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Excel - 将以 Z 结尾的 ISO8601 日期转换为适合输入“带时区的时间戳”类型的 Postgres 字段的格式

[英]Excel - convert ISO8601 date ending in Z into format suitable for entry into Postgres field with type 'timestamp with time zone'

原文 2019-12-05 23:27:40 5 1 excel/ csv/ datetime/ iso8601

I have a CSV file that I'm opening in Excel.我有一个要在 Excel 中打开的 CSV 文件。 It has dates in this format:它具有以下格式的日期：

2019-09-15T00:11:57.4030000Z

I want this:我要这个：

2019-09-15 00:11:57.403+00

I think I may be able to use the "format cells > custom" option in Excel, but what would do I need to specify as the 'type' for the format?我想我可以在 Excel 中使用“格式化单元格 > 自定义”选项，但是我需要指定什么作为格式的“类型”？ I tried using this to get most of it, but it doesnt' work:我尝试使用它来获取大部分内容，但它不起作用：

yyyy-mm-dd hh:mm:ss.000

And when I try to apply some of the pre-existing types built into Excel, they don't seem to change the date format - it's as if the original date format isn't being recognised as valid.当我尝试应用 Excel 中内置的一些预先存在的类型时，它们似乎没有更改日期格式 - 就好像原始日期格式未被识别为有效一样。

1 个解决方案

My OP asks how to do this in Excel, however I found it easier to parse the file with a regex.我的 OP 询问如何在 Excel 中执行此操作，但是我发现使用正则表达式解析文件更容易。 I used the excellent dnGrep tool for this, with the following pattern, that groups the date and time into substitution groups, then uses them in the replacement as $1 and $2 :为此，我使用了出色的dnGrep工具，采用以下模式，将日期和时间分组到替换组中，然后在替换中将它们用作$1和$2 ：

search - ([0-9]{4}[-][0-9]{2}[-][0-9]{2})[T]([0-9]{2}:[0-9]{2}:[0-9]{2}.[0-9]{3})[0-9]{4}Z搜索 - ([0-9]{4}[-][0-9]{2}[-][0-9]{2})[T]([0-9]{2}:[0-9]{2}:[0-9]{2}.[0-9]{3})[0-9]{4}Z

replace - $1 $2+00更换 - $1 $2+00