对Kendo Grid列/ DataSource进行多重排序 - 动态设置排序(Multiple sort on Kendo Grid columns / DataSource - set sorting dynamically)
我想要完成的是当用户在kendo网格中对列进行排序时应用“自动”辅助列排序。
所以在这个JS小提琴示例中 ,如果用户按“值”排序,它也将按“名称”排序。 请注意,0会一起排序,但名称不是按字母顺序排列的。 我希望它们是按字母顺序排列的(次要排序)。
这是尝试重写数据源排序以实现此目的。 我正在使用用户的原始排序并在“SortedName”上添加其他排序。 根据记录的排序数组,它似乎很接近,但仍然无法正常工作。
关于如何实现这一目标的任何其他想法?
注意:我不希望允许用户按多列排序。 我正在使用它的真实世界示例可能有多达50多列(不幸的是),因此多种排序可能会令人困惑/不直观。 而且我希望它能在幕后完成而无需额外的用户交互。
覆盖kendo数据源sort()的示例代码:
dataSource.originalSort = dataSource.sort; dataSource.sort = function () { // take the user's sort and apply sorting on an additional column // the sort array should look like this: [ { field: "Value", dir: "asc" }, // this is what the user sorted by { field: "SortedName", dir: "asc" }, // and I'm adding this ] return dataSource.originalSort.apply(this, arguments); }
What I'm trying to accomplish is to apply an "automatic" secondary column sort when a user sorts a column in a kendo grid.
So in this JS fiddle example, if a user sorts by "Value", it'll also sort by "Name". Note that the 0s are sorted together, but the names aren't alphabetical. I'd like them to be alphabetical (the secondary sort).
Here's an attempt at overriding the datasource sorting to accomplish this. I'm taking the user's original sort and the adding an additional sort on "SortedName". Based on the sorting array that's logged, it seems to be close but is still not working.
Any other ideas on how to accomplish this?
Note: I don't want to allow users to sort by multiple columns. The real world example I'm using this for can have up to 50+ columns (unfortunately), so multiple sort can get confusing / unintuitive. And I'd like it to be done behind the scenes without extra user interaction.
Example code for overriding kendo datasource sort():
dataSource.originalSort = dataSource.sort; dataSource.sort = function () { // take the user's sort and apply sorting on an additional column // the sort array should look like this: [ { field: "Value", dir: "asc" }, // this is what the user sorted by { field: "SortedName", dir: "asc" }, // and I'm adding this ] return dataSource.originalSort.apply(this, arguments); }
原文:https://stackoverflow.com/questions/38757379
最满意答案
SELECT o.CustomerId, COUNT(DISTINCT o.OrderId) AS OrderCount, COUNT(oi.OrderItemId) AS OrderItemCount, COUNT(oi.OrderItemId) / COUNT(DISTINCT o.OrderId) avg FROM OrderItem oi INNER JOIN Order o ON o.OrderId = oi.OrderId WHERE o.CategoryId = 52 -- website sales GROUP BY o.CustomerId order by COUNT(oi.OrderItemId) / COUNT(DISTINCT o.OrderId) desc
只需将加入添加到客户
SELECT o.CustomerId, COUNT(DISTINCT o.OrderId) AS OrderCount, COUNT(oi.OrderItemId) AS OrderItemCount, COUNT(oi.OrderItemId) / COUNT(DISTINCT o.OrderId) avg FROM OrderItem oi INNER JOIN Order o ON o.OrderId = oi.OrderId WHERE o.CategoryId = 52 -- website sales GROUP BY o.CustomerId order by COUNT(oi.OrderItemId) / COUNT(DISTINCT o.OrderId) desc
Just add in the join to customer
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