JavaScript 中的三元运算符条件为真时才赋值

Question

Is it possible to do something like this in JavaScript?

max = (max < b) ? b;

In other words, assign value only if the condition is true. If the condition is false, do nothing (no assignment). Is this possible?

Answer 1

Don't use the ternary operator then, it requires a third argument. You would need to reassign max to max if you don't want it to change ( max = (max < b) ? b : max ).

An if-statement is much more clear:

if (max < b) max = b;

And if you need it to be an expression, you can (ab)use the short-circuit-evaluation of AND :

(max < b) && (max = b)

Btw, if you want to avoid repeating variable names (or expressions?), you could use the maximum function :

max = Math.max(max, b);

Answer 2

An expression with ternary operator must have both values, ie for both the true and false cases.

You can however

max = (max < b) ? b : max;

in this case, if condition is false, value of max will not change.

Answer 3

You can just set max to itself if the condition is false.

max = (max < b) ? b : max;

Or you can try using the && operator:

(max < b) && (max = b);

Or to keep your code simple, just use an if .

if(max < v) max = b;

Answer 4

我认为三元更合适试试这个

(max < b) ? max = b : '';

Answer 5

没有不是三元运算符的特定运算符，但您可以像这样使用它：

max = (max < b) ? b : max;

Answer 6

我认为更好的方法可能是

max = Math.max(max, b)

Answer 7

你可以这样做：

(max < b) ? max = b : ''

Answer 8

you can try:

(max < b) && (max = b);

look at this example:

 let max = 10; let b = 15; (max < b) && (max = b)// this will be true console.log("max=", max); let maxx = 10 let bb = 5; (maxx < bb) && (maxx = bb)// this will be false console.log("maxx=", maxx);

Answer 9

The ternary operator is used where we have a minimum of two possible outcomes.

let no = 10;
let max = 20;

max = no > max ? no : max

if you only want to use if, not else, then I recommend you to use if not the ternary operator.

let no = 10;
let max = 20;
    
if (no > max){
   max = no
}

if you want to use ternary operator, but don't want to use else part, then you may use this, but this won't be useful in every problem

let no = 10;
let max = 20;

no > max ? max = no : 0

here we are assigning value to a max variable only when the condition is true, otherwise, we are not doing anything

Answer 10

Performace Results

As most of the answers pretty much tell you almost every possible method, here are some performance results for the four most common methods over 40 million iterations and an average of 10 runs each.

Using Ternary

For a = b > a? b: a a = b > a? b: a the average time was for iterations over 10 runs
Whereas for (b > a) && (a = b) the average time was for iterations over 10 runs

Using Math.max

for a = Math.max(a, b) the average time was for iterations over 10 runs

Using an if statement

for if (b > a) a = b the average time was for iterations over 10 runs

Answer 11

look:

(max < b) ? max = b : null;

it's going to work out, I think you wanna avoid using a colon, unfortunately, it won't happen

Answer 12

There was no example of ES6, we can use in this way:

let list = [{id: "abc", name: "test1"}, {id: "xyz", name: "test2"}]
let selectedData = {};
      list.some((exp) => {
        return (
          exp.id == "xyz" &&
          ((selectedData = exp), true)
        );
      })
console.log(selectedData);

---

Added description after a negative vote: Just adding a description to explain my solution. If you have an array of objects and you want if only one condition satisfies then we can use some . It will set the selectedData when the id == "xyz" . As asked in the question, this will assign value only if the condition is true otherwise selectedData will be empty.

Answer 13

Instead of the ternary operator , you could use Logical AND assignment (&&=) .

Basic examples

 let output = false; output &&= true; console.log(output)

Reassignment is not possible because output has been defined as false .

 const b = 10; const max = 1; let sum = max < b; sum &&= 'new value'; console.log(sum);

Returns a string, because the variable sum has been defined as true .