从nodejs运行gradlew

试图从nodeJS运行gradlew：

const spawn = require('child_process').spawn; const path = require('path'); function run() { var path_dir = path.resolve('./../movies/VanillaApp/android'); var options = { cwd: path_dir }; const ls = spawn('gradlew', ['assembleRelease'], options); ls.stdout.on('data', (data) => { console.log(`stdout: ${data}`); }); ls.stderr.on('data', (data) => { console.log(`stderr: ${data}`); }); ls.on('close', (code) => { console.log(`child process exited with code ${code}`); }); } module.exports = { run: run }; 

terminal等价：

 > ./gradlew assembleRelease 

得到：

 events.js:154 throw er; // Unhandled 'error' event ^ Error: spawn gradlew ENOENT at exports._errnoException (util.js:856:11) at Process.ChildProcess._handle.onexit (internal/child_process.js:178:32) at onErrorNT (internal/child_process.js:344:16) at _combinedTickCallback (node.js:377:13) at process._tickCallback (node.js:401:11) at Function.Module.runMain (module.js:449:11) at startup (node.js:141:18) at node.js:933:3 

更新：

我已经设法与child_process.exec运行，但不与产卵。

Answer 1

我只是自己偶然发现了这个问题，并使用–project-dir命令行选项设置了spawnw ：

 var taskDone = this.async(); grunt.util.spawn({ cmd: "my/relative/path/gradlew", args: ["clean", "--project-dir", "my/relative/path"], opts: { stdio: "inherit" } }, function (error, result) { if (error) { taskDone(false); } else { taskDone(); } });