我编写了一个递归运行的函数来查找名称包含给定世界的文件。我不明白承诺是如何运作的,尽管努力了,但是找不到用承诺写这个函数的方法。
我尝试在findPath函数中返回一个promise但由于extractFiles调用了findPath,我无法使用它。我试图创建一份承诺清单并返回所有承诺但不能成功。
那我怎么能用promises写这些函数呢?
const fs = require('fs');
const path = require('path');
function findPath(targetPath, targetWord, done) {
if (!fs.existsSync(targetPath)) return;
fs.readdir(targetPath, (err, allPaths) => {
if (err) done(err, null);
for (aPath of allPaths) {
aPath = path.join(targetPath, aPath);
extractFiles(aPath, targetWord, done);
}
});
function extractFiles(aPath, targetWord, done) {
fs.lstat(aPath, (err, stat) => {
if (err) done(err, null);
if (stat.isDirectory()) {
findPath(aPath, targetWord, done);
}
else if (aPath.indexOf(targetWord) >= 0) {
let fileName = aPath.split('.')[0];
done(null, fileName);
}
});
}
}
findPath('../modules', 'routes', file => {
console.log(file);
});
答案 0 :(得分:0)
首先,为了使“核心”代码更具可读性,我将宣传fs函数
const promisify1p = fn => p1 => new Promise((resolve, reject) => {
fn(p1, (err, result) => {
if(err) {
reject(err);
} else {
resolve(result);
}
});
});
const readdirAsync = promisify1p(fs.readdir);
const lstatAsync = promisify1p(fs.lstat);
然后,就像你对任何其他承诺一样链接承诺
const fs = require('fs');
const path = require('path');
function findPath(targetPath, targetWord) {
const readPath = target =>
readdirAsync(target)
.then(allPaths =>
Promise.all(allPaths.map(aPath => extractFiles(path.join(target, aPath))))
.then(x => x.filter(x=>x)) // remove all the "false" entries - i.e. that don't match targetWord
.then(x => [].concat.apply([], x)) // flatten the result
);
const extractFiles = aPath =>
lstatAsync(aPath).then(stat => {
if (stat.isDirectory()) {
return readPath(aPath);
} else if (aPath.includes(targetWord)) {
return aPath.split('.')[0];
}
return false;
});
return readPath(targetPath);
}
findPath('../modules', 'routes')
.then(results => {
// do things with the results - which is an array of files that contain the targetWord
})
.catch(err => console.error(err));
根本不多。