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
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);
}
}
}
其他回答
非递归版本
您并没有说要递归地执行,所以我假设您只需要目录的直接子级。
示例代码:
const fs = require('fs');
const path = require('path');
fs.readdirSync('your-directory-path')
.filter((file) => fs.lstatSync(path.join(folder, file)).isFile());
加载fs:
const fs = require('fs');
异步读取文件:
fs.readdir('./dir', function (err, files) {
// "files" is an Array with files names
});
读取文件同步:
var files = fs.readdirSync('./dir');
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
}
下面是一个仅使用本机fs和路径模块的简单解决方案:
// sync version
function walkSync(currentDirPath, callback) {
var fs = require('fs'),
path = require('path');
fs.readdirSync(currentDirPath).forEach(function (name) {
var filePath = path.join(currentDirPath, name);
var stat = fs.statSync(filePath);
if (stat.isFile()) {
callback(filePath, stat);
} else if (stat.isDirectory()) {
walkSync(filePath, callback);
}
});
}
或异步版本(改用fs.readder):
// async version with basic error handling
function walk(currentDirPath, callback) {
var fs = require('fs'),
path = require('path');
fs.readdir(currentDirPath, function (err, files) {
if (err) {
throw new Error(err);
}
files.forEach(function (name) {
var filePath = path.join(currentDirPath, name);
var stat = fs.statSync(filePath);
if (stat.isFile()) {
callback(filePath, stat);
} else if (stat.isDirectory()) {
walk(filePath, callback);
}
});
});
}
然后您只需调用(同步版本):
walkSync('path/to/root/dir', function(filePath, stat) {
// do something with "filePath"...
});
或异步版本:
walk('path/to/root/dir', function(filePath, stat) {
// do something with "filePath"...
});
不同之处在于节点在执行IO时如何阻塞。考虑到上面的API是相同的,您可以只使用异步版本来确保最大性能。
然而,使用同步版本有一个优点。在遍历完成后立即执行一些代码更容易,就像在遍历后的下一条语句中一样。对于异步版本,您需要一些额外的方法来知道何时完成。也许首先创建所有路径的映射,然后枚举它们。对于简单的build/util脚本(与高性能web服务器相比),您可以使用同步版本而不会造成任何损坏。
我最近为此开发了一个工具,它可以做到这一点。。。它异步获取目录并返回项目列表。您可以获取目录、文件或两者,首先是文件夹。如果不想获取整个文件夹,也可以对数据进行分页。
https://www.npmjs.com/package/fs-browser
这是链接,希望它能帮助到某人!