如果字符串文字中未定义可选链接变量,则打印空字符串后备的简洁方法

发布于 2025-01-20 18:33:03 字数 1778 浏览 3 评论 0原文

假设我有这个字符串文字用例:

function getClassNames(props){
    my_classNames = `class-nameA class-nameB ${props?.one} ${props?.two} class-nameC`;
    return my_classNames;
}

console.log(getClassNames({one: 'class-name-one', two: 'class-name-two'}))
  // prints: class-nameA class-nameB class-name-one class-name-two class-nameC
console.log(getClassNames({two: 'class-name-two'}))
  // prints: class-nameA class-nameB undefined class-name-two class-nameC
console.log(getClassNames())
  // prints: class-nameA class-nameB undefined undefined class-nameC

问题是当任何 props 的键未定义时,它会打印 undefined。我希望它打印空字符串。

解决此问题的一种解决方案是使用短路或评估,如下所示:

function getClassNames(props){
    my_classNames = `class-nameA class-nameB ${props?.one || ''} ${props?.two || ''} class-nameC`;
    return my_classNames;
}

console.log(getClassNames())
  // prints: class-nameA class-nameB   class-nameC

或者更好的解决方案(以避免额外的空格),使用之前的解决方案和短路与评估像这样:

function getClassNames(props){
    my_classNames = `class-nameA class-nameB${props?.one && ' '+props?.one || ''}${props?.two && ' '+props?.two || ''}class-nameC`;
    return my_classNames;
}

console.log(getClassNames())
  // prints: class-nameA class-nameB class-nameC

为了使其更简洁,我可以将以前的解决方案中的重复项/重复包装到函数中:

function getClassNames(props){
    my_classNames = `class-nameA class-nameB${concatWithSpace(props?.one)}${concatWithSpace(props?.two)}class-nameC`;
    return my_classNames;
}

function concatWithSpace(str){
    return str && ' '+str || '';
}

console.log(getClassNames())
  // prints: class-nameA class-nameB class-nameC

是否有更简洁的内置方式(语法)来打印空字符串(而不是未定义)对于可选链变量而不使用前面的我提到的“不太简洁”的解决方案?

Let's say I have this string literal use case:

function getClassNames(props){
    my_classNames = `class-nameA class-nameB ${props?.one} ${props?.two} class-nameC`;
    return my_classNames;
}

console.log(getClassNames({one: 'class-name-one', two: 'class-name-two'}))
  // prints: class-nameA class-nameB class-name-one class-name-two class-nameC
console.log(getClassNames({two: 'class-name-two'}))
  // prints: class-nameA class-nameB undefined class-name-two class-nameC
console.log(getClassNames())
  // prints: class-nameA class-nameB undefined undefined class-nameC

The problem is when any of props' keys is undefined, it prints undefined. I want it to print empty string instead.

One solution to this is to use short-circuit OR evaluation like this:

function getClassNames(props){
    my_classNames = `class-nameA class-nameB ${props?.one || ''} ${props?.two || ''} class-nameC`;
    return my_classNames;
}

console.log(getClassNames())
  // prints: class-nameA class-nameB   class-nameC

Or a better solution (to avoid extra spaces), to use previous solution with short-circuit AND evaluation like this:

function getClassNames(props){
    my_classNames = `class-nameA class-nameB${props?.one && ' '+props?.one || ''}${props?.two && ' '+props?.two || ''}class-nameC`;
    return my_classNames;
}

console.log(getClassNames())
  // prints: class-nameA class-nameB class-nameC

To make it more concise, I can wrap duplicates/repetitions in previous solution to a function:

function getClassNames(props){
    my_classNames = `class-nameA class-nameB${concatWithSpace(props?.one)}${concatWithSpace(props?.two)}class-nameC`;
    return my_classNames;
}

function concatWithSpace(str){
    return str && ' '+str || '';
}

console.log(getClassNames())
  // prints: class-nameA class-nameB class-nameC

Is there a more concise built-in way (syntax) to print empty string (instead of undefined) for optionally-chained variables without using the previous "not-so concise" solutions I mentioned?

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评论(2

等待圉鍢 2025-01-27 18:33:03

不要使用模板字符串,而是使用数组并对其进行一些处理:

function getClassNames(props){
    const my_classNames = ['class-nameA', 'class-nameB', props?.one, props?.two, 'class-nameC'];
    return my_classNames.filter(Boolean).join(' ');
}

这也适用于开头或结尾的可选部分或空数组。

Don't use a template string, but an array and do some processing on it:

function getClassNames(props){
    const my_classNames = ['class-nameA', 'class-nameB', props?.one, props?.two, 'class-nameC'];
    return my_classNames.filter(Boolean).join(' ');
}

This also works properly with optional parts at the begin or end, or an empty array.

情何以堪。 2025-01-27 18:33:03

也许破坏props并分配默认值?

像这样:(

function getClassNames({ one = '', two = '' }){
    my_classNames = `class-nameA class-nameB ${one} ${two} class-nameC`;
    return my_classNames;
}

编辑)

或,您可以为模板文字编写助手函数(标记的模板

/* types textContent(stringList: TemplateStringsArray, ...valueList: unknown[]) */
function classString(stringList, ...valueList) {
    return stringList.reduce((text, string, i) => (
        text + string + (valueList[i] || '')
    ), '').replace(/ +/g, ' '); // (optional) replace multiple spaces with single
}

var one = undefined;
var two = undefined;
var three = "something";

console.log(classString`class-nameA class-nameB ${one} ${two} ${three} class-nameC`);

maybe destructuring props and assigning default values?

like this:

function getClassNames({ one = '', two = '' }){
    my_classNames = `class-nameA class-nameB ${one} ${two} class-nameC`;
    return my_classNames;
}

(EDIT)

or, you can write a helper function for template literals (tagged template)

/* types textContent(stringList: TemplateStringsArray, ...valueList: unknown[]) */
function classString(stringList, ...valueList) {
    return stringList.reduce((text, string, i) => (
        text + string + (valueList[i] || '')
    ), '').replace(/ +/g, ' '); // (optional) replace multiple spaces with single
}

var one = undefined;
var two = undefined;
var three = "something";

console.log(classString`class-nameA class-nameB ${one} ${two} ${three} class-nameC`);
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