我的package.json
没有启动脚本,所以npm启动不起作用。
我的package.json看起来像
{
"name": "ionic-hello-world",
"author": "Ionic Framework",
"homepage": "http://ionicframework.com/",
"private": true,
"scripts": {
"build": "ionic-app-scripts build",
"watch": "ionic-app-scripts watch",
"serve:before": "watch",
"emulate:before": "build",
"deploy:before": "build",
"build:before": "build",
"run:before": "build"
},
"dependencies": {
"@angular/common": "2.2.1",
"@angular/compiler": "2.2.1",
"@angular/compiler-cli": "2.2.1",
"@angular/core": "2.2.1",
"@angular/forms": "2.2.1",
"@angular/http": "2.2.1",
"@angular/platform-browser": "2.2.1",
"@angular/platform-browser-dynamic": "2.2.1",
"@angular/platform-server": "2.2.1",
"@ionic/storage": "1.1.7",
"@agm/core": "1.0.0-beta.0",
"es6-promise": "3.0.2",
"es6-shim": "^0.35.0",
"ionic-angular": "2.0.0",
"ionic-native": "^2.2.14",
"ionicons": "3.0.0",
"ng2-cordova-oauth": "0.0.6",
"rxjs": "5.0.0-beta.12",
"sw-toolbox": "3.4.0",
"zone.js": "0.6.26"
},
"devDependencies": {
"@ionic/app-scripts": "1.0.0",
"typescript": "2.1.5"
},
"cordovaPlugins": [
"cordova-plugin-device",
"cordova-plugin-console",
"cordova-plugin-whitelist",
"cordova-plugin-splashscreen",
"cordova-plugin-statusbar",
"ionic-plugin-keyboard",
"cordova-plugin-inappbrowser"
],
"cordovaPlatforms": [
"ios",
{
"platform": "ios",
"version": "",
"locator": "ios"
}
],
"description": "third-party-auth-ionic2-tutorial: An Ionic project"
}
启动脚本应包含什么内容?
2条答案
按热度按时间mum43rcc1#
据我所知,Ionic项目不使用package.json中的启动脚本,因此不需要
npm start
。使用
ionic start
创建新项目,并使用ionic serve
在开发期间运行项目,如Getting Started Guide中所述。在需要的时候还有其他命令来构建、部署和打包您的项目,请确保检查离子文档以了解如何以“离子方式”执行这些操作
3mpgtkmj2#
运行命令
离子积分启用cordova --add
更新config.xml文件和package.json文件
用于构建目的