string 仅当字符串不为 null 或为空时才使用分隔符连接字符串

Join strings with a delimiter only if strings are not null or empty

提问by froadie

This feels like it should be simple, so sorry if I'm missing something here, but I'm trying to find a simple way to concatenate only non-null or non-empty strings.

这感觉应该很简单，如果我在这里遗漏了一些东西，很抱歉，但我试图找到一种简单的方法来连接非空或非空字符串。

I have several distinct address fields:

我有几个不同的地址字段：

var address;
var city;
var state;
var zip;

The values for these get set based on some form fields in the page and some other js code.

这些值是根据页面中的某些表单字段和其他一些 js 代码设置的。

I want to output the full address in a div, delimited by comma + space, so something like this:

我想在 a 中输出完整地址div，以逗号 + 空格分隔，所以是这样的：

$("#addressDiv").append(address + ", " + city + ", " + state + ", " + zip);

Problem is, one or all of these fields could be null/empty.

问题是，这些字段中的一个或所有字段可能为空/空。

Is there any simple way to join all of the non-empty fields in this group of fields, without doing a check of the length of each individually before adding it to the string?

是否有任何简单的方法可以连接这组字段中的所有非空字段，而无需在将其添加到字符串之前单独检查每个字段的长度？

回答by georg

Consider

考虑

var address = "foo";
var city;
var state = "bar";
var zip;

text = [address, city, state, zip].filter(Boolean).join(", ");
console.log(text)

.filter(Boolean)(which is the same as .filter(x => x)) removes all "falsy" values (nulls, undefineds, empty strings etc). If your definition of "empty" is different, then you'll have to provide it, for example:

.filter(Boolean)（与 相同.filter(x => x)）删除所有“虚假”值（空值、未定义、空字符串等）。如果您对“空”的定义不同，那么您必须提供它，例如：

 [...].filter(x => typeof x === 'string' && x.length > 0)

will only keep non-empty strings in the list.

只会在列表中保留非空字符串。

--

——

(obsolete jquery answer)

（过时的jquery答案）

var address = "foo";
var city;
var state = "bar";
var zip;

text = $.grep([address, city, state, zip], Boolean).join(", "); // foo, bar

回答by aga

Yet another one-line solution, which doesn't require jQuery:

另一个单行解决方案，不需要jQuery：

var address = "foo";
var city;
var state = "bar";
var zip;

text = [address, city, state, zip].filter(function (val) {return val;}).join(', ');

回答by Tim Santeford

Lodashsolution: _.filter([address, city, state, zip]).join()

Lodash解决方案：_.filter([address, city, state, zip]).join()

回答by tikhpavel

Just:

只是：

[address, city, state, zip].filter(Boolean).join(', ');

回答by Alexander Abakumov

@aga's solutionis great, but it doesn't work in older browsers like IE8 due to the lack of Array.prototype.filter()in their JavaScript engines.

@aga 的解决方案很棒，但由于其 JavaScript 引擎中缺少Array.prototype.filter()，因此它在 IE8 等较旧的浏览器中不起作用。

For those who are interested in an efficient solution working in a wide range of browsers (including IE 5.5 - 8) and which doesn't require jQuery, see below:

对于那些对在各种浏览器（包括 IE 5.5 - 8）中工作且不需要 jQuery 的高效解决方案感兴趣的人，请参见下文：

var join = function (separator /*, strings */) {
    // Do not use:
    //      var args = Array.prototype.slice.call(arguments, 1);
    // since it prevents optimizations in JavaScript engines (V8 for example).
    // (See https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/arguments)
    // So we construct a new array by iterating through the arguments object
    var argsLength = arguments.length,
        strings = [];

    // Iterate through the arguments object skipping separator arg
    for (var i = 1, j = 0; i < argsLength; ++i) {
        var arg = arguments[i];

        // Filter undefineds, nulls, empty strings, 0s
        if (arg) {
            strings[j++] = arg;
        }
    }

    return strings.join(separator);
};

It includes some performance optimizations described on MDN here.

它包括MDN描述的一些性能优化这里。

And here is a usage example:

这是一个使用示例：

var fullAddress = join(', ', address, city, state, zip);

回答by Arun P Johny

Try

尝试

function joinIfPresent(){
    return $.map(arguments, function(val){
        return val && val.length > 0 ? val : undefined;
    }).join(', ')
}
$("#addressDiv").append(joinIfPresent(address, city, state, zip));

Demo: Fiddle

演示：小提琴

回答by nicb

$.each([address,city,state,zip], 
    function(i,v) { 
        if(v){
             var s = (i>0 ? ", ":"") + v;
             $("#addressDiv").append(s);
        } 
    }
);`