Question

在javascript,我有:

var inf = id +  |  + city ;

如果补贴或城市无效,则无效。

如果id夫或城市无效,则是否有任何动摇的表述方式。

我知道,你可以做以下工作:

var inf = (id ?? "") + (city ?? "");

javascript中的任何类似方法?

Answer 1

var inf = (id == null ?    : id) +  |  + (city == null ?    : city)

Answer 2

如何:

var inf = [id, city].join( | );

EDIT: You can remove the "blank" parts before joining, so that if only one of id and city is null, inf will just contain that part and if both are null inf will be empty.

var inf = _([id, city]).compact().join( | ); // underscore.js
var inf = [id, city].compact().join( | ); // sugar.js
var inf = [id, city].filter(function(str) { return str; }).join( | ); // without helpers

Answer 3

总长枪,但试图:

var inf = (id || "") + "|" + (city || "");

Answer 4

function nullToStr(str) {
    return !str || 0 === str.length ?    : str;
}

使用:

console.log(nullToStr(null);

        function nullToStr(str) {
    return !str || 0 === str.length ?    : str;
}

console.log(`The value which was null = ${nullToStr(null)}`);
console.log(`The value which was null = ${nullToStr(undefined)}`);
console.log(`The value which was null = ${nullToStr(  )}`);
console.log(`The value which was null = ${nullToStr( asdasd )}`);

我认为,这将有助于而且更容易加以利用。很显然,它基于这一深处发现的其他答案。

Answer 5

var inf = (id && city) ? (id+"|"+city) : "";

Answer 6

Equivalent to c# var inf = (id ?”) + (city ?”; (if id and city are unable)

var inf = (id ||   ) + (city ||   );

参见:。 Short-Circuit Evaluation。可否赔偿不是javascript中的问题(在js中,所有变量都是无效的),但必须分配<代码>id和city (但不需要价值,如var id,城市。

Answer 7

if (id != ""){
inf = id;
if (city != ""){ inf += " | " + city ;}
}
else 
inf= city;

友情链接