[英]Gulp: target to debug mocha tests
I have a set of gulp.js targets for running my mocha tests that work like a charm running through gulp-mocha . 我有一组gulp.js目标用于运行我的摩卡测试,其工作方式类似于通过gulp-mocha运行的魅力。 Question: how do I debug my mocha tests running through gulp?
问题:如何调试通过gulp运行的mocha测试? I would like to use something like node-inspector to set break points in my src and test files to see what's going on.
我想使用像node-inspector这样的东西在我的src和测试文件中设置断点,看看发生了什么。 I am already able to accomplish this by calling node directly:
我已经能够通过直接调用node来完成此任务:
node --debug-brk node_modules/gulp/bin/gulp.js test
But I'd prefer a gulp target that wraps this for me, eg: 但是我更喜欢一个为我包装的gulp目标,例如:
gulp.task('test-debug', 'Run unit tests in debug mode', function (cb) {
// todo?
});
Ideas? 想法? I want to avoid a
bash
script or some other separate file since I'm trying to create a reusable gulpfile
with targets that are usable by someone who doesn't know gulp. 我想避免使用
bash
脚本或其他单独的文件,因为我正在尝试创建一个可重用的gulpfile
,其目标可供不知道gulpfile
的人使用。
Here is my current gulpfile.js
这是我目前的
gulpfile.js
// gulpfile.js
var gulp = require('gulp'),
mocha = require('gulp-mocha'),
gutil = require('gulp-util'),
help = require('gulp-help');
help(gulp); // add help messages to targets
var exitCode = 0;
// kill process on failure
process.on('exit', function () {
process.nextTick(function () {
var msg = "gulp '" + gulp.seq + "' failed";
console.log(gutil.colors.red(msg));
process.exit(exitCode);
});
});
function testErrorHandler(err) {
gutil.beep();
gutil.log(err.message);
exitCode = 1;
}
gulp.task('test', 'Run unit tests and exit on failure', function () {
return gulp.src('./lib/*/test/**/*.js')
.pipe(mocha({
reporter: 'dot'
}))
.on('error', function (err) {
testErrorHandler(err);
process.emit('exit');
});
});
gulp.task('test-watch', 'Run unit tests', function (cb) {
return gulp.src('./lib/*/test/**/*.js')
.pipe(mocha({
reporter: 'min',
G: true
}))
.on('error', testErrorHandler);
});
gulp.task('watch', 'Watch files and run tests on change', function () {
gulp.watch('./lib/**/*.js', ['test-watch']);
});
With some guidance from @BrianGlaz I came up with the following task. 在@BrianGlaz的一些指导下,我想出了以下任务。 Ends up being rather simple.
结束相当简单。 Plus it pipes all output to the parent's
stdout
so I don't have to handle stdout.on
manually: 另外,它将所有输出管道传输到父级的
stdout
因此我不必手动处理stdout.on
:
// Run all unit tests in debug mode
gulp.task('test-debug', function () {
var spawn = require('child_process').spawn;
spawn('node', [
'--debug-brk',
path.join(__dirname, 'node_modules/gulp/bin/gulp.js'),
'test'
], { stdio: 'inherit' });
});
You can use Node's Child Process
class to run command line commands from within a node app. 您可以使用Node的
Child Process
类从节点应用程序中运行命令行命令。 In your case I would recommend childprocess.spawn() . 在你的情况下,我会建议childprocess.spawn() 。 It acts as an event emitter so you can subscribe to
data
to retrieve output from stdout
. 它充当事件发射器,因此您可以订阅
data
以从stdout
检索输出。 In terms of using this from within gulp, some work would probably need to be done to return a stream that could be piped to another gulp task. 就在gulp中使用它而言,可能需要做一些工作来返回可以通过管道传输到另一个gulp任务的流。
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