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MongoDB 通过搜索找到用户返回所有用户

[英]MongoDB find user by search returning all users

I am trying to query a user by their name field, and I am testing it in insomnia.我正在尝试通过他们的姓名字段查询用户,并且我正在对其进行失眠测试。 I have my user schema like so:我有这样的用户架构:

const userSchema = mongoose.Schema({
  id: { type: String },

  name: {
    type: String,
    required: true,
  },
  companion: {
    type: String,
    required: false
  },
  bio: { 
    type: String,
    required: false
  },
  email: {
    type: String,
    required: true,
  },
  password: {
    type: String,
    required: true,
    min: 5
  },
  userImage: {
    type: String,
    required: false,
  },
});

And I have my route for /search which runs the getUserBySearch function: router.get('/search', getUserBySearch)我有我的 /search 路线,它运行 getUserBySearch function: router.get('/search', getUserBySearch)

getUserBySearch logic: getUserBySearch逻辑:

export const getUserBySearch = async (req, res) => {
                         // tried req.query and req.query.search
  const { searchQuery } = req.params
  try {
              // make the search query not case sensitive
    const user = new RegExp(searchQuery, `i`)

    //find the user's name using the name field
    const userFound = await User.find({name: user})
    res.json({data: userFound})

  } catch (error) {
    res.status(404).json({message: error.message})
  }
}

Tested in insomnia under the route: http://localhost:3001/user/search?searchQuery=test在路由下失眠测试: http://localhost:3001/user/search?searchQuery=test

I should only be receiving the users whose name field include test ;我应该只接收name字段包含test的用户; however I get back 200 response with ALL of the users in the DB.但是我得到了数据库中所有用户的 200 响应。 how can I only retrieve the users related to my search query?我怎样才能只检索与我的搜索查询相关的用户?

In your case http://localhost:3001/user/search?searchQuery=test , you should be accessing the ?searchQuery=test as req.query.searchQuery :在您的情况下http://localhost:3001/user/search?searchQuery=test ,您应该访问?searchQuery=test作为req.query.searchQuery

const { searchQuery } = req.query;
const user = new RegExp(searchQuery, `i`);

Since, you were wrongly reading the query value, you had undefined as a value for user constant, hence getting all users in the DB.由于您错误地读取了查询值,因此您undefineduser常量的值,因此将所有用户都放入数据库中。

exports.searchRecipe = async(req,res) => {
  try {
    let searchTerm = req.body.searchTerm;
    let user = await user.find( { $text: { $search: searchTerm, $diacriticSensitive: true } });
    res.render('search', {user} );
  } catch (error) {
    res.satus(500).send({message: error.message || "Error Occured" });
  } 
}

and in the search form try this,#在搜索表单中试试这个,#

<form method="POST" action="/search">
            <input type="search" name="searchTerm" class="form-control" placeholder="Search..." aria-label="Search">
        </form>

try like this,试试这样

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