如何在 gulp 中运行 bash 命令？
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How to run bash commands in gulp?
提问by houhr
I want to add some bash commands at the end of gulp.watchfunction to accelerate my development speed. So, I am wondering if it is possible. Thanks!
我想在gulp.watch函数末尾添加一些bash命令来加快我的开发速度。所以，我想知道是否有可能。谢谢！
采纳答案by Erik
Use https://www.npmjs.org/package/gulp-shell.
使用https://www.npmjs.org/package/gulp-shell。
A handy command line interface for gulp
一个方便的 gulp 命令行界面
回答by Mangled Deutz
I would go with:
我会去：
var spawn = require('child_process').spawn;
var fancyLog = require('fancy-log');
var beeper = require('beeper');
gulp.task('default', function(){
gulp.watch('*.js', function(e) {
// Do run some gulp tasks here
// ...
// Finally execute your script below - here "ls -lA"
var child = spawn("ls", ["-lA"], {cwd: process.cwd()}),
stdout = '',
stderr = '';
child.stdout.setEncoding('utf8');
child.stdout.on('data', function (data) {
stdout += data;
fancyLog(data);
});
child.stderr.setEncoding('utf8');
child.stderr.on('data', function (data) {
stderr += data;
fancyLog.error(data));
beeper();
});
child.on('close', function(code) {
fancyLog("Done with exit code", code);
fancyLog("You access complete stdout and stderr from here"); // stdout, stderr
});
});
});
Nothing really "gulp" in here - mainly using child processes http://nodejs.org/api/child_process.htmland spoofing the result into fancy-log
这里没有真正“吞咽” - 主要使用子进程http://nodejs.org/api/child_process.html并将结果欺骗到fancy-log
回答by David Lemon
The simplest solution is as easy as:
最简单的解决方案很简单：
var child = require('child_process');
var gulp = require('gulp');
gulp.task('launch-ls',function(done) {
child.spawn('ls', [ '-la'], { stdio: 'inherit' });
});
It doesn't use node streams and gulp pipes but it will do the work.
它不使用节点流和吞咽管道，但它会完成工作。
