How to Test If a Variable Does Not Equal Either of Two Values

How do I test if a variable does not equal either of two values?

Think of ! (negation operator) as "not", || (boolean-or operator) as "or" and && (boolean-and operator) as "and". See Operators and Operator Precedence.

Thus:

if(!(a || b)) {
  // means neither a nor b
}

However, using De Morgan's Law, it could be written as:

if(!a && !b) {
  // is not a and is not b
}

a and b above can be any expression (such as test == 'B' or whatever it needs to be).

Once again, if test == 'A' and test == 'B', are the expressions, note the expansion of the 1st form:

// if(!(a || b)) 
if(!((test == 'A') || (test == 'B')))
// or more simply, removing the inner parenthesis as
// || and && have a lower precedence than comparison and negation operators
if(!(test == 'A' || test == 'B'))
// and using DeMorgan's, we can turn this into
// this is the same as substituting into if(!a && !b)
if(!(test == 'A') && !(test == 'B'))
// and this can be simplified as !(x == y) is the same as (x != y)
if(test != 'A' && test != 'B')

How to test that variable is not equal to multiple things?

The while bit could be refactored a little to make it a little bit cleaner by checking if the element is within a list of choices like so

while choice not in [1, 2, 3]:

This is checking if the value of choice is not an element in that list

Simpler way to check if variable is not equal to multiple string values?

For your first code, you can use a short alteration of the answer given by
@ShankarDamodaran using in_array():

if ( !in_array($some_variable, array('uk','in'), true ) ) {

or even shorter with [] notation available since php 5.4 as pointed out by @Forty in the comments

if ( !in_array($some_variable, ['uk','in'], true ) ) {

is the same as:

if ( $some_variable !== 'uk' && $some_variable !== 'in' ) {

... but shorter. Especially if you compare more than just 'uk' and 'in'.
I do not use an additional variable (Shankar used $os) but instead define the array in the if statement. Some might find that dirty, i find it quick and neat :D

The problem with your second code is that it can easily be exchanged with just TRUE since:

if (true) {

equals

if ( $some_variable !== 'uk' || $some_variable !== 'in' ) {

You are asking if the value of a string is not A or Not B. If it is A, it is definitely not also B and if it is B it is definitely not A. And if it is C or literally anything else, it is also not A and not B. So that statement always (not taking into account schrödingers law here) returns true.

How do I check if a variable is not equal to multiple things in C++?

How do I check if a variable is not equal to multiple things

Is the only option to just do:

input != '1' && input != '2' && input != '3' etc etc

In the general case, for an arbitrary set of values: No, that is not the only option, but it is the simplest. And simplest is often best, or at least good enough.

If you dislike the redundant repetition of input !=, a variadic template can be used to generate the expression. I've written an example of this in another question: https://stackoverflow.com/a/51497146/2079303

In specific cases, there may be better alternatives. There exists std::isdigit for example for exactly the particular case in your example code.

In order to check if a variable is (not) equal to mutliple things which are not known until runtime, the typical solution is to use a set data structure, such as std::unordered_set.

How do I test if a variable does not equal a lot of variables - Angular

Create an array to hold the values you want to check against. And use the includes method.

In TS:

   let arr = [17,18,19];

In HTML:

*ngIf="!arr.includes(value.data)"

How do I test if a JS variable does equal either of two values?

I thought your original post was pretty straight forward and thought it was a perfectly fine answer to your problem.

Having said that you can use RegEx like @le_m mentioned. I just tested this in the console and this works.

The '//' tell the interpreter that we will be starting a regular expression. The parenthesis allow us to declare a subexpression. A subexpression that will be evaluated independently of other expressions.

The 'i' stands for insensitive. In this case it isn't entirely necessary but I think it might be useful for you down the road. The 'i' will allow you to avoid casing issues. Also, should you pass in longer strings that contain 'display', or 'type' in a sentence, this regex will also find those.

var type = 'display';
if (/(type|display)/i.test(type)) { 
  console.log('word found')
}

How to test multiple variables for equality against a single value?

You misunderstand how boolean expressions work; they don't work like an English sentence and guess that you are talking about the same comparison for all names here. You are looking for:

if x == 1 or y == 1 or z == 1:

x and y are otherwise evaluated on their own (False if 0, True otherwise).

You can shorten that using a containment test against a tuple:

if 1 in (x, y, z):

or better still:

if 1 in {x, y, z}:

using a set to take advantage of the constant-cost membership test (i.e. in takes a fixed amount of time whatever the left-hand operand is).

Explanation

When you use or, python sees each side of the operator as separate expressions. The expression x or y == 1 is treated as first a boolean test for x, then if that is False, the expression y == 1 is tested.

This is due to operator precedence. The or operator has a lower precedence than the == test, so the latter is evaluated first.

However, even if this were not the case, and the expression x or y or z == 1 was actually interpreted as (x or y or z) == 1 instead, this would still not do what you expect it to do.

x or y or z would evaluate to the first argument that is 'truthy', e.g. not False, numeric 0 or empty (see boolean expressions for details on what Python considers false in a boolean context).

So for the values x = 2; y = 1; z = 0, x or y or z would resolve to 2, because that is the first true-like value in the arguments. Then 2 == 1 would be False, even though y == 1 would be True.

The same would apply to the inverse; testing multiple values against a single variable; x == 1 or 2 or 3 would fail for the same reasons. Use x == 1 or x == 2 or x == 3 or x in {1, 2, 3}.

JavaScript: Simple way to check if variable is equal to one of two or more values?

You can stash your values inside an array and check whether the variable exists in the array by using [].indexOf:

if([5, 6].indexOf(x) > -1) {
  // ...
}

If -1 is returned then the variable doesn't exist in the array.

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