JavaScript如何用reference数组替换blank数组的下划线占位符

实现逻辑 #

• 先将reference数组所有元素拼接为完整的待插入字符串
• 遍历blank数组，将其中唯一的_占位符替换为待插入字符串
• 将替换完成的blank数组所有元素拼接为完整字符串返回

完整代码示例（保持原有的全局blank变量写法） #

const reference = ['this is', 'a beautiful', 'car'];
const blank = ['I know', '_'];

function fillBlank(reference) {
    const ref = reference.join(' ');
    if(blank.length) {
        // 遍历替换占位符后拼接返回
        return blank.map(item => item === '_' ? ref : item).join(' ');
    } else {
        return ref;
    }
}

console.log(fillBlank(reference)) // 输出：I know this is a beautiful car

不同占位符位置的测试结果 #

• 当blank = ['_', 'I know']时，输出：this is a beautiful car I know
• 当blank = ['I know', '_', 'right?']时，输出：I know this is a beautiful car right?

通用性优化方案 #

建议将blank作为函数入参传入，避免依赖外部全局变量，提升复用性：

function fillBlank(reference, blank) {
    const ref = reference.join(' ');
    return blank.length ? blank.map(item => item === '_' ? ref : item).join(' ') : ref;
}

// 调用示例
const reference = ['this is', 'a beautiful', 'car'];
console.log(fillBlank(reference, ['I know', '_']))
console.log(fillBlank(reference, ['_', 'I know']))
console.log(fillBlank(reference, ['I know', '_', 'right?']))

内容的提问来源于stack exchange，提问作者Sara Ree