Typescript - Check combinations of parameters are null

Question

I wrote this function that construct an object based if some values are null or not.

protected temporalFilter({ range }: Filter): W {
    const temporalKey = ['created', 'bootedSince', 'firstComunication'];
    if (temporalKey.includes(range.key)) {
      if (range.from && !range.to) {
        return {
          [range.from as keyof T]: { gte: range.from }
        } as unknown as W;
      }
      else if (!range.from && range.to) {
        return {
          [range.to as keyof T]: { lte: range.to }
        } as unknown as W;
      }
      else if (range.from && range.to) {
        return {
          AND: [
            { [range.key as keyof T]: { lte: range.to } },
            { [range.key as keyof T]: { gte: range.from } }
          ]
        } as unknown as W;
      }
      else return {} as unknown as W;
    }
    else return null;
  }

But, there is a compact way to do that or some refactoring that can I do?

pls share reproducable example. Where did you get Filter. I assume it is from elastic search? — captain-yossarian from Ukraine
– captain-yossarian from Ukraine, Commented Aug 31, 2021 at 10:39

Newbie · Accepted Answer · 2021-08-31 12:03:52Z

2

This is the most compact yet readable form I can think of:

temporalFilter({range}: Filter): W {
  if (['created', 'bootedSince', 'firstComunication'].includes(range.key)) {
    const f = range.from ? { [range.from]: {gte: range.from} } : undefined;
    const t = range.to ? { [range.to]: {lte: range.to} } : undefined;
    return ((f && t) ? { AND: [f, t] } : (f || t)) as unknown as W;
  } else {
    return null;
  }
}

I suggest you fix your typing; the usage of so many unknown in a single function suggests poor typing. Furthermore, this { [range.from]: {gte: range.from} } it is not wrong, but it is strange to have the same value as filtering condition and as key, perhaps you meant { [range.key]: {gte: range.from} }

edited Aug 31, 2021 at 12:03

answered Aug 31, 2021 at 11:03

Newbie

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4 Comments

th3g3ntl3man Over a year ago

Thank you for your suggestions. I put as unknow as W because I have this error the type is not assignable to W. How can I resolve?

Newbie Over a year ago

The error you are getting is due to problems within Filter and Wif you can provide the complete context (eg Filter definition and what W is), i can address you better.

Newbie Over a year ago

Btw, I edited my answer to address your current typing issues in the same way you did

th3g3ntl3man Over a year ago

W is a type generated from Prisma and Filter as defined as follow: export type Filter = { key?: any, field?: any, range?: { key?: any, from?: any, to?: any, }, };

sasidhar · Accepted Answer · 2021-08-31 12:34:34Z

0

Something like this should work right? I did not test this, just a thought on how to use conditional spreading:

 return {
     ...(!!range.from && {[range.from]: {gte: range.from}}),
     ...(!!range.to && {[range.to]: {lte: range.to}})
    }

edited Aug 31, 2021 at 12:34

answered Aug 31, 2021 at 10:44

sasidhar

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2 Comments

darklightcode Over a year ago

Unfortunately this doesn't cover for the null case, write a complete answer. You have some typos, you closed the parentheses with brackets, other than that, i like alot your solution so far.

sasidhar Over a year ago

I am not sure what you meant by null case, the includes condition can be tested separately and return null when necessary. Is there some other case I am missing? @darklightcode

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