我有一个名为“alldetails”的集合,其中包含一些集合的详细信息
{
"name" : "Test1",
"table_name" : "collection1",
"column_name" : "column1"
},
{
"name" : "Test2",
"table_name" : "collection2",
"column_name" : "column2"
},
{
"name" : "Test3",
"table_name" : "collection3",
"column_name" : "column3"
}
我有 collection1,collection2 和 collection3 分别有 column1,column2,colum3
我必须从“alldetails”中获取所有名称,并且必须根据列名获取其他表的最小值和最大值。
所以我想要下面的输出
{name: ["Test1","Test2","Test3"],
date: [{min_date: "2018-12-01", max_date: "2018-12-31", name: "Test1"},
{min_date: "2018-12-01", max_date: "2018-12-31", name: "Test2"},
{min_date: "2018-12-01", max_date: "2018-12-31", name: "Test3"}]
}
我尝试了以下代码,因为非阻塞它不等待响应。
alldetails.find({}, { _id: 0 }).then(async function(result) {
let result_data = {};
let resolvedFinalArray = {};
let array = [];
result_data["name"]= [];
result_data["date"] = [];
resolvedFinalArray = await Promise.all(result.map(async value => {
result_data["name"].push(value.name)
getResult(value.table_name,value.column_name,function(response){
result_data["date"].push({min_date: response.minvalue, max_date: response.maxvalue, name:value.name})
});
}));
setTimeout(function()
{
console.log(resolvedFinalArray);
}, 3000);
});
请建议我一个解决方案。