Error out on TypeScript error

[guillaumejenkins]

I would like my gulp task to fail if there were TypeScript compilation errors. I couldn't find a way to do this. If it's not already supported, could you consider adding an option for it?

[jrieken]

The build-function accepts an error handler which is called for each compilation issue. You should use that make your outer task fail.

[guillaumejenkins]

I see; thanks for the help. However, in the usage example, you create the compiler outside of the gulp task:

    var tsb = require('gulp-tsb');

    // create and keep compiler
    var compilation = tsb.create({
        target: 'es5',
        module: 'commonjs',
        declaration: false
    });

    gulp.task('build', function() {
        return gulp.src('src/**/*.ts')
            .pipe(compilation()) // <- new compilation
            .pipe(gulp.dest(''));
    });

This means we don't have access to the gulp task's callback, which is what we need to invoke in order to make the task fail. Could it be possible to have the IncrementalCompiler take the error callback as argument instead of the create function? Then we can do something like:

    gulp.task('build', function(gulpCb) {
        return gulp.src('src/**/*.ts')
            .pipe(compilation(gulpCb)) // <- Pass the gulp callback (or any other error handler) directly to the compilation 
            .pipe(gulp.dest(''));
    });

Otherwise, we need to recreate the compiler inside the gulp task every time we run the task:

    gulp.task('build', function(gulpCb) {
        var compilation = tsb.create({ ... }, /*verbose*/ true, /*json*/ false, /*onError*/ gulpCb);
        return gulp.src('src/**/*.ts')
            .pipe(compilation()) // <- new compilation
            .pipe(gulp.dest(''));
    });

@jrieken Is there any downside to calling tsb.create() every time we run a task? Do we lose anything by not reusing the same compiler?