NodeJs：从具有类型定义的文件中读取并作为 class 定义写入文件

Question

我有一个带有类型定义的打字稿文件。我需要找到特定的类型名称并将其写入另一个文件，但作为 class。例如：

type exampleOne = {
    atrA: string
    atrB: number
}
type exampleTwo = {
    atrA: number
    atrB: string
    atrC: string
}

并将 exampleTwo 写入另一个文件：

class exampleTwo {
    atrA: number
    atrB: string
    atrC: string
}

我有这个想法，但不知道如何实现：

读取文件
找到我想要的类型名称
select 从我要查找的行的开头，直到下一个结束括号
将单词 'type' 替换为 'class' 并取消等号
写入另一个文件

Answer 1

您可能正在寻找这个：https://www.typescriptlang.org/docs/handbook/typescript-in-5-minutes.html

在这种情况下，也许您正在尝试输入类似这样的内容？

class exampleTwo {
  atrA: number;
  atrB: string;
  atrC: string;

constructor(atrA: number, atrB: string, atrC: string)
  this.atrA = atrA;
  this.atrB = atrB;
  this.atrC = atrC;
}

const newExample = new exampleTwo(2, 'hello', 'world');

Answer 2

由于我没有找到解决这个问题的方法，所以我最终通过在bash中编写自己的脚本来解决这个问题。我可以使用 javascript 但我更喜欢 bash:

# Take as argument a name for reference
NEW=

# Verify argument is passed
if [ $# -ne 1 ]; then
    echo "One argument is required."
    exit 1
fi

# Location where the types are
PRISMA_URL='./node_modules/.prisma/client/index.d.ts'

# Location to save the file
NEW_DTO_CREATE_URL="./src/dtos/create-"${NEW}".dto.ts"

# Verify if already exists that file
if [ -f ${NEW_DTO_CREATE_URL} ]; then
    echo "Already exists"
else

  # Get the line number where starts the type I need
  LINE_N=$(grep -nm1 "export type ${NEW}CreateManyInput = {" ${PRISMA_URL} | grep -Po '^[^:]+')
  echo "export class Create${NEW^}Dto {" >> ${NEW_DTO_CREATE_URL}
  # Read all the line content on that line number
  LINE=$(head -n ${LINE_N} ${PRISMA_URL} | tail -1)

  # Start the loop. While LINE isn't a closed bracket...
  while [[ ${LINE} != *"}"* ]]; do
      ((LINE_N+=1))
      LINE=$(head -n ${LINE_N} ${PRISMA_URL} | tail -1)
      echo "${LINE}" >> ${NEW_DTO_CREATE_URL}
  done
fi

NodeJs：从具有类型定义的文件中读取并作为 class 定义写入文件

NodeJs: read from file with type definitions and write to file as class definition

javascript

node.js

typescript

prisma