I'm wondering how to combine and (&&) with or (||) in JavaScript.
I want to check if either both a and b equal 1 or if both c and d equal 1.
I've tried this:
if (a == 1 && b == 1 || c == 1 && d == 1)
{
//Do something
}
But it doesn't work.
How can I write this condition correctly?
&& precedes ||. == precedes both of them.
From your minimal example I don't see, why it doesn't achieve your desired effect. What kind of value types do a–d have? JavaScript might have some non-obvious type coercion going on. Maybe try comparing with === or convert to numbers explicitly.
Side note: many lint tools for C-like languages recommend to throw in parentheses for readability when mixing logical operators.
Operator Precedence can be overridden by placing the expression between parenthesis.
if ((+a == 1 && +b == 1) || (+c == 1 && +d == 1)) // Use brackets to group them
{
// your code
}
This will prevent you from such cases like if(0&&0 || 1&&1) .
Well now that I've finished telling everybody else (except David) why their answers are wrong, let me give them the same chance to hassle me.
Your existing code as shown should already do what you seem to be describing. But is it possible that when you say:
"I want to check if either both a and b equals 1 or if both c and d equals 1."
...your use of the word "either" mean that you want to check if one and only one of the following conditions is true:
both a and b are 1, but c and d are not both 1
both c and d are 1, but a and b are not both 1
That is, you want one pair of variables to be 1, but you don't want all four variables to be 1 at the same time?
If so, that is an exclusive OR operation, which in JS is the ^ operator:
if ((a == 1 && b == 1) ^ (c == 1 && d == 1)) {
Note that unlike with a logical OR || operator, you have to add parentheses around the AND && parts of the expression because ^ has higher precendence. (^ is actually a bitwise operator, but it will work for you here since the operands you'd be using with it are all booleans.)
place some extra brackets to differentiate the and n or conditions
if ((a == 1 && b == 1) || (c == 1 && d == 1))
Related
The normal way to write an if-statement is
if (a == b) {a=1}
but in eg. Perl it is possible to write the same as
a=1 if (a == b)
Question
Is a similar syntax possible with EcmaScript?
There is not a specific statement to do that. You still have several options though
if (a == b) {
a = 1
}
if (a == b) a = 1 // No parenthesis
a == b ? a = 1 : null // Using ternary operator
a = a == b ? 1 : a // Using ternary operator
What I'm trying to is run some code if the number of input[type="checkbox"] is checked equal to the value of 2 or 3.
Here is an example of my JQuery:
if ($('input:checkbox:checked').length == 2 or 3) {
// run some code
}
Im not sure how to program it to understand whether the value is 2 or 3.
Any ideas?
Use the or operator ||
if ($('#id').length === 2 || $('#id').length === 3) {
// run some code
}
Please read up on Javascript logical operators. This is basic stuff.
Logical operators are used to determine the logic between variables or values.
Given that x = 6 and y = 3, the table below explains the logical operators:
Operator Description Example
========================================================
&& and (x < 10 && y > 1) is true
|| or (x === 5 || y === 5) is false
! not !(x === y) is true
I have a if condition like so:
$(document).ready(function(){
var name = 'Stack';
var lastname = 'Overflow';
if( name == 'Stack' && lastname == 'Overflow' )
alert('Hi Stacker!');
});
So my alert is fired...
If I put my condition inside a brackets like this:
$(document).ready(function(){
var name = 'Stack';
var lastname = 'Overflow';
if( (name == 'Stack') && (lastname == 'Overflow') ){
alert('Hi Stacker!');
}
});
My alert is also fired...
My question is: When and why I should use parentheses inside my if condition? Thank you!
There is no much difference in your example.
You can read that for reference
Operator Precedence (JavaScript)
You use it to force associations, in your example this won't change anything. But consider this case:
A and B or C
is a lot different from
A and (B or C)
Here you would be preventing the expression from being understood as (A and B) or C which is the natural way javascript and maths do things.
Because of operator precedence :
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Operator_Precedence
you'd better read this if you want more details
But basically == (equality) weights more than && (logical-and) , so A == B is evaluated before C && D given
C <- A == B
and
D <- E == F
so adding parenthesis to C or D dont matter,in that specific situation.
Brackets can be used, if multiple conditions needs to be checked.
For ex: User is also a Stacker if full name is 'StackOverflow' then add another condition with 'Or' operator.
if(((name == 'Stack') && (lastname == 'Overflow')) || FullName =='StackOverflow')
As you can see, name and last name are placed inside one bracket meaning that it gives a result either true or false and then it does OR operation with FullName condition.
And having brackets around name, lastname and fullname fields are optional since it doesn't make any difference to the condition. But if you are checking other condition with FullName then group them into bracket.
Brackets are never required in such situations.
Of course, try to read first solution (without) and second solution (with). Second one it's clear, fast-readable and easy to mantain.
