我在以下模块fetch.js中苦苦挣扎,该模块调用一个在多个文件中生成输出的子进程。我希望此模块返回包含该输出的JSON。
var util = require('util');
const path = require('path');
const fs = require('fs');
const path1 = path.resolve(__dirname, 'out', 'file1');
const path2 = path.resolve(__dirname, 'out', 'file2');
let out1, out2;
function generate() {
var spawn = require('child_process').spawn;
var prog = spawn('prog', [<arguments>]);
prog.on('exit', function (code) {
if (code != 0) {
console.log('Compilation failed.');
} else {
out1 = fs.readFileSync(path1, 'utf8');
out2 = fs.readFileSync(path2, 'utf8');
}
})
prog.stderr.on('data', function (data) {
console.log(data);
})
return {out1, out2}
}
module.exports = new Promise(function(resolve, reject) {
resolve(generate());
});
我主要执行以下操作
const promise = require('./fetch');
promise.then((o1, o2) => {
console.log('O = ' + typeof o1);
})
我知道问题是发生的返回发生在out1和out2获取它们的值。如何纠正? TIA
答案 0 :(得分:0)
您可以从generate
函数本身返回promise,并导出generate
函数。
const path = require('path');
const fs = require('fs');
function generate() {
return new Promise((res, rej) => {
const spawn = require('child_process').spawn;
const prog = spawn('prog', []);
prog.on('exit', function (code) {
if (code != 0) {
rej('Compilation failed');
} else {
const path1 = path.resolve(__dirname, 'out', 'file1');
const path2 = path.resolve(__dirname, 'out', 'file2');
const out1 = fs.readFileSync(path1, 'utf8');
const out2 = fs.readFileSync(path2, 'utf8');
res({ out1, out2 })
}
})
prog.stderr.on('data', function (data) {
console.log(data);
})
})
}
module.exports = generate;
现在,当您使用它时,可以将其用作:
const fetch = require('./fetch');
fetch()
.then(data=>{
....
});