【发布时间】:2017-03-08 15:00:11
【问题描述】:
我有 2 个目录:
应用
es5 javascript files compiled by gulp tasksrc
typescript source files
我想在 src/ 目录中的任何更改上从 app/ 重新启动 nodemon make start 脚本。但它不起作用 - 应用程序刚刚启动,但 nodemon 不会对更改进行任何重新加载。 (貌似没看src/)
这是我的gulpfile:
var gulp = require("gulp-help")(require("gulp")),
ts = require("gulp-typescript"),
nodemon = require("gulp-nodemon");
/***********************************************************************************
* Build es5 javascript files from typescript sources
***********************************************************************************/
gulp.task("compile", function() {
var project = ts.createProject({
"target": "ES5",
"module": "commonjs",
"moduleResolution": "node",
"sourceMap": true,
"emitDecoratorMetadata": true,
"experimentalDecorators": true,
"removeComments": true,
"noImplicitAny": false
});
return gulp.src("./src/**/*.ts")
.pipe(project())
.pipe(gulp.dest("./app/"))
});
/***********************************************************************************
* Call "build" task and start nodemon with watch (autorestart on changes found)
***********************************************************************************/
gulp.task("default", ["compile"], function() {
var stream = nodemon({
script: "app/",
watch: "src",
tasks: ["compile"],
env: { "DEBUG": "Application,Request,Response" }
});
return stream;
});
【问题讨论】:
-
你可以使用 gulp.watch, imo
标签: javascript node.js typescript gulp nodemon