我把每一个问题都放进去了。我试图查询一个mysql数据库两次，第二次，从第一次的每个结果多次，但我无法解决如何等待第二次查询的结果，然后再继续
我想要这样的输出：
{
"data":[
{
"name":"Title result",
"images":[
{
"id":1,
"place_id":705,
"path_image":"http://3.bp.blogspot.com/-iwF-ImFpzvk/T6fKhC6F7YI/AAAAAAAAARA/FyKpNcDsP8M/s1600/asd2e1.jpg"
},
{
"id":2,
"place_id":705,
"path_image":"https://asrt.bp.com/data/photo/2014/07/22/sddrfr2.jpg",
}
]
}
]
}但我只能这样：
{
"data":[
{
"name":"Title result",
"images":[]
}这是我的密码：
return new Promise((resolve, reject) => {
const { connection, errorHandler } = deps;
let arrayData = [];
let imageData = [];
connection.query(
"SELECT * FROM places WHERE id = 705",
(error, rows, results) => {
rows.forEach((row) => {
connection.query(
"SELECT * FROM place_gallery WHERE place_id = 705",
(error, rows, results) => {
imageData = rows;
}
)
arrayData.push({ name: row.title, images: imageData })
});
if (error) {
errorHandler(error, "failed", reject);
return false;
}
resolve({ data: arrayData });
}
);
})
},如何解决这个问题？
1条答案
按热度按时间wrrgggsh1#
试试这个，另一种创建dbcall函数的方法，您可以使用
util.promisify()```const dbcall = (query) => {
return new Promise((resolve, reject) => {
connection.query(
query,
(error, rows, results) => {
if (error) return reject(error);
return resolve(rows);
});
});
};
const somefunc = async () => {
const {
connection,
errorHandler
} = deps;
let arrayData = [];
try {
const rows = await dbcall("SELECT * FROM places WHERE id = 705");
rows.forEach(async (row) => {
const imageData = await dbcall("SELECT * FROM place_gallery WHERE place_id = 705");
arrayData.push({
name: row.title,
images: imageData
});
});
} catch (error) {
console.log(error);
}
return arrayData;
}