在JavaScript中测试变量是否为null，而不会导致未定义的错误

Question

I am trying to display the results of various javascript null checking techniques. codepen

However when my variable is actually null It breaks and I get no html output. I get an error in the console "y is undefined". I know it is undefined but I am trying to get it to go through the loop to give you each of the tests results. I can't seem to figure out what could be wrong. Any help would be great.

//var y = "im y";

var test1 = function () {
if (typeof (y) === 'undefined') {
    return "test 1 y is undefined";
} else {
    return "test 1 y is defined";
}
};

var test2 = function () {
if (!y) {
    return "test 2 y is undefined";
} else {
    return "test 2 y is defined";
}
};

var test3 = function () {
if (y === null) {
    return "test 3 y is undefined";
} else {
    return "test 3 y is defined";
}
};

var test4 = function () {
if (y == null) {
    return "test 4 y is undefined";
} else {
    return "test 4 y is defined";
}
};

var test5 = function () {
if (typeof (y) === undefined) {
    return "test 5 y is undefined";
} else {
    return "test 5 y is defined";
}
};

var a = [test1, test2, test3, test4, test5];
var b = [test1(), test2(), test3(), test4(), test5()];
var somehtml = [];

$.each(a, function (index) {
somehtml.push('<pre>' + a[index] + '</pre>');
var x = b[index];
somehtml.push('<p>' + x + '</p>');
});
$("div#stuff").html(somehtml.join(""));

Answer 1

simple check like below

if (y) {
alert("defined");
return "test 1 y is defined";    

} else {
alert("not defined");
 return "test 1 y is undefined";   
}

Answer 2

First of all, undefined != null. Second, undefined != 'undefined'.

Your code examples run as expected. I suggest you read about javascript type coercion and truthy/falsy values . Then start using "use strict"; at the top of your javascript files to avoid confusion.

Answer 3

Sorry f I am not getting it properly.I understood your issue from the comments only. .You are getting the error in console because you are running the other tests like

if (!y) {

and

if (y === null) {

even when y is undefined , which will trigger error in your console.

Your logic should be like, if the variable is found undefined then dont do the other tests and else do them.

the test

if (typeof (y) === 'undefined') {

will be true only if the variable is undefined and null values will pass through

updated according to below comments

pls see this fiddle . This has the idea I conveyed. Also one more suggestion would be you can think of chaining the functions instead of triggering them in a loop.