我正在尝试以递归方式从目录中获取所有文本文件(即搜索所有子文件夹):
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,我只得到第一个文件的名称。
如果我从函数中取出返回值并更新全局变量,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?
最佳答案
嗯,concat
不是就地突变,而是返回一个新数组,所以我想说你应该这样做
pathNames = pathNames.concat(getPathNames(pathName));
关于javascript - 无法从目录中递归获取所有文本文件名,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54018650/