JavaScript 为空
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Javascript isnull
提问by Phillip Senn
This is a really great function written in jQuery to determine the value of a url field:
这是一个用 jQuery 编写的非常棒的函数，用于确定 url 字段的值：
$.urlParam = function(name){
var results = new RegExp('[\?&]' + name + '=([^&#]*)').exec(window.location.href);
return results[1] || 0;
}
// example.com?someparam=name&otherparam=8&id=6
$.urlParam('someparam'); // name
$.urlParam('id'); // 6
$.urlParam('notavar'); // null
http://snipplr.com/view/11583/retrieve-url-params-with-jquery/
http://snipplr.com/view/11583/retrieve-url-params-with-jquery/
I would like to add a condition to test for null, but this looks kind of klunky:
我想添加一个条件来测试 null，但这看起来有点笨拙：
if (results == null) {
return 0;
} else {
return results[1] || 0;
}
Q: What's the elegant way to accomplish the above if/then statement?
问：完成上述 if/then 语句的优雅方式是什么？
回答by Brad
return results == null ? 0 : (results[1] || 0);
回答by Rich
return results == null ? 0 : ( results[1] || 0 );
回答by Dan Davies Brackett
the most terse solution would be to change return results[1] || 0;to return (results && results[1]) || 0.
最简洁的解决方案是更改return results[1] || 0;为return (results && results[1]) || 0.
回答by streetparade
You could try this:
你可以试试这个：
if(typeof(results) == "undefined") {
return 0;
} else {
return results[1] || 0;
}
回答by KingRider
Why not try .test()? ... Try its and best boolean (true or false):
为什么不试试.test()？... 尝试它和最好的布尔值（真或假）：
$.urlParam = function(name){
var results = new RegExp('[\?&]' + name + '=([^&#]*)');
return results.test(window.location.href);
}
Tutorial:http://www.w3schools.com/jsref/jsref_regexp_test.asp
回答by Fatih
I'm using this function
我正在使用这个功能
function isNull() {
for (var i = 0; i < arguments.length; i++) {
if (
typeof arguments[i] !== 'undefined'
&& arguments[i] !== undefined
&& arguments[i] != null
&& arguments[i] != NaN
&& arguments[i]
) return arguments[i];
}
}
test
测试
console.log(isNull(null, null, undefined, 'Target'));
回答by Teun D
return (results||0) && results[1] || 0;
The && operator acts as guard and returns the 0 if results if falsy and return the rightmost part if truthy.
&& 运算符充当守卫，如果结果为假则返回 0，如果为真则返回最右边的部分。
回答by Dan Ochiana
All mentioned solutions are legit but if we're talking about elegance then I'll pitch in with the following example:
所有提到的解决方案都是合法的，但如果我们谈论的是优雅，那么我将使用以下示例进行介绍：
//function that checks if an object is null
var isNull = function(obj) {
return obj == null;
}
if(isNull(results)){
return 0;
} else {
return results[1] || 0;
}
Using the isNull function helps the code be more readable.
使用 isNull 函数有助于代码更具可读性。
回答by Dan Ochiana
I prefer the style
我更喜欢这种风格
(results || [, 0]) [1]
回答by pixeline
if (typeof(results)!='undefined'){
return results[1];
} else {
return 0;
};
But you might want to check if results is an array. Arrays are of type Object so you will need this function
但是您可能想检查结果是否是一个数组。数组是 Object 类型，所以你需要这个函数
function typeOf(value) {
var s = typeof value;
if (s === 'object') {
if (value) {
if (value instanceof Array) {
s = 'array';
}
} else {
s = 'null';
}
}
return s;
}
So your code becomes
所以你的代码变成
if (typeOf(results)==='array'){
return results[1];
}
else
{
return 0;
}
