Rails Simple Blogger Heroku应用程序：预编译资产失败

Question

我已经尝试了书中所有内容，但仍然遇到相同的错误。 令人惊讶的是，我在其他任何地方都没有找到这个确切的错误。

remote:        error Couldn't find a package.json file in "/tmp/build_0e5b0c14c98db4a63ddfc87fc6d11490"

Here is the full error on Pastebin

Here is the repository for the (rather simple) blogger project

我什至尝试在本地运行rake assets:precompile，并且完全没有错误。 我运行了RAILS_ENV=production bundle exec rake assets:precompile，被推送到我的仓库中，然后尝试部署，什么也没有。

我通过成堆的StackOverflow问题大吃一惊，并尝试使用谷歌搜索我的错误，给我遇到的每个编辑/命令都尝试了一次，但是没有运气。

编辑：以下是我的package.json文件的内容

{
  "name": "blogger",
  "private": true,
  "dependencies": {
    "@rails/actioncable": "^6.0.0",
    "@rails/activestorage": "^6.0.0",
    "@rails/ujs": "^6.0.0",
    "@rails/webpacker": "4.2.2",
    "turbolinks": "^5.2.0"
  },
  "version": "0.1.0",
  "devDependencies": {
    "webpack-dev-server": "^3.10.1"
  }
}

Answer 1

Rails 6的基本配置：-

运行yarn install，或仅运行yarn。它将创建一个package.json文件：-

package.json'

此文件包含您的Webpack所需的软件包。

有关更多参考，请阅读：-

#yarn docs

#webpacker

如果要在生产环境中进行预编译，请在以下文件中更改配置：-

config/webpacker.yml

production:
  <<: *default

  # Production depends on precompilation of packs prior to booting for performance.
  compile:true ## just make this compile true

  # Extract and emit a css file
  extract_css: true

  # Cache manifest.json for performance
  cache_manifest: true

然后执行命令：-

RAILS_ENV=production rake assets:clobber 
RAILS_ENV=production rake assets:precompile

Answer 2

我的package.json文件在我的Github存储库中不存在，尽管它确实存在在我的本地系统中。

我没有任何待提交的待提交文件，我的.gitignore没有显示任何json文件的排除。 我不确定为什么没有将其上传到我的存储库。

所以我不得不通过运行以下命令来强制将其添加到登台区域： string path = @"C:\Data\Countries.xml"; if (!File.Exists(path)) { File.Create(path).Dispose(); using (TextWriter tw = new StreamWriter(path)) { tw.WriteLine("The very first line!"); tw.Close(); } } else if (File.Exists(path)) { using (TextWriter tw = new StreamWriter(path)) { tw.WriteLine("The next line!"); tw.Close(); } } //for (int i = 0; i < count; ++i) //{ // System.IO.File.Delete(@"C:\Data\C#\Example.txt"); //} } [Serializable()] public class Country { [System.Xml.Serialization.XmlElement("Population")] public string Population { get; set; } [System.Xml.Serialization.XmlElement("Name")] public string Name { get; set; } } [Serializable()] [System.Xml.Serialization.XmlRoot("CountriesCollection")] public class CountriesCollection { [XmlArray("Countries")] [XmlArrayItem("Country", typeof(Country))] public Country[] Country { get; set; } } private void btnDelete_Click(object sender, EventArgs e) { //DataSet ds = new DataSet(); //ds.ReadXml(@"C:\Users\Curso\Desktop\xmls\qq.xml"); // dataGridView1.DataSource = ds.Tables[0]; /* public static T DeserializeFromXml<Empleado>(@"C:\Users\Curso\Desktop\xmls\qq.xml"); { T result; XmlSerializer ser = new XmlSerializer(typeof(T)); using (TextReader tr = new StringReader(xml)) { result = (T)ser.Deserialize(tr); } return result; }*/ /* XmlSerializer ser = new XmlSerializer(typeof(Car)); Car myObject; using (XmlReader reader = XmlReader.Create(@"C:\Data\A\cars.xml")) { myObject = (Car)ser.Deserialize(reader); }*/ CountriesCollection countries = null; string path = @"C:\Data\A\cars.xml"; XmlSerializer serializer = new XmlSerializer(typeof(CountriesCollection)); StreamReader reader = new StreamReader(path); countries = (CountriesCollection)serializer.Deserialize(reader); reader.Close(); foreach (Country country in countries.Country) { Debug.WriteLine("País: " + country.Name + " Población: " + country.Population); } /*foreach(Country country in countries) { Debug.WriteLine("Countries:: " + country); }*/ Console.WriteLine("{" + string.Join(", ", countries.Country.ToString()) + "}"); （记入@San's answer to a different question）。

然后，我简单地进行了更改并成功推送到Heroku！