[英]What is the right way ? With javascript and mysql
我有一個名為“cars”的mysql表,我在其中存儲:“品牌”,“模型”,“顏色”。 我想得到這樣的數據:
[
{
brand: "audi",
cars: [
{
"model": "coupe",
"color": "red",
},
{
"model": "a3",
"color": "blue",
},
]
},
{
brand: "renault",
cars: [
{
"model": "modus",
"color": "white",
},
{
"model": "clio",
"color": "green",
},
]
},
...
]
所以,我正在做的是第一個mysql查詢,我按品牌分組,然后我迭代結果以獲得每個品牌的所有汽車:
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)
})
}
})
我正在展示的代碼正在運行,但我不認為對mysql查詢進行迭代是一件好事。
我做了很多研究,我真的不知道該怎么做。
是否可以通過一個單獨的mysql查詢獲得此結果? 或者我應該從“cars”表中獲取所有行,然后使用JS填充格式化數據? 或者我應該繼續使用這個mysql查詢迭代?
謝謝
你可以這樣做:
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()
這樣的事情應該可以解決問題。 它的客戶端轉型雖然:
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})
}
})
我們可以通過使用GROUP_CONCAT
, CONCAT
mysql函數來實現單個查詢本身。 一旦你得到結果汽車鑰匙將JSON字符串。 您可以使用JSON.parse方法進行轉換
select brand,CONCAT("[",GROUP_CONCAT(CONCAT("{'model':'",model,"','color':'",color,"'}")),"]") as cars from cars GROUP BY brand;
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