alasql

Question

<script src="https://cdn.jsdelivr.net/alasql/0.3/alasql.min.js"></script>


var ProductList = JSON.parse('[{"ProductID":1,"ProductName":"Shoes"},{"ProductID":2,"ProductName":"Chocolate"}]');


alasql("CREATE TABLE Products (ProductID INT, ProductName string)");

alasql.tables.Products = ProductList; // when inspect, the table is present

var x = alasql("SELECT * FROM Products order by ProductID"); //throws error

Answer 1

我解决了。

db.tables.Products.data - 我缺少数据。

不需要解析。

var db = new alasql.Database();

db.exec("CREATE TABLE Products (ProductID INT, ProductName string)");

var ProductList = [{"ProductID":1,"ProductName":"Shoes"},{"ProductID":2,"ProductName":"Chocolate"}];

db.tables.Products.data = ProductList;

var res = db.exec("SELECT * FROM Products");

Answer 2

根据您分享的图片，我看到您没有提供json数据源。

您可以通过...修复它

var x=alasql("Select * FROM ?  ORDER BY ProductID",[Products])

alasql - 未定义的数据源编号 0，简单 Javascript 对象作为源然后查询，给出此错误

alasql - Data source number 0 in undefined ,simple Javascript object as source then query, gives this error