Question

我有一个为实例分配属性的构造函数：

#!/bin/bash

# Updates all svn externals in a remote repository
# argument 1 is the server
# argument 2 is the path to the externals directory relative to the server root
#
# Note: all externals in the repository must be relative to the server root.
# See `svn help pset`
#
# Example:
# If you have externals in https://example.com/svn/project1/trunk, then:
#
# this-script https://example.com /svn/project1/trunk
#

server=$1
repo=$2

# Get the commited properties
svn checkout $server$repo wc --depth=empty
svn pget svn:externals wc > wc.externals

# Remove empty lines
sed -i '/^$/d' wc.externals

# for each external in existing properties:
while read line; do

  # tokenize
  IFS='@ ' tokens=( $line )

  # extract interesting stuff from the tokens
  extrepo=${tokens[0]}
  rev=${tokens[1]}
  dir=${tokens[2]}

  # Query the external repository for the last change
  latest=$(svn info $server$extrepo | sed -ne 's/^Last Changed Rev: //p')

  # Write a new set of externals based on latest info
  echo "$extrepo@$latest $dir" >> wc.newexternals 

done <wc.externals

# Get the differences between the new and old externals
details=$(diff -y --suppress-common-lines wc.externals wc.newexternals)

# If differences exist (and the diff was valid)
retval=$?
if [ $retval -eq 1 ]; then

  # Set the new properties
  svn pset svn:externals -F wc.newexternals wc

  # Generate the SVN log
  echo "Automatic externals update" > wc.log
  echo $details >> wc.log

  # Commit the changes
  svn ci -F wc.log wc
fi

# Cleanup
rm -rf wc wc.externals wc.newexternals wc.log

然后我可以创建一个这样的实例：

class BaseModel {
    constructor (args = {}) {
        for (let key in args) {
            this[key] = args[key]
        }
    }
}

class User extends BaseModel {
    name: string
}

我现在想用class-transformer代替此基本功能：

let user = new User({name: 'Jon'})

我的代码库在很多地方都使用第一种方法，因此我想在构造函数中实现新方法，这样我就不会破坏旧代码：

let user = plainToClass(User, {name: 'Jon'})

如何获取构造函数中的类类型？我不能使用User，因为这是基本模型，其他类也可能从其扩展。

Answer 1

我想它是new.target。 Doc
要获取构造函数，请使用new.target.prototype.constructor，它的属性为name-获取类名。

class BaseModel {
    typeName: string;

    constructor (args = {}) {
        this.typeName = new.taget.constructor.name;
    }
}

class User extends BaseModel {
}

const user = new User();
const typeName = user.typeName; // this would return 'User'

在构造函数中使用plainToClass

1 个答案: