if statement - multiple conditions in an if - else in javascript - Stack Overflow

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I'm trying to make a program that only accepts a few values. So, if "e" variable is not 1 or 2 or 3, says that the number is not correct, but if the value is equal to those numbers, then the else part is run.

All of this may sound very begginer level and easy to implement, and it is, but I ran the code and EVERY vaule I set to "e" runs the if part.

Here is code:

var e;

e=parseFloat(prompt("Input e",""));

 if(e!=1 || e!=2 || e!=3)
  {
  alert("put again E");
  }

 else
  {
  //whatever
  }

I'm trying to make a program that only accepts a few values. So, if "e" variable is not 1 or 2 or 3, says that the number is not correct, but if the value is equal to those numbers, then the else part is run.

All of this may sound very begginer level and easy to implement, and it is, but I ran the code and EVERY vaule I set to "e" runs the if part.

Here is code:

var e;

e=parseFloat(prompt("Input e",""));

 if(e!=1 || e!=2 || e!=3)
  {
  alert("put again E");
  }

 else
  {
  //whatever
  }

• javascript
• if-statement
• conditional-statements

Share Improve this question Follow edited Apr 14, 2015 at 7:02 n-dru 9,44022 gold badges3030 silver badges4343 bronze badges asked Apr 14, 2015 at 7:00 MasterGeekMXMasterGeekMX 6311 gold badge11 silver badge77 bronze badges 3

• Try doing parseInt instead of parseFloat and check for their equality instead of != – Abhinav Gauniyal Commented Apr 14, 2015 at 7:03
• 4 For which value of e is (e!=1 || e!=2 || e!=3) false? Think about it. – Biffen Commented Apr 14, 2015 at 7:03
• So many answers lol check this code : jsfiddle/v6h7gvs3 – Abhinav Gauniyal Commented Apr 14, 2015 at 7:06

Add a ment  | 

5 Answers 5

Sorted by: Reset to default 8

In English you said "not 1 or 2 or 3", but that is written as !(e == 1 || e == 2 || e == 3); or, you could go with the logically equivalent "not 1, and not 2, and not 3", expressed as e != 1 && e != 2 && e != 3.

What you wrote is "not 1 or not 2 or not 3". If the value is 1, then it is not 2 (and also not 3), so "not 1 or not 2 or not 3" is still true. In fact, it is true for any value, because at least two of those (if not all three) will be true.

It's because || means or.

if(e!=1 || e!=2 || e!=3)

If you input e = 1 you will have

if(false OR true OR true)

which of course, evaluates to true.

You want &&, which means and, resulting in:

if(e!=1 && e!=2 && e!=3)

if you want to maintain your code structure. Or you could take the advice of the others, and put your "else" code into the "if" block and use ==.

Since e can't have the value of 1,2 and 3 simultaneously, your condition is always going to evaluate to true. your version reads

if the value is different from 1 or different from 2 or different from 3 then do this.

So you will need to change it to something that reads more like

if the value is not either 1 or 2 or 3 then

if(!(e == 1 || e == 2 || e == 3)){...}

or you could do

if(e != 1 && e != 2 && e != 3){...}

which would read

if the value isn't 1 and isn't 2 and isn't 3

The result of those two options would be the same.

You should use == rather than != like:

 if(e==1 || e==2 || e==3)//then re enter value

With your if, you mean if e is neither of 1,2,3 then ask user to reenter the value.

The reason is simple. If for example you input the value 1 the first part of your condition returns false, but the other 2 parts return true, then the condition can be read like this:

if( false || true || true ) {
 ...
}

So no matter what input, there will always be 2 true values against a false value. To get what you want, use && instead of ||.

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