"键入'字符串 |string[]' 不可分配给类型 'string' [英] "Type 'string | string[]' is not assignable to type 'string'

问题描述

我正在使用 node + typescript 和集成的 swagger 进行 API 调用.我大摇大摆地提出以下要求

I am using node + typescript and integrated swagger for API Calls. I have a following request in swagger

http://localhost:3033/employees/search/?username=test

这里我要查找 username = test 的记录.

所以我使用 URL 模块解析查询字符串，如下所示，我在控制台中获取用户名.

So I parse the query string using URL module, like below and I am getting the username in my console.

var params = URL.parse(req.url, true).query;
console.log(params); //gives me [Object: null prototype] { username: 'test' } in my console

但每当我尝试将 params.username 分配给如下常量时

But whenever I try to assign params.username to a constant like below

const username:string =params.username;

它给了我以下错误

键入'字符串 |string[]' 不可分配给类型 'string'.类型 'string[]' 不能分配给类型 'string'.

Type 'string | string[]' is not assignable to type 'string'. Type 'string[]' is not assignable to type 'string'.

我的完整代码如下

import * as URL from 'url';

public getUserByName = async (req: Request, res: Response, next: NextFunction) => {
var params = URL.parse(req.url, true).query;
const username:string =params.username;

try {
  const findOneUserData: Employee = await this.userService.findUserByName(username);
  res.status(200).json({ data: findOneUserData, message: 'findOne' });
} catch (error) {
  next(error);
}

}

推荐答案

好的，所以我得到了我犯的错误.第一个是@ford04 在评论中提到的.我直接将 Object 分配给字符串.我在我的问题中改变了它.第二个是通过 toString() 函数将解析的查询字符串转换为字符串.

Okay ,so I got the mistake that I made. The first one was what @ford04 has mentioned in the comment. I was directly assigning Object to the string. I changed it in my question. The second was converted the parsed query-string to string by toString() function.

public getUserByName = async(req: Request, res: Response, next: NextFunction) => {
  var params = URL.parse(req.url, true).query;
  console.log(params.username);
  const username: string = params.username.toString(); //this is the key

  try {
    const findOneUserData: Employee = await this.userService.findUserByName(username);
    res.status(200).json({
      data: findOneUserData,
      message: 'findOne'
    });
  } catch (error) {
    next(error);
  }
}

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