如何在Gulp中运行外部文件？(How do I run an external file within Gulp?)

如何在Gulp中运行外部文件？(How do I run an external file within Gulp?)

目前我正在使用“gulp-run”插件来运行.bat文件。 该插件现已被弃用，我正在寻找现在执行.bat的最佳方法。

当前代码：

var gulp = require('gulp'); var run = require('gulp-run'); module.exports = function() { run('c:/xxx/xxx/runme.bat').exec(); };

根据@cmrn建议的解决方案 ：

var exec = require('child_process').exec; var batchLocation = 'c:/xxx/xxx/runme.bat'; gulp.task('task', function (cb) { exec(batchLocation, function (err, stdout, stderr) { console.log(stdout); console.log(stderr); cb(err); }); })

At the moment I'm using the "gulp-run" plugin to run a .bat file. That plugin has now been deprecated and I'm looking for the best way to execute the .bat now.

Current code:

var gulp = require('gulp'); var run = require('gulp-run'); module.exports = function() { run('c:/xxx/xxx/runme.bat').exec(); };

Solution as per @cmrn suggestion:

var exec = require('child_process').exec; var batchLocation = 'c:/xxx/xxx/runme.bat'; gulp.task('task', function (cb) { exec(batchLocation, function (err, stdout, stderr) { console.log(stdout); console.log(stderr); cb(err); }); })

最满意答案

如果您需要将脚本作为gulp流的一部分运行（即在pipe() ），您可以使用gulp-exec 。 如果没有，您可以使用下面复制的child_process.exec -exec README中描述的child_process.exec 。

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);
  });
})
If you need to run the script as part of a gulp stream (i.e. in pipe()) you can use gulp-exec. If not, you can just use child_process.exec as described in the gulp-exec README, copied below. 
 
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);
  });
})