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我正在嘗試編寫一個吞嚥任務來執行程序,並等待任何按鍵來退出程序。如何正確結束吞噬任務
到目前爲止,我有這樣的:
gulp.task('runDevelopment', function (done) {
console.log('\n\n---\nPress any key to exit\n---\n');
var api = spawn('node', ['node_modules/api-gateway-server/server.js', 'etc/swagger.yaml']);
api.stdout.on('data', function (data) {
console.log(data.toString());
});
api.stderr.on('data', function (data) {
console.log(data.toString());
});
process.stdin.on('data', function() {
api.kill('SIGKILL');
done();
});
});
這將運行該程序,並殺死它作爲預期,但一飲而盡決不會退卻。我看到這在控制檯:
20:04 $ gulp
[20:06:54] Using gulpfile ~/Development/swiki/gulpfile.js
[20:06:54] Starting 'documentCopy'...
[20:06:54] Starting 'documentZip'...
[20:06:54] Starting 'runDevelopment'...
---
Press any key to exit
---
[20:06:54] Finished 'documentCopy' after 52 ms
[20:06:54] Finished 'documentZip' after 41 ms
[20:06:54] Starting 'package'...
[20:06:54] Finished 'package' after 4.9 μs
API Gateway server listening on port 7111
[20:06:57] Finished 'runDevelopment' after 3.06 s
[20:06:57] Starting 'run'...
[20:06:57] Finished 'run' after 2.72 μs
[20:06:57] Starting 'default'...
[20:06:57] Finished 'default' after 1.16 μs
Terminated: 15
✘-TERM ~/Development/swiki [master|…6]
20:07 $
一口給我Terminated: 15
後,才killall gulp
在另一端。
我怎樣才能讓它正常工作?