可以缩短此代码吗？

Question

This is an equation to check whether 1 of the 3 elements has value. If yes, all elements must have a value:

if(   (a!='' || b!='' || c!='')   &&   (a=='' || b=='' || c=='')   )
    alert('Please fill all elements');

Answer 1

You could use the modulo operator, making this a solution for 4, 5, or more variables as well -- just adapt the final number:

if ((!a+!b+!c)%3) 

The ! (a boolean negation) makes !a a boolean result, based on the fact that an empty string is falsy, while non-empty strings are truthy, so you get true or false in these cases respectively.

Booleans are coerced to numbers (0 or 1) when added up. So the sum is either 0, 1, 2, or 3.

The % operator gives the remainder after division by 3, which for the possible values is the original value except for 3: in that case the result is 0. So we get a non-zero value for, and only for, the cases where we want to show a validation error.

In the context of an if condition this value is coerced to a boolean value, and non-zero numerical values are truthy.

Dynamic number of inputs

If you have an array of inputs, then you can use a similar pattern with filter :

// Assume arr = [a,b,c];
if ( arr.filter(Boolean).length%arr.length )

Note how the Boolean function is used to essentially do the same as the double negation did (actually, it would be the same if I had used !! instead of ! earlier on -- now it gives the opposite boolean value, but that works just as well).

Afterthoughts

Although this answers your question, you should really ask yourself if it is worth it to reduce code that much: readable code is a much more important asset than shorter code, certainly when execution time is not influenced by it.

Answer 2

This may not seem shorter for 3 variables, but it definitely is for a greater number:

 let a = "test", b = "", c = ""; let arr = [a, b, c] let filteredArr = arr.filter(item => item === '') if (filteredArr.length > 0 && filteredArr.length < arr.length) { alert('Please fill all elements'); } 

Answer 3

You can do

if([a,b,c].find(entry => entry == '').length && [a,b,c].find(entry => entry !== '').length) {
     alert('Please fill all elements');
}

IMO is more readable as well, it first checks if any is empty, if so it check if any has data, and if both is true, asks to fill all elements

Answer 4

 if(!( !!a === !!b && !!b === !!c))

That should do it. Basically this fails if all three conditions are equally truthy or falsy.

Answer 5

So, some elements are empty, but not all elements are empty?

const emptyCount = (a=='') + (b=='') + (c=='');
if (emptyCount > 0 && emptyCount != 3) {
    alert('Please fill all elements');
}

Answer 6

Why don't you just do like

if (a === "" || b === "" || c === ""){
  alert('Please fill all elements');
}