When I use Node.js to compile a single TypeScript file from project, which has dependencies on other files I get multiple "Could not find symbol .." errors. Despite these errors JS file is created correctly. There are no explicit references in TypeScript files.
Is there a way to suppress these errors? Or tell compiler where to look for dependencies, but not recompile all files?
The reason I want to do this is performance. My goal is to use Grunt to watch for TypeScript files. On change it would compile them to JS, do some transformations, minimize, bundle and run unit tests. However, only a single step to compile all 160 typescript files to JS takes around 8 seconds, which is painfully slow for development process. Compiling only single changed file should speed it up.
There isn't an option at the moment. I've been planning on adding it to grunt-ts for /// <reference's https://github.com/basarat/grunt-ts/issues/34 but the same can apply for node.
Related
I am writing code for webtask.io and it seems as if I can only upload one file to run. I would like to organize my code into multiple files with different modules but then have them be compiled into one for upload. I have tried gulp with gulp-concat but this just concatenates the files without preserving any execution logic. I would assume there is a simple way to do this.
Webpack works well, especially with the webpack-node-externals' package: https://www.npmjs.com/package/webpack-node-externals.
Just set the --bundle argument.
$ wt create index.js --bundle
I'm having a problem using Gulp to compile a RequireJS project properly. What I need to do is have gulp create a single distribution file that only includes the file necessary to have the application run.
In our application we are following a modular approach breaking out major pieces of functionality into different repos. So while developing my piece I have RequireJS including angular and many other vendor libraries that are common to all of the projects in the application. However when I go to move my piece into the larger application I no longer need these files in the final output since those dependencies also exist in that application (and having those extra libraries also makes the final distribution file over 300K).
I've tried creating another main.js (called gulp-main.js) file that only includes the dependencies that I need but when I run the gulp process it fails. I don't get an error but it seems to be failing because I'm not including the required dependencies for the project to build successfully. Below is the config object that is being passed to the RequireJS optimize method.
var config = {
baseUrl: 'app/',
mainConfigFile: 'app/main.js',
out: 'dist/app/output.js',
name: 'main'
};
Any ideas on what I could do to either remove the unnecessary vendor files or even split them into a single vendor and a single non-vendor file would really be appreciated. I have already tried using the modules array option but that does not produce the results that I am after since it seems to create a single file for each item defined not a single compiled JS file with all scripts contained within.
Thanks in advance.
When you don't want some file in your final output. add " ! " in Your gulp task's src
example :
gulp.src(['./app/*.js', '!./node_modules/**']) // '!./vendor-libraries-dest to igonore'
I have a TypeScript project that compiles down to a single JS file in /dist.
I have Jasmine tests, also written in TypeScript, that reference the various classes they test with references like:
/// <reference path="../../../src/script/classUnderTest.ts" />
When I compile the tests the TypeScript compiler pulls in all the referenced classes and I get a single JS file containing the test code and the code under test.
This actually works fine for just running the tests, but now I'd like code coverage too. From what I can tell, to get Istanbul to work I need to have the code under test separate from the test code. Also it would be nice to be testing exactly the JS file that will be live.
So, how can I get the type safety and autocomplete benefits of "/// reference" whilst using my compiled JS file when the tests are actually run?
(Or am I barking up the wrong tree entirely?)
Note, I am building this on a Mac, so from what I've read Chutzpah is not currently an option. I'm also currently using only npm scripts to do builds. I'd prefer to not bring in grunt or gulp unless it's absolutely necessary.
So, how can I get the type safety and autocomplete benefits of "/// reference" whilst using my compiled JS file when the tests are actually run?
Don't using TypeScript's --outFile option (few other reasons). Instead use modules e.g. --module commonsjs and a module loader e.g. webpack.
More
Browser quickstart : https://basarat.gitbooks.io/typescript/content/docs/quick/browser.html
Modules:
https://basarat.gitbooks.io/typescript/content/docs/project/modules.html
Are there any known issues with mixing nodejs modules (require) with typescript definition files (d.ts) multiple times over files within a module?
My scenario is that I have a module namespace per folder (much like I would in C#), then I basically compile them all via tsc to an outputted my-module.js. However I keep getting really odd errors like Could not find type HTMLElement but lots of people have pointed out that tsc includes the typescript lib file by default which contains all those types.
I have noticed a few people having odd errors when they are including the same d.ts files over multiple files which are all compiled with the --out flag to get it all into one file, so could this be causing my issues?
An example of my usage would be:
///<reference path="path/to/knockout.d.ts" />
import ko = require("knockout");
This would then be put in each file which requires knockout js, which is at least 10 files in the module i'm trying to compile currently. It just bombs out constantly saying knockout.d.ts cannot find the type HTMLElemet, Element, Document etc.
If you are using external modules (which you are if you have a top level "import" - as shown above), then you can't use the --out switch to combine multiple source files. It is a limitation that with external modules that one source file = one module. With source that is not in an external module (i.e. contributes to the 'global' scope), you can combine input source to one output JavaScript file using --out.
I have no idea about the "could not find HTMLElement" issues. If you can provide a repro (and outline which version you are using) I can take a look.
In an attempt to separate my .ts source files from the .js output, I'm trying to implement a formal file-watcher in TypeScript and it seems as though the ability to specify an output path for a single file does not exist.
The flow, just to be clear: I begin watching my entire src directory, make a change to, say, src/views/HomeView.ts, and i want node to pick up that the file has been changed, and move the compiled version to public/js/views/HomeView.js.
Using tsc myfile.ts --out myfile.js it travels through all of the modules and compiles each in the same path that the .ts file exists, without placing the final file(s) in the properly specified path. It does however create an empty file where I would like it to end up.
I'm wondering:
1) Is it possible to use the --out parameter and only compile that one file? I do not want it to traverse imports and compile every single file, but merely do syntax / error checking at compile-time (and that's only a secondary requirement, not necessary); and
2) Is there a bug in the compiler that prevents it from properly combining / creating files? Again, the final output in the --out path directive is empty, but a file is indeed created.
Any help would be appreciated.
* Update *
While I don't want to close this question as I do believe it's still an issue, I went ahead and implemented the core TypeScript compiler to achieve what I needed to do, bypassing tsc altogether. Please see https://github.com/damassi/TypeScript-Watcher for more information and usage.
When you use the --out parameter, you get a single file with the compiler walking the dependencies and working out the correct order for the final file.
For example...
tsc --out final.js app.ts
Will find any dependencies in app.ts and compile them all too. After it works out the correct order it will save all of the JavaScript in final.js. If this JavaScript file is empty, it is normally indicative of a compiler error.
It is not possible to use the --out parameter to generate a JavaScript file for just the TypeScript file you specify, unless that file has no dependencies.