Here is the code snippet:
var prjId = navObj.itemId || navObj
Does this mean that prjId is either equal to navObj.itemId or navObj? What does it then mean that a variable is equal to navigation object?
Thank you in advance for answers!
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This question is similar to: What does the || (logical OR, double vertical line) operator do with non-boolean operands?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem.dumbass– dumbass2024-11-12 12:39:25 +00:00Commented Nov 12, 2024 at 12:39
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5 Answers
This is equivalent to the following:
var prjId;
if(navObj.itemId)
prjId = navObj.itemId;
else
prjId = navObj;
Comments
If navObj.itemId is set to false or has not been defined at all,
prjId = navObj;
otherwise:
prjId = navObj.itemId;.
answered Feb 7, 2014 at 16:15
display-name-is-missing
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2 Comments
Jage
@p.s.w.g has the more correct answer. If navObj.itemId was set to false, it would be defined and yet would set prjId to navObj.
display-name-is-missing
@Jage correct. Didn't think about that. Updated answer to cover that corner.
No. The || operator first tries to convert navObj.itemId to a Boolean value.
It will be converted to
trueif it is already the Boolean value,true, a number other than0orNaN, a non-empty string, or an object that is notnullorundefined. These are known as "truthy" values.It will be converted to
falseif it is already the Boolean valuefalse,0,NaN, an empty string,nullorundefined. These are known as "falsey" values.
If navObj.itemId is "truthy", navObj.itemId is assigned to prjId, otherwise navObj is assigned to prjId.
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3 Comments
Mauro Sampietro
I'm not js expert but that's not a ternary operator. To me it would mean that prjId is true if one of those condition is true. isn't it?
Jason M. Batchelor
No, as I understand it, p.s.w.g's answer is not even discussing a ternary, but the /quality/ ("truthy" vs. "falsey") if you will of the items on either side of the
|| operator. If the /quality/ of the left side is truthy, then the variable is set to the /value/ of the left side. If not, the right side is used.
p.s.w.g
@user711061 Not exactly. The
|| operator in JS is a kind of simplified version of the ?: operator. x || y is equivalent to x ? x : y. The result is x or y, not the truth values associated with them.It simply means that if the left operand of the logical or operator (||) is a truthy value then return it otherwise return the right operand.
The following values are always falsy:
- false
- 0
- empty string
- null
- undefined
- NAN (not a number)
So if navObj.itemId doesn't evaluate to anything of the above, then it will be assigned to the prjId variable.
This is widely used when we have optional parameters in a function, for example. It is a way of specifying the default value for an optional parameter. But of course that's not the only use of it.
Comments
It sets prjId to the property of itemId on navObj, if it exists (evaluates to 'truthy'), if it doesn't exist (or it evaluates to 'falsy'), prjId gets set to navObj.
