如果满足条件，则停止 Gulp 任务

2022-01-12 00:00:00 node.js javascript gulp

我正在尝试这样做，因此如果未指定 --theme 标志，它会停止 gulp 任务并想知道以 DRY 方式执行此任务的最佳方法.

I am trying to make it so if a --theme flag isn't specified it stops the gulp task and wondering the best way to do it in a DRY way.

如果未指定 --theme，我希望每个单独的任务停止，并且如果未满足，我希望默认任务停止.

I would like each individual task to stop if a --theme isn't specified and also have the default task stop if it isn't met.

到目前为止，我已经尝试了一些没有运气的事情.

I have tried a few things with no luck so far.

谢谢，

gulp.task('test', function() {

    if(typeof(args.theme) == 'undefined' || args.theme === true) {
        console.log(msg.noTheme);
        return; // end task
    }

    // run rest of task...

});

gulp.task('test-2', function() {

    if(typeof(args.theme) == 'undefined' || args.theme === true) {
        console.log(msg.noTheme);
        return; // end task
    }

    // run rest of task...

});

gulp.task('default', ['test-1', 'test-2']);

推荐答案

您可以简单地停止脚本:

You can simply stop the script with:

process.exit()

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