我试图使用Node.js获取目录中所有文件的名称列表。我希望输出文件名数组。我该怎么做?
当前回答
开箱即用
如果您想要一个具有开箱即用的目录结构的对象,我强烈建议您检查目录树。
假设你有这样的结构:
photos
│ june
│ └── windsurf.jpg
└── january
├── ski.png
└── snowboard.jpg
const dirTree = require("directory-tree");
const tree = dirTree("/path/to/photos");
将返回:
{
path: "photos",
name: "photos",
size: 600,
type: "directory",
children: [
{
path: "photos/june",
name: "june",
size: 400,
type: "directory",
children: [
{
path: "photos/june/windsurf.jpg",
name: "windsurf.jpg",
size: 400,
type: "file",
extension: ".jpg"
}
]
},
{
path: "photos/january",
name: "january",
size: 200,
type: "directory",
children: [
{
path: "photos/january/ski.png",
name: "ski.png",
size: 100,
type: "file",
extension: ".png"
},
{
path: "photos/january/snowboard.jpg",
name: "snowboard.jpg",
size: 100,
type: "file",
extension: ".jpg"
}
]
}
]
}
自定义对象
否则,如果要使用自定义设置创建目录树对象,请查看以下代码段。在这个代码沙盒上可以看到一个活生生的例子。
// my-script.js
const fs = require("fs");
const path = require("path");
const isDirectory = filePath => fs.statSync(filePath).isDirectory();
const isFile = filePath => fs.statSync(filePath).isFile();
const getDirectoryDetails = filePath => {
const dirs = fs.readdirSync(filePath);
return {
dirs: dirs.filter(name => isDirectory(path.join(filePath, name))),
files: dirs.filter(name => isFile(path.join(filePath, name)))
};
};
const getFilesRecursively = (parentPath, currentFolder) => {
const currentFolderPath = path.join(parentPath, currentFolder);
let currentDirectoryDetails = getDirectoryDetails(currentFolderPath);
const final = {
current_dir: currentFolder,
dirs: currentDirectoryDetails.dirs.map(dir =>
getFilesRecursively(currentFolderPath, dir)
),
files: currentDirectoryDetails.files
};
return final;
};
const getAllFiles = relativePath => {
const fullPath = path.join(__dirname, relativePath);
const parentDirectoryPath = path.dirname(fullPath);
const leafDirectory = path.basename(fullPath);
const allFiles = getFilesRecursively(parentDirectoryPath, leafDirectory);
return allFiles;
};
module.exports = { getAllFiles };
然后,您可以简单地执行以下操作:
// another-file.js
const { getAllFiles } = require("path/to/my-script");
const allFiles = getAllFiles("/path/to/my-directory");
其他回答
这是一个TypeScript,可选递归,可选错误日志和异步解决方案。可以为要查找的文件名指定正则表达式。
我使用了fs extra,因为这是对fs的一个简单的超集改进。
import * as FsExtra from 'fs-extra'
/**
* Finds files in the folder that match filePattern, optionally passing back errors .
* If folderDepth isn't specified, only the first level is searched. Otherwise anything up
* to Infinity is supported.
*
* @static
* @param {string} folder The folder to start in.
* @param {string} [filePattern='.*'] A regular expression of the files you want to find.
* @param {(Error[] | undefined)} [errors=undefined]
* @param {number} [folderDepth=0]
* @returns {Promise<string[]>}
* @memberof FileHelper
*/
public static async findFiles(
folder: string,
filePattern: string = '.*',
errors: Error[] | undefined = undefined,
folderDepth: number = 0
): Promise<string[]> {
const results: string[] = []
// Get all files from the folder
let items = await FsExtra.readdir(folder).catch(error => {
if (errors) {
errors.push(error) // Save errors if we wish (e.g. folder perms issues)
}
return results
})
// Go through to the required depth and no further
folderDepth = folderDepth - 1
// Loop through the results, possibly recurse
for (const item of items) {
try {
const fullPath = Path.join(folder, item)
if (
FsExtra.statSync(fullPath).isDirectory() &&
folderDepth > -1)
) {
// Its a folder, recursively get the child folders' files
results.push(
...(await FileHelper.findFiles(fullPath, filePattern, errors, folderDepth))
)
} else {
// Filter by the file name pattern, if there is one
if (filePattern === '.*' || item.search(new RegExp(filePattern, 'i')) > -1) {
results.push(fullPath)
}
}
} catch (error) {
if (errors) {
errors.push(error) // Save errors if we wish
}
}
}
return results
}
function getFilesRecursiveSync(dir, fileList, optionalFilterFunction) {
if (!fileList) {
grunt.log.error("Variable 'fileList' is undefined or NULL.");
return;
}
var files = fs.readdirSync(dir);
for (var i in files) {
if (!files.hasOwnProperty(i)) continue;
var name = dir + '/' + files[i];
if (fs.statSync(name).isDirectory()) {
getFilesRecursiveSync(name, fileList, optionalFilterFunction);
} else {
if (optionalFilterFunction && optionalFilterFunction(name) !== true)
continue;
fileList.push(name);
}
}
}
如果上面的许多选项看起来太复杂,或者不是您想要的,这里是另一种使用node-dir的方法https://github.com/fshost/node-dir
npm install node-dir
下面是一个somple函数,用于列出在子目录中搜索的所有.xml文件
import * as nDir from 'node-dir' ;
listXMLs(rootFolderPath) {
let xmlFiles ;
nDir.files(rootFolderPath, function(err, items) {
xmlFiles = items.filter(i => {
return path.extname(i) === '.xml' ;
}) ;
console.log(xmlFiles) ;
});
}
我的单行代码:
const fs = require("fs")
const path = 'somePath/'
const filesArray = fs.readdirSync(path).filter(file => fs.lstatSync(path+file).isFile())
如果有人:
只想列出项目本地子文件夹中的文件名(不包括目录)
✅ 无其他依赖项✅ 1功能✅ 规格化路径(Unix与Windows)
const fs = require("fs");
const path = require("path");
/**
* @param {string} relativeName "resources/foo/goo"
* @return {string[]}
*/
const listFileNames = (relativeName) => {
try {
const folderPath = path.join(process.cwd(), ...relativeName.split("/"));
return fs
.readdirSync(folderPath, { withFileTypes: true })
.filter((dirent) => dirent.isFile())
.map((dirent) => dirent.name.split(".")[0]);
} catch (err) {
// ...
}
};
README.md
package.json
resources
|-- countries
|-- usa.yaml
|-- japan.yaml
|-- gb.yaml
|-- provinces
|-- .........
listFileNames("resources/countries") #=> ["usa", "japan", "gb"]
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