just have 2 question regarding JS conditional operator, is the below 2 expression valid?
1.
if(isUser && isUser === true || isGuest && isGuest === true){
//...
}
I am wondering do I have to add () to make it like and still have the same functioning:
if((isUser && isUser === true) || (isGuest && isGuest === true)){
//...
}
const items = list.orderList && list.orderList.isUser === true || list.orderList.isGuest ? list.items : [];
I am wondering do I have to add () to make it like and functioning the same as above conditional operator:
const items = list.orderList && (list.orderList.isUser === true || list.orderList.isGuest === true) ? list.items : [];
As per Operator Precedence in the MDN docs, logical AND takes precedence over logical OR. Therefore,
expression1 || expression2 && expression3
will evaluate to
expression1 || (expression2 && expression3)
Therefore,
isUser && isUser === true || isGuest && isGuest === true
naturally evaluates to
(isUser && isUser === true) || (isGuest && isGuest === true)
anyway, so you do not need parentheses..
But since, in your second example, you want to evaluate OR then AND, you do need parentheses for it to evaluate the way you require, as
list.orderList && list.orderList.isUser === true || list.orderList.isGuest
will evaluate to
(list.orderList && list.orderList.isUser === true) || list.orderList.isGuest
Related
let a = null, b = null;
if ((a !== null) && a.active && ((a === null) ? (a === b) : true)) {
}
After babel transpile above condition is transformed into
if (a !== null && a.active && a === null ? a === b : true) {
}
and it is always returning true because ternary operator has less precedency than comparison operators
any solution not to omit parentheses in this kind of conditions?
So I m building one game which is drenching the field according to selected colour and I have 4 different conditions:
let condition1 = drenchX + 1 === mixedX && drenchY === mixedY && this.mixedFields[i].style.backgroundColor === this.service.clickedColor;
let condition2 = drenchX === mixedX && drenchY + 1 === mixedY && this.mixedFields[i].style.backgroundColor === this.service.clickedColor;
let condition3 = drenchX - 1 === mixedX && drenchY === mixedY && this.mixedFields[i].style.backgroundColor === this.service.clickedColor;
let condition4 = drenchX === mixedX && drenchY - 1 === mixedY && this.mixedFields[i].style.backgroundColor === this.service.clickedColor;
I need to get all possible combinations of this conditions, all 4 of them can be true, or 3 or 2 or just one.
This works fine but I think it is too long.
My if statement looks like this:
if((condition1 && condition2 && condition3 && condition4) || (condition1 && condition2 && condition3) || (condition1 && condition2 && condition4) || (condition1 && condition3 && condition4) || (condition2 && condition3 && condition4) || (condition1 && condition2) || (condition1 && condition3) || (condition1 && condition4) || (condition2 && condition3) || (condition2 && condition4) || (condition3 && condition4) || condition1 || condition2 || condition3 || condition4) {
// some action
}
It all works perfectly but I m looking for a shorter solution and I m still a beginner in this. So if anyone know a better way to do this that would be great, my statement is just too long and it doesn't looks good at all.
Btw, project is in Angular. Maybe to create some function for this?
If a && b is true, then a || b is also true. That said it means your condition can just be
if (condition1 || condition2 || condition3 || condition4)
If I understand correctly, the if-statement is checking whether any of the conditions are met. Therefore, this should do the trick:
if(condition1 ||
condition2 ||
condition3 ||
condition4)
{
//some action
}
Thanks everyone for advices, but I found a way to simplify it. I stored conditions into array and used for-in to loop thru all possible conditions.
let conditionalArray: boolean[] = [condition1, condition2, condition3, condition4];
for(let c in conditionalArray){
if(conditionalArray[c]) {
//....
}
}
If I compare "a" and "b", that should be false.
If I compare "a" and "a", that should be true.
If I compare "" and null, that should be true.
I could write my own method, but thought there was perhaps a JavaScript shortcut.
