There is code:

function search(list, q){
  var result = {};
  for(let id in list)(
    (!q.id    || (id == q.id)) &&
    (!q.name  || (list[id].name.search(q.name) > -1)) &&
    result[id] = list[id]
  );

  return result;
}

I get this error:

Uncaught ReferenceError: Invalid left-hand side in assignment    script.js:4

Why "&&" is wrong?

There is code:

function search(list, q){
  var result = {};
  for(let id in list)(
    (!q.id    || (id == q.id)) &&
    (!q.name  || (list[id].name.search(q.name) > -1)) &&
    result[id] = list[id]
  );

  return result;
}

I get this error:

Uncaught ReferenceError: Invalid left-hand side in assignment    script.js:4

Why "&&" is wrong?

• javascript
• angularjs
• ecmascript-6

Share Improve this question Follow edited Dec 6, 2017 at 6:43 Vala Khosravi 2,57033 gold badges2525 silver badges5454 bronze badges asked Dec 1, 2017 at 3:30 Turar AbuTurar Abu 36811 gold badge33 silver badges1212 bronze badges 5

• are you missing if before the first set of parentheses? – Dan O Commented Dec 1, 2017 at 3:32
• 1 result[id] = list[id] should be result[id] === list[id] as it is a condition – Aravind Commented Dec 1, 2017 at 3:33
• @Aravind that is not correct. – Pointy Commented Dec 1, 2017 at 3:38
• @DanO "if" must not be important, because all logical operator in ( ), not { } – Turar Abu Commented Dec 1, 2017 at 3:43
• @TurarAbu see my answer: the problem is that your assignment expression at the end of the && list is not in parentheses. – Pointy Commented Dec 1, 2017 at 3:43

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2 Answers 2

Sorted by: Reset to default 7

The problem is that the assignment operator, =, is a low-precedence operator, so it's being interpreted in a way you don't expect. If you put that last expression in parentheses, it works:

  for(let id in list)(
    (!q.id    || (id == q.id)) &&
    (!q.name  || (list[id].name.search(q.name) > -1)) &&
    (result[id] = list[id])
  );

There seems to be a typo in your code:

result[id] = list[id] should be result[id] == list[id] or result[id] === list[id](if you're doing a strict parison)