"bbb/aaa/ss.pdf" 如何获取到文件名 ss.pdf

split('/') 分割字符串 

function getFileName(filePath) {
    // 使用正斜杠 '/' 分割路径，并获取最后一个元素作为文件名
    let parts = filePath.split('/')
    // console.log(parts) // ['bbb', 'aaa', 'ss.pdf']
    let fileName = parts[parts.length - 1]
    return fileName
}

let filePath = "bbb/aaa/ss.pdf"
let fileName = getFileName(filePath)
console.log(fileName) // 输出: ss.pdf

function getFileName(filePath) {
    return filePath.split('/').pop()
}
let filePath = "bbb/aaa/ss.pdf"
let fileName = getFileName(filePath)
console.log(fileName) // 输出: ss.pdf

substr

function getFileName(filePath) {
    return filePath.substr(filePath.lastIndexOf('/')+1)
}
let filePath = "bbb/aaa/ss.pdf"
let fileName = getFileName(filePath)
console.log(fileName) // 输出: ss.pdf