Question

我试图以递归方式（即搜索所有子文件夹）从目录中提取所有txt文件：

let fs = require("fs");

function getPathNames(dirName) {
    let pathNames = [];
    for (let fileName of fs.readdirSync(dirName)) {
        let pathName = dirName + "/" + fileName;
        if (fs.statSync(pathName).isDirectory())
            pathNames.concat(getPathNames(pathName));
        else if (pathName.endsWith(".txt"))
            pathNames.push(pathName);
    }
    return pathNames;
}

但是，当我调用getPathNames(".")时，我只会得到第一个文件的名称。

如果我从函数中取出返回值，而是更新一个全局变量，则效果很好：

let fs = require("fs");

let pathNames = [];

function getPathNames(dirName) {
    for (let fileName of fs.readdirSync(dirName)) {
        let pathName = dirName + "/" + fileName;
        if (fs.statSync(pathName).isDirectory())
            getPathNames(pathName);
        else if (pathName.endsWith(".txt"))
            pathNames.push(pathName);
    }
}

有人发现第一种方法有什么问题吗？

谢谢！

Answer 1

好吧，concat并不是就位突变，而是返回一个新数组，所以我想你应该这样做

pathNames = pathNames.concat(getPathNames(pathName));

无法从目录中递归获取所有文本文件名

1 个答案: