Shorthand JS Condition

Question

I need a bit of assistance with this shorthand condition. My attempt at it so far is becoming a bit of a challenge and cant seem to make it more readable. I believe it was short circuited from minifying.

  if (a === !0 || perSearch.rates.fy2) {
    e();
  } else if ( perSearch.rates.fy1.multiple || perSearch.rates.fy2.multiple ){
    calculateRates();
  } else {
    perSearch.rates.fy1.multiple;
  }

From this terse expression:

a === !0 || (perSearch.rates.fy2 ? perSearch.rates.fy1.multiple || perSearch.rates.fy2.multiple ? e() : calculateRates() : perSearch.rates.fy1.multiple ? e() : calculateRates())

Answer 1

Your expression corresponds to

if (a === !0) {
} else {
    if (perSearch.rates.fy2) {
        if (perSearch.rates.fy1.multiple || perSearch.rates.fy2.multiple) {
            e();
        } else {
            calculateRates();
        }
    } else {
        if (perSearch.rates.fy1.multiple) {
            e();
        } else {
            calculateRates();
        }
    }
}

which can be simplified to

if (a !== true) {
    if (perSearch.rates.fy1.multiple || (perSearch.rates.fy2 && perSearch.rates.fy2.multiple)) {
        e();
    } else {
        calculateRates();
    }
}

Answer 2

This should be it:

if (a !== false) {
    if (perSearch.rates.fy2) {
        if (!perSearch.rates.fy1.multiple) {
            if (perSearch.rates.fy2.multiple) {
                e()
            }
            else {
                calculateRates()
            }
        }
    }
    else {
        if (perSearch.rates.fy1.multiple) {
            e()
        }
        else {
            calculateRates()
        }
    }
}

Answer 3

Is this a return of some sort (return a || result), or a large conditional (a is true or this other code evaluates true)?. Either way I would tend to attack it in stages, block code, then abstract, repeat until the code looks manageable.

Block

Block out your conditionals to make it easier to read.

(
  a === !0 
  || (
    perSearch.rates.fy2 
      ? (perSearch.rates.fy1.multiple || (perSearch.rates.fy2.multiple ? e() : calculateRates())) 
      : (perSearch.rates.fy1.multiple ? e()   : calculateRates())
  )
)

Abstract

Abstract out some any big easy repetitions of logic.

const x = rates => ( (rates) ? e() : calculateRates() );

(
  a === true || (perSearch.rates.fy2)
    ? ((perSearch.rates.fy1.multiple) || x(perSearch.rates.fy2.multiple))
    : x(perSearch.rates.fy1.multiple)
)

Continue

Work into the code to separate out the conditionals.

const calc = rates => ((rates) ? e() : calculateRates());
const compare = rates => {
  let fy1 = (rates.hasOwnProperty('fy1')) ? rates.fy1 : false;
  let fy2 = (rates.hasOwnProperty('fy2')) ? rates.fy2 : false;

  if (fy2) {
    if (fy1.multiple) {
      return fy1.multiple;
    } 
    return calc(fy2.multiple);
  } else {
    return calc(fy1.multiple);
  }
}

a === true || compare(perSearch.rates); 

Edit (more continue!)

Looking at this again I think it would benefit from some early returns.

Look at conditional for simplification.

• If fy2 and if fy1.multiple {return fy1.multiple}
• If not fy2 {return fy1.multiple}

• If fy2 and not fy1.multiple {return fy2.multiple}

  const calc = rates => ((rates) ? e() : calculateRates());
  const compare = rates => {
    let fy1 = (rates.hasOwnProperty('fy1')) ? rates.fy1 : false;
    let fy2 = (rates.hasOwnProperty('fy2')) ? rates.fy2 : false;
  
    // consolidate conditions leading to same place.
    if (!fy2 || (fy2 && fy1 && fy1.multiple)) {
      return calc(fy1.multiple);
    } 
    return calc(fy2.multiple);
  }
  
  a === true || compare(perSearch.rates);