"user 3", bookingid:12, product: "ui" },
    { name:"user 4", bookingid:13, product: "ef" },
];
let searchValue = '12,13,15';

let set = new Set(searchValue.split(',').map(Number)); // for faster lookup

bookingArr.forEach(item => set.delete(item.bookingid));

let notmatchedSearchValue = Array.from(set).join(',');

console.log({ searchValue, notmatchedSearchValue });
// expected result notmatchedsearchValue = '15'

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

A more versatile approach would involve a reducer function that includes a filter-process independent of a specific array item property name. This reducer is capable of both filtering array items based on a custom property name (key) and providing the final string-value of missing (non-matching) values ... something along the lines of ...

function collectMatchingItemAndMissingValue(collector, item) {
  const { key, searchValues, missingValues, matchingItems } = collector;
  const value = item[key];

  if (searchValues.has(value)) {

    matchingItems.push(item);
    missingValues.delete(value);    
  }
  return collector;
}

function getMatchingItemsAndMissingValues(arr, key, searchValues, castValue) {
  searchValues = new Set(
    searchValues.split(',').map(castValue)
  );
  const {
    missingValues,
    matchingItems,
  } = arr.reduce(collectMatchingItemAndMissingValue, {
    key,
    searchValues,
    missingValues: new Set([...searchValues]),
    matchingItems: [],
  });
  return {
    matchingItems,
    missingValues: [...missingValues].join(', '),
  };
}

const bookingArr = [
  { name:"user 1", bookingid:10, product: "ab" },
  { name:"user 1", bookingid:10, product: "cd" },
  { name:"user 2", bookingid:11, product: "ui" },
  { name:"user 1", bookingid:10, product: "ef" },
  { name:"user 3", bookingid:12, product: "ui" },
  { name:"user 4", bookingid:13, product: "ef" },
];
const searchValues = '12, 13, 15';
const castValue = str => Number(str.trim());
const {

  matchingItems,
  missingValues,

} = getMatchingItemsAndMissingValues(bookingArr, 'bookingid', searchValues, castValue);

console.log({ matchingItems, missingValues, searchValues });

console.log({
  ...getMatchingItemsAndMissingValues(
    bookingArr,
    'product',
    'xy, ui, mn, qr, ef',
    str => str.trim(),
  ),
  searchValues: 'xy, ui, mn, qr, ef'
});

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

Answer 3

Answer №2

This scenario presents a different challenge compared to your previous inquiry. In this case, it is necessary to create a Set from the elements in the bookingArr array rather than from the searchVal.

let bookingArr = [
  { name: "user 1", bookingid: 10, product: "ab" },
  { name: "user 1", bookingid: 10, product: "cd" },

  { name: "user 2", bookingid: 11, product: "ui" },
  { name: "user 1", bookingid: 10, product: "ef" },

  { name: "user 3", bookingid: 12, product: "ui" },
  { name: "user 4", bookingid: 13, product: "ef" },
];

let searchValue = "12,13,15";
let set = new Set(bookingArr.map((b) => b.bookingid));
let res = searchValue
  .split(",")
  .map(Number)
  .filter((s) => !set.has(s))
  .join();
console.log(res);

Answer 4

This scenario presents a different challenge compared to your previous inquiry. In this case, it is necessary to create a Set from the elements in the bookingArr array rather than from the searchVal.

let bookingArr = [
  { name: "user 1", bookingid: 10, product: "ab" },
  { name: "user 1", bookingid: 10, product: "cd" },

  { name: "user 2", bookingid: 11, product: "ui" },
  { name: "user 1", bookingid: 10, product: "ef" },

  { name: "user 3", bookingid: 12, product: "ui" },
  { name: "user 4", bookingid: 13, product: "ef" },
];

let searchValue = "12,13,15";
let set = new Set(bookingArr.map((b) => b.bookingid));
let res = searchValue
  .split(",")
  .map(Number)
  .filter((s) => !set.has(s))
  .join();
console.log(res);

Answer 5

Answer №3

Here is a possible solution to your problem:

for (let item of items) {
    list.delete(item.id);
}

Afterwards, the list will only contain the items that were not deleted.

If you wish to keep the original values in the list, you can create a copy of the list using let copyList = new Set(list);, and then operate on copyList instead of list in the loop.

For a more concise approach, you can use the following one-liner, inspired by @Barmar:

[...list].filter(x => !items.some(y => y.id == x)).join(",");

Answer 6

Here is a possible solution to your problem:

for (let item of items) {
    list.delete(item.id);
}

Afterwards, the list will only contain the items that were not deleted.

If you wish to keep the original values in the list, you can create a copy of the list using let copyList = new Set(list);, and then operate on copyList instead of list in the loop.

For a more concise approach, you can use the following one-liner, inspired by @Barmar:

[...list].filter(x => !items.some(y => y.id == x)).join(",");

Answer 7

Answer №4

If you're open to creating a shallow copy of the new set as an array, we can utilize the array some() method along with negation ! to identify any IDs from that set which are not present in bookingArr.

let bookingArr = [
    { name:"user 1", bookingid:10, product: "ab" },
    { name:"user 1", bookingid:10, product: "cd" },

    { name:"user 2", bookingid:11, product: "ui" },
    { name:"user 1", bookingid:10, product: "ef" },

    { name:"user 3", bookingid:12, product: "ui" },
    { name:"user 4", bookingid:13, product: "ef" },
];
let searchValue = "12,13,15";
let set = new Set(searchValue.split(",").map(Number)); // for faster lookup
let res = bookingArr.filter(x => set.has(x.bookingid));
// console.log(res);
let notMatchedSearchValues = [...set].filter(y => !bookingArr.some(x => y === x.bookingid));

notMatchedSearchValues.forEach(id => console.log(`notMatchedSearchValue: ${id}`));

Answer 8

If you're open to creating a shallow copy of the new set as an array, we can utilize the array some() method along with negation ! to identify any IDs from that set which are not present in bookingArr.

let bookingArr = [
    { name:"user 1", bookingid:10, product: "ab" },
    { name:"user 1", bookingid:10, product: "cd" },

    { name:"user 2", bookingid:11, product: "ui" },
    { name:"user 1", bookingid:10, product: "ef" },

    { name:"user 3", bookingid:12, product: "ui" },
    { name:"user 4", bookingid:13, product: "ef" },
];
let searchValue = "12,13,15";
let set = new Set(searchValue.split(",").map(Number)); // for faster lookup
let res = bookingArr.filter(x => set.has(x.bookingid));
// console.log(res);
let notMatchedSearchValues = [...set].filter(y => !bookingArr.some(x => y === x.bookingid));

notMatchedSearchValues.forEach(id => console.log(`notMatchedSearchValue: ${id}`));

Is there a way to obtain a non-matching string in JavaScript?

Answer №1

Answer №2

Answer №3

Answer №4

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