From the command line I can start the tsc compiler like this:
../../node_modules/.bin/tsc I want to incorporate this into a node build script.
There is a typescript compiler for node but it seems much more work to set up rather than just shelling out. You have to pull in all the right files etc.
I have this code :
fs.emptyDirSync(paths.appBuild); const json = ts.parseConfigFileTextToJson(tsconfig, ts.sys.readFile(tsconfig), true); const { options } = ts.parseJsonConfigFileContent(json.config, ts.sys, path.dirname(tsconfig)); options.configFilePath = paths.tsConfig; options.outDir = outDir; options.src = src; options.noEmitOnError = true; options.pretty = true; options.sourceMap = process.argv.includes('--source-map'); let rootFile = path.join(process.cwd(), 'src/index.tsx'); if (!fs.existsSync(rootFile)) { rootFile = path.join(process.cwd(), 'src/index.ts'); } const host = ts.createCompilerHost(options, true); const prog = ts.createProgram([rootFile], options, host); const result = prog.emit(); But This will miss files that are not imported in the RootFile.
How can I simply shell out to the tsc exe from node?
1 Answers
Answers 1
You could use child_process.exec:
const path = require('path'); const { exec } = require('child_process'); const tscPath = path.join(__dirname, '../../node_modules/.bin/tsc'); const tsc = exec(`${tscPath} ${process.argv.slice(2).join(' ')}`); tsc.stdout.on('data', data => console.log(data)); tsc.stderr.on('data', data => console.error(data)); tsc.on('close', code => console.log(`tsc exited with code ${code}`)); 
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