Of course if you have to change precedence (in this case you just have an AND condition, but what if you need and AND and an OR? 1 and 2 or 3 priority changes between "(1 and 2) or 3" - "1 and (2 or 3)"
Brackets () are used to group statements.
Ex - if you have an expression '2 + 3 * 5'.
There are two ways of reading this: (2+3)*5 or 2+(3*5).
To get the correct o/p based on operator precedence, you have to group the correct expression like * has higher precedence over +, so the correct one will be 2+(3*5).
first bracket requires for complex condition to ensure operator precedence correctly. lets say a example
var x=1,y=1,z=0;
if(x==0 && y==1 || z==0)
{
//always true for any value of x
}
what will happen here
(0 && 1 ||1) ----> (0 ||1)----->1
&& has high precedence over ||
if(x==0 && (y==1 || z==0)) alert('1');
{
//correct way
}
if you do not use (y==1 || z==0) bracket the condition always will be true for any value of x.
but if you use (..) the condition return correct result.
Conditional statements can be grouped together using parenthesis. And is not only limited to if statements. You can run the example below in your Chrome Developer Tools.
Example 1:
Console Execution
false && false || true
// true
flow
false && false || true
| | |
|________| |
| |
| |
false or true
| |
|_________________|
|
|
true
Example 2:
Console Execution
false && (false || true)
// false
flow
false && (false || true)
| |
|______________|
|
|
false
Helpful resources for playing around with JSInterpreter and AST's:
https://neil.fraser.name/software/JS-Interpreter/
https://esprima.org/demo/parse.html#
Consider the statement
if( i==10 || j == 11 && k == 12 || l == 13)
what you would want is if either i is 10 or j is 11 And either k is 12 or l is 13 then the result shouldbe true, but say if i is 10, j is 11, k is 10 and l is 13 the condition will fail, because the fate of equation is decided at k as aoon as && comes in picture. Now if you dont want this to happen the put it like this
if( (i==10 || j == 11) && (k == 12 || l == 13))
In this ORs will be executed first and the result will be true.
a = 1;
b = "1";
if (a == b && a = 1) {
console.log("a==b");
}
The Javascript code above will result in an error in the if statement in Google Chrome 26.0.1410.43:
Uncaught ReferenceError: Invalid left-hand side in assignment
I think this is because the variable a in the second part of the statement &&, a=1 cannot be assigned. However, when I try the code below, I'm totally confused!
a = 1;
b = "1";
if (a = 1 && a == b) {
console.log("a==b");
}
Why is the one statement right but the other statement wrong?
= has lower operator precendence than both && and ==, which means that your first assignment turns into
if ((a == b && a) = 1) {
Since you can't assign to an expression in this way, this will give you an error.
The second version is parsed as a = (1 && a == b); that is, the result of the expression 1 && a == b is assigned to a.
The first version does not work because the lefthand side of the assignment is not parsed as you expected. It parses the expression as if you're trying to assign a value to everything on the righthand side--(a == b && a) = 1.
This is all based on the precedence of the various operators. The problem here stems from the fact that = has a lower precedence than the other operators.
Because the order of operations is not what you expect. a == b && a = 1 is equivalent to (a == b && a) = 1 which is equivalent to false = 1.
If you really want to do the assignment, you need to use parentheses around it: a == b && (a = 1).
In if (a = 1 && a == b),
The operations to be first performed is 1 && a == b. 1 && the result of a == b is performed. The result of this && operation is assigned to a.
It is slightly hard to explain but I want to do something that looks like this:
if(a === 4 && b === true && c === "words" || "numbersandwords")DoSomething();
but it ends running without it matching the first operators. I want to know how to have the last operator except 2 different inputs while still making sure the other criteria are met before running.
You just need to use parentheses, e.g.:
if(a == 4 && b == true && (c == "words" || c == "numbersandwords")) { DoSomething(); }
Just use a few brackets to separate your or parts and the and parts, and add the c === before the last string. Without that equality part at the end, the 'numbersandwords' string always equates to true.
if(a === 4 && b === true && (c === "words" || c === "numbersandwords")){
DoSomething();
}
In JavaScript, like other languages, every operator (like && and ||) has a precendence that determines the order in which it's evaluated. && has higher precedence than ||, so it's evaluated first. Therefore, every term on the left is anded together. Even if they are all false, however, the overall result is true because it's ored with "numbersandwords", which evaluates to true (as does everything except 0, -0, null, false, NaN, undefined, or the empty string). The first thing you need to do is actually compare something (presumably c) to it. Then you can change the order of evaluation using parentheses, which has higher precedence than anything else:
if(a === 4 && b === true && (c === "words" || c === "numbersandwords")) DoSomething();
Alternatively, you can break the test up into several if statements if you may want to eventually do something slightly different based on the value of c (or it just better expresses your intent):
if(a === 4 && b === true)
{
if(c === "words" || c === "numbersandwords")
{
DoSomething();
}
}