Error: gulp-typescript: A project cannot be used in two compilations at the same time. Create multiple projects with createProject instead

[chenzhutian]

Expected behavior:
No error
When I run watch, I got the following error

Actual behavior:

[22:39:13] Error: gulp-typescript: A project cannot be used in two compilations at the same time. Create multiple projects with createProject instead.
at project (/Users/zhutian/Code/billofchen/node_modules/gulp-typescript/release/project.js:36:19)
at componentTS (/Users/zhutian/Code/billofchen/gulpfile.js:50:15)
at bound (domain.js:415:14)
at runBound (domain.js:428:12)
at asyncRunner (/Users/zhutian/Code/billofchen/node_modules/async-done/index.js:55:18)
at processTicksAndRejections (internal/process/task_queues.js:75:11)

Your gulpfile:

const tsProject = tsCompiler.createProject(path.join(__dirname, 'tsconfig-build.json'))

async function componentTS() {
    return tsProject.src()
        .pipe(sourcemaps.init())
        .pipe(tsProject())
        .pipe(sourcemaps.write())
        .pipe(gulp.dest(distPath))
}


function parallelWatch() {
    gulp.watch(
        [path.join(srcPath, '**/*.ts'), '!' + path.join(srcPath, '**/__test__/**')],
        { ignoreInitial: false },
        componentTS
    )
}

const watch = gulp.series(cleanDist, parallelWatch)

tsconfig.json

Include your tsconfig, if related to this issue.

{
    "compilerOptions": {
        "target": "ES6",
        "lib": ["ES6", "DOM"],
        "strictNullChecks": true,
        "noImplicitAny": false,
        "noUnusedLocals": false,
        "noUnusedParameters": false,
        "module": "CommonJS",
        "moduleResolution": "node",
        "allowJs": false,
        "experimentalDecorators": true,
        "noImplicitThis": true,
        "noImplicitReturns": true,
        "alwaysStrict": true,
        "inlineSourceMap": true,
        "inlineSources": true,
        "noFallthroughCasesInSwitch": true,
        "incremental": true,
        "strict": true,
        "removeComments": true,
        "pretty": true,
        "esModuleInterop": true,
        "skipLibCheck": true,
        "allowSyntheticDefaultImports": true,
        "strictPropertyInitialization": true,

    },
    "include": ["./src/**/*.ts", "node_modules/miniprogram-api-typings/index.d.ts"]
}

[tillig]

I got around this by creating the tsProject inside the Gulp task, which the docs say not to do... but it was the only thing I could find that stopped this error from happening. I only have the one build task that uses the tsProject and watch just calls that.

Note my Gulpfile is also in TypeScript, unclear if that makes a difference.

function build(): NodeJS.ReadWriteStream {
  const tsProject = typescript.createProject(path.join(taskName, 'tsconfig.json'));
  return tsProject
    .src()
    .pipe(eslint.default())
    .pipe(eslint.format())
    .pipe(eslint.failAfterError())
    .pipe(tsProject(typescript.reporter.longReporter()))
    .js
    .pipe(gulp.dest((file) => {
      return file.base ?? process.cwd();
    }));
}

function watch(): FSWatcher {
  return gulp.watch(path.join('src', '**', '*.ts'), build);
}

exports.watch = gulp.series(build, watch);