Javascript 从 gulp 运行 shell 命令
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Running a shell command from gulp
提问by Ben Aston
I would like to run a shell command from gulp, using gulp-shell. I see the following idiom being used the gulpfile.
我想从 gulp 运行一个 shell 命令，使用gulp-shell. 我看到 gulpfile 使用了以下习语。
Is this the idiomatic way to run a command from a gulp task?
这是从 gulp 任务运行命令的惯用方法吗？
var cmd = 'ls';
gulp.src('', {read: false})
.pipe(shell(cmd, {quiet: true}))
.on('error', function (err) {
gutil.log(err);
});
回答by ddprrt
gulp-shellhas been blacklisted. You should use gulp-execinstead, which has also a better documentation.
gulp-shell已经被列入黑名单。您应该改用gulp-exec，它也有更好的文档。
For your case it actually states:
对于您的情况，它实际上指出：
Note: If you just want to run a command, just run the command, don't use this plugin:
注意：如果你只是想运行一个命令，只运行命令，不要使用这个插件：
var exec = require('child_process').exec;
gulp.task('task', function (cb) {
exec('ping localhost', function (err, stdout, stderr) {
console.log(stdout);
console.log(stderr);
cb(err);
});
})
回答by AutoSponge
The new way to do this that keeps console output the same (e.g., with colors):
保持控制台输出相同的新方法（例如，使用颜色）：
see: https://nodejs.org/api/child_process.html#child_process_child_process_spawn_command_args_options
请参阅：https: //nodejs.org/api/child_process.html#child_process_child_process_spawn_command_args_options
var gulp = require('gulp');
var spawn = require('child_process').spawn;
gulp.task('my-task', function (cb) {
var cmd = spawn('cmd', ['arg1', 'agr2'], {stdio: 'inherit'});
cmd.on('close', function (code) {
console.log('my-task exited with code ' + code);
cb(code);
});
});回答by birnbaum
With gulp 4your tasks can directly return a child process to signal task completion:
使用gulp 4，您的任务可以直接返回子进程以表示任务完成：
'use strict';
var cp = require('child_process');
var gulp = require('gulp');
gulp.task('reset', function() {
return cp.execFile('git checkout -- .');
});
回答by azerafati
You could simply do this:
你可以简单地这样做：
const { spawn } = require('child_process');
const gulp = require('gulp');
gulp.task('list', function() {
const cmd = spawn('ls');
cmd.stdout.on('data', (data) => {
console.log(`stdout: ${data}`);
});
return cmd;
});
