CODEWITHSUNDEEP
javascriptif-statementswitch-statement
loid
loid

Reputation: 27

switch, else if & or (Beginner Q)

I have a function which appears to always be returning true on the first condition irrelevant of input. I have been reading about using switch versus else if as I have 16 conditions to check and I want to ensure I have working 'best practice'

Can I achieve the same thing using both options:

function lsaupdateTotals() {
  var x = variablename.value;
  var y = variablename2.value;


  if (x = 1) || (y = y > 1 && y < 280) {
      rdlsa = 7.45;
    } else if (x = 2 || (y = y > 281 && y < 460)) {
      rdlsa = 11.65;

      / or switch: /

      switch (x) {
        case 1:
        case y > 1:
        case y < 280:
          y = 7.45;
          break;
      }

Upvotes: 0

Views: 59

Answers (3)

Adam Coster
Adam Coster

Reputation: 1472

You've got a few issues in your code:

  • You're using = instead of == or === to check equality in your if statements. A single = sign always means "set equal to", not "is equal to?".
  • In your if statement your parens make things a bit ambiguous. It's possible that this works just fine, but wrapping the entire question in parens is guaranteed to work as intended while also being completely clear.

Rewriting your if statement according to the above:

  if (x == 1 || (y == y > 1 && y < 280)) {
    rdlsa = 7.45;
  }
  else if (x == 2 || (y == y > 281 && y < 460)) {
    rdlsa = 11.65;
  }

(EDIT: Note that the y == y > 1 part is almost definitely not doing what you want it to. That's asking "is y the same thing as y > 1?")

In your switch, think of the value in each case as being a place holder for what you're putting into it. So in your example, using y>1 doesn't make sense to evaluate against x, because it's asking if x is *equal to y>1, but y>1 is always true or false and is independent of x.

Upvotes: 2

Minal Shah
Minal Shah

Reputation: 1486

you are getting the conditions wrong. Please replace your code with the below lines

function lsaupdateTotals() {
var x = variablename.value;
var y = variablename2.value;


if ((x == 1) || (y > 1 && y < 280)) {
  rdlsa = 7.45;
  } else if ((x == 2) || (y > 281 && y < 460)) {
  rdlsa = 11.65;
 }

  / or switch: /

  switch (x) {
    case 1:
    case y > 1:
    case y < 280:
      y = 7.45;
      break;
  }

Upvotes: 1

Raffaele
Raffaele

Reputation: 767

There are several problem in your code:

  1. In javascript, to compare 2 numbers (or strings) you have to use the syntax ===, so if (x = 1) should became if (x === '1') (as I'm expecting x is a string).
  2. The if condition should be in a parenthesis: if (x = 1) || (y = y > 1 && y < 280) { => if ((x === 1) || (y === y > 1 && y < 280)) {
  3. It's not clear what you mean with y = y > 1 (or y === y > 1) in first if (second parenthesis)
  4. In switch/case syntax you cannot use y>1, please refer to switch/case syntax (internet is full of documentation)
  5. When you put an assignment into the if (using = instead of ===) the if statement consider true the condition if the value after the = is not null, 0, empty string, false or undefined, for this reason when you write if(x=1){ the condition is always true.

Upvotes: 4

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