In the code below, because s is null d = "test" but if s = "hello" then d would = "hello".
Is this correct as it works? what is the correct way to use ||
var s = null;
var d = s || "test";
alert(d);
|| is "or" ; and understanding what happens here is a bit trickery
var a=false;
var b=true;
result=a||b
will give "result" true (as b is true). What happens is:
if you had
var a=true;
var b="test";
result=a||b
result will yield true; as no other value needs to be checked by the logic of "||"
with
var a=null;
var b="test";
result=a||b;
a will be checked first - it is null, which converts to "false". b is "test", which is non-null, and converts to "true". so the value of b will be assigned.
And yes, this is a correct way to use || ; the feature you use is also called short-circuit evaluation (as it stops evaluating the boolean expression as early as possible)
This works, but if s evaluates to a 'falsy' value, you'll get your default, which might not be what you intended. A more robust, but wordy idiom is
d = (typeof s === "undefined") ? "test" : s;
Yes it is correct unless s is allowed to be blank or 0 which are also falsy values
var s = null;
var d = s || "test";
var s = 0;
var d = s || "test";
var s = "";
var d = s || "test";
All will result in d being "test"
||
is a logical operator. When s is not null then the condition of (s) is true so d is assigned the value of s, otherwise it is assigned 'test'
||
is the OR
operator in javascript
so a||b
means a OR b
in simple terms
explanation of question you have asked is that id you simply do somethings like these in js you will ultimately get in the else block
if(null)
if(undefined)
so s||"test"
will mean which ever is not null or undefined which in this case is test
yes correct, || symbols just does the job of OR. when the first condition is true its gonna return that one.. else it will move to the next... simple as it is...
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