什么是正确的方法？ 用javascript和mysql

Question

I have a mysql table, named "cars", where I store : "brand", "model", "color". I would like to get the data like this :

[
    {
        brand: "audi",
        cars: [
            {
                "model": "coupe",
                "color": "red",
            },
            {
                "model": "a3",
                "color": "blue",
            },
        ]
    },
    {
        brand: "renault",
        cars: [
            {
                "model": "modus",
                "color": "white",
            },
            {
                "model": "clio",
                "color": "green",
            },
        ]
    },
    ...
]

So, what I'm doing is a first mysql query where I group by brand, then I iterate the result to get all the cars for each brand :

const query = "SELECT brand FROM cars GROUP BY brand"
mysql.query(query, values, (err, result) => {
    for (let i = 0; i < result.length; i++) {
        const query2 = "SELECT model, color FROM cars WHERE brand = ?"
        const values2 = [result[i].brand]
        mysql.query(query2, values2, (err2, result2) => {
            result[i].cars = result2
            callback(result)
        })
    }
})

The code that I'm showing is working, but I don't think that doing an iteration for a mysql query is a good thing.

I've done a lot of research, and I don't really know what to do.

Is it possible to get this result with one single mysql query ? Or should I get all rows from the "cars" table and then do JS stuff to format the data ? Or should I continue using this mysql query iteration ?

Thank you

Answer 1

You can do something like:

const carsByBrand=[],brands={};  
function find(){
 const query = "SELECT model, color FROM cars WHERE brand IN ('renault','bmw')"
 mysql.query(query, values, (err, result) => {
    /*assuming value  to be like: [
            {
              brand:audi,
                "model": "coupe",
                "color": "red",
            },
            ]
       */

  values.forEach((car)=>{
      if(brands[car.brand]){
        carsByBrand[brands[car.brand]].cars.push(car)
      }else{
        let index=carsByBrand.push({
          brand:car.brand,
          cars:[car]
        })
        brands[car.brand]=index;
      }
    })    
  }
 })

}
find()

Answer 2

Something like this should do the trick. Its client side transformation though:

let carArr = []

const query = "SELECT brand, model, color from cars"
mysql.query(query, values, (err, result) => {
    for (let i = 0; i < result.length; i++) {
        // Get the index of the brand in the carArr array
        let brandIndex = carArr.map(c => c.brand).indexOf(result.brand)

        // If the brand is not already in carArr, enter a new value for it
        if(brandIndex == -1)
            carArr.push({"brand": result.brand, "cars": [{"model": result.model, "color": result.color}]})

        // If the brand already exists, add the car type to this specific barnd
        else
            carArr[brandIndex].cars.push({"model": result.model, "color": result.color})
    }
})

Answer 3

We can achieve by single query itself by using GROUP_CONCAT , CONCAT mysql function. once you got result cars key will JSON string. you can convert by using JSON.parse method

select brand,CONCAT("[",GROUP_CONCAT(CONCAT("{'model':'",model,"','color':'",color,"'}")),"]") as cars from cars GROUP BY brand;