Lua string.gsub text between pattern

Question

How would I extract in Lua a text between a pattern. For example

s="this is a test string. <!2014-05-03 23:12:08!> something more"

I would need only the date/time as result: 2014-05-03 23:12:08
print(string.gsub(s, "%<!.-%!>")) doesn't work
I would need all the text WITHOUT the date/time like: "this is a test string. something more"

If you can be sure that no other < or > occur inside the text; use the %b matching: s:gsub( '%b<>', '' ) — hjpotter92
– hjpotter92, Commented May 19, 2014 at 7:41

Yu Hao · Accepted Answer · 2014-05-19 03:26:35Z

8

The pattern "<!.-!>" works, but you need to use string.match to get the date/time part:

 print(string.match(s, "<