Kusto：如何将列表列转换为行答案

【问题标题】：Kusto:How to convert list columns to rowsKusto：如何将列表列转换为行
【发布时间】：2021-07-08 09:04:19
【问题描述】：

我在表格中有一列，列类型是字符串。

["2021-03-18T08:49:08.4670000Z","2021-03-18T13:41:59.3230000Z","2021-03-18T13:46:59.2390000Z","2021-03-18T15:00:13.1960000Z","2021-03-18T16:19:47.0000000Z","2021-03-18T16:26:49.8480000Z","2021-03-18T17:06:49.7910000Z","2021-03-18T18:19:44.8760000Z","2021-03-18T19:34:44.8710000Z","2021-03-18T19:48:01.7160000Z","2021-03-18T20:23:01.6540000Z","2021-03-18T20:56:49.7960000Z","2021-03-18T22:21:49.7880000Z","2021-03-18T23:44:47.6500000Z","2021-03-19]

我想将此列表保存为行；喜欢：

Start_time
2021-03-18T08:49:08.4670000Z
2021-03-18T13:41:59.3230000Z
2021-03-18T13:46:59.2390000Z

我尝试了一些东西，但它不会起作用。

【问题讨论】：

标签： kql

【解决方案1】：

你应该把它变成动态的并使用 mv-expand:

datatable(Values:string)['["2021-03-18T08:49:08.4670000Z","2021-03-18T13:41:59.3230000Z","2021-03-18T13:46:59.2390000Z","2021-03-18T15:00:13.1960000Z","2021-03-18T16:19:47.0000000Z","2021-03-18T16:26:49.8480000Z","2021-03-18T17:06:49.7910000Z","2021-03-18T18:19:44.8760000Z","2021-03-18T19:34:44.8710000Z","2021-03-18T19:48:01.7160000Z","2021-03-18T20:23:01.6540000Z","2021-03-18T20:56:49.7960000Z","2021-03-18T22:21:49.7880000Z","2021-03-18T23:44:47.6500000Z","2021-03-19"]']
| extend Values =todynamic(Values)
| mv-expand Values to typeof(datetime)

请注意，您示例中的最后一个值缺少结束语。作为旁注，最好首先将列键入为动态。

【讨论】：

感谢您的解决方案。但这不适用于我的情况。

【解决方案2】：

我尝试了以下方式

将动态列转换为字符串

| mv-expand start_time =split(start_time, ",") 到 typeof(string) | mv-expand end_time =split(end_time, ",") 到 typeof(string)
将字符串转换为日期时间。

project todatetime(start_time), todatetime(end_time), Events; q2

【讨论】：