Edit: I was thinking something like this:
areDbSame(s1, s2) {
if (s1 === null) s1 = "";
if (s2 === null) s2 = "";
return s1 === s2;
}
Edit2: Settled on this version:
areDbSame(s1, s2) {
return (s1 === null ? "" : s1) === (s2 === null ? "" : s2);
}
Just before you test the equality of your string, you could do a simple one line enforcement, by converting to '' in the case of null. For example (if you also don't care about undefined, false, etc):
// testString becomes the one you are testing
var testString = myString || '';
If you only want to ensure null is blank
var testString = (myString === null) ? '' : myString;
Then you can simply do your string comparisons using testString, and not worry about the null equalities.
IMO this is the cleanest answer because it doesn't convolute the original equality testing of javascript strings. It is the same as saying, let's split the problem up into two parts.
1) When should my string be considered blank, and
2) Now I can just check for regular string equality.
function areEqualStrings(a, b) {
var otherEqualValues = ['', null];
if(typeof a === 'string' && typeof b === 'string') {
return a === b;
} else if(otherEqualValues.indexOf(a) > -1 && otherEqualValues.indexOf(b) > -1) {
return !a === !b;
} else {
return false;
}
}
When coercing JavaScript values, !null is true and !'' is true, so those would result in being equal.
Here's the test (screenshotted from my console):
This function should do it. It type checks first and short circuits otherwise.
function stringCompare(a, b) {
if (((a === null || typeof a === 'string') ||
(b === null || typeof b === 'string')) &&
((a === '' && b === null) ||
(b === '' && a === null) ||
(a === b))) {
return true;
}
return false;
}
No it hasn`t. The two first cases you can do naturally using operator =.
The third case it is impossible because "" is considered a empty string and null has any type. So they never can be true naturally. To do this, you have to write your own method.
Just to be clear. You can use operators = (equal) to do comparison:
== equal to
`x == 8 false
x == 5 true
x == "5" true
=== equal value and equal type
x === 5 true
x === "5" false
Hope it helps
I have two if statements that'd like to turn into a ternary but I don't know how.
if (val === true && optval === 'car' || val === true && optval === 'truck' )view_list.style.display = 'none';
if (val === true && optval === 'buss' || val === true && optval === 'van')view_list.style.display = '';
Apart in hacks I won't mention, a ternary is used when you want to return a value in all cases.
If val isn't true, then you do nothing.
Therefore this can't be used here.
view_list.style.display =
(val === true && optval === 'car' ||
val === true && optval === 'truck' ) ?
'none' :
((val === true && optval === 'buss' ||
val === true && optval === 'van') ?
'' :
view_list.style.display ) ;
or more succinctly :
view_list.style.display =
(val === true && ( optval === 'car' || optval === 'truck' ) ) ?
'none' :
((val === true && ( optval === 'buss' || optval === 'van' ) ?
'' :
view_list.style.display );
The parentheses around the second conditional are optional.
Of course this is not really a good approach, and this answer should not be seen as an endorsement to this style of programming.
Because nobody has said it yet, switch
if (val === true)
switch (optval) {
case 'car':
case 'truck':
view_list.style.display = 'none';
break;
case 'buss':
case 'van':
view_list.style.display = '';
break;
}
Not what you were asking for, but as #dystroy said, this is not the best case for a ternary.
But you can still make it a bit more elegant. How about
var styleMapping = {
'car': 'none',
'truck': 'none',
'buss': '',
'van': ''
};
// Apparently val is always required in your example
if (val) {
view_list.style.display = styleMapping[optval];
}
This also has the benefit that refactoring and adding new styles is very easy without having to write much code. Just add one line in the mapping and you're done. And of course it's much more readable and closer to what you intend to do.
You don't really care about boolean logic, you just want to have a different style based on what optval is set to.
So, I'm looking into writing a slightly more complex operation with logic operators in an if-else statement. I know I can do parentheses, and I know it's the better way of doing this, but I've gotten curious and so I'm going to ask. If I were to do something like this:
if (firstRun == true || selectedCategory != undefined && selectedState != undefined) {
//Do something
} else {
//Do something else
}
How will that be operated without the use of parentheses? I know there is an order of operations for logic operators, similar to PEMDAS, right? I'm curious if it'll be ran something like this:
firstRun == true || (selectedCategory != undefined && selectedState != undefined)
or maybe if the 'OR' operator takes precedence instead and it ends up going like:
(firstRun == true || selectedCategory != undefined) && selectedState != undefined
The full list would be nice, if you can find it somewhere, of the order of operations for this. Thanks!
My rule of thumb, which covers basically 99% of all use cases for conditional statements, is:
Grouping: ()
Member access . or [...]
Not: !
Comparison, e.g. < , >= , === , !=, ...
Logical AND &&
Logical OR ||
MDN gives you the exhaustive breakdown: JavaScript Operator Precedence
So for your example:
(firstRun == true || selectedCategory != undefined && selectedState != undefined)
equals
(firstRun == true) || ((selectedCategory != undefined) && (selectedState != undefined))
For anything more complex than the above mentioned cases, I would look into refactoring the code for readability's sake anyway!
There is a pretty good rule of thumb to this. Think of these operators as of mathematical ones:
AND is multiplication (eg. 0 * 1 = 0 => FALSE)
OR is adding (eg. 0 + 1 = 1 => TRUE)
When you remember this, all you have to know is that multiplication always comes before addition.
See this chart for precedence.
I'm not going to explain what happens because the next guy reading your code will think: "WTF? Does that do what it should?"
So the better solution is to wrap the terms in parentheses even if you know the precedence, applied it correctly and the code works
This follows the old wisdom that you shouldn't do everything you can just because you can do it. Always keep an eye on the consequences.
See Operator precedence.
&& is before ||, so your expression is equivalent to:
firstRun == true || (selectedCategory != undefined && selectedState != undefined)
It will be the first:
firstRun == true || (selectedCategory != undefined && selectedState != undefined)
As a general rule, in most programming languages, AND has higher precedence.
While logical operator precedence is not actually defined in the ECMAScript Specification, MDN does a pretty good job of it and even has a separate page for logical operators.
My concern I suppose, since logical operator precedence is not actually defined in the ECMAScript specification, each individual browser vendor can potentially be different (I'm talking to you, Internet Explorer!), so your mileage may vary.
In the event anyone wants to test this across different browsers, here's a test case fiddle: http://jsfiddle.net/HdzXq/
$(document).ready(function() {
function log(test) {
$('div#out').append($('<p />').text(test));
}
function testOperatorPrecedence() {
log('(false || false && false) === ' + (false || false && false).toString());
log('(false || false && true) === ' + (false || false && true).toString());
log('(false || true && false) === ' + (false || true && false).toString());
log('(false || true && true) === ' + (false || true && true).toString());
log('(true || false && false) === ' + (true || false && false).toString());
log('(true || false && true) === ' + (true || false && true).toString());
log('(true || true && false) === ' + (true || true && false).toString());
log('(true || true && true) === ' + (true || true && true).toString());
log('----------------------------');
log('(false || (false && false)) === ' + (false || (false && false)).toString());
log('(false || (false && true )) === ' + (false || (false && true)).toString());
log('(false || (true && false)) === ' + (false || (true && false)).toString());
log('(false || (true && true )) === ' + (false || (true && true)).toString());
log('(true || (false && false)) === ' + (true || (false && false)).toString());
log('(true || (false && true )) === ' + (true || (false && true)).toString());
log('(true || (true && false)) === ' + (true || (true && false)).toString());
log('(true || (true && true )) === ' + (true || (true && true)).toString());
}
testOperatorPrecedence();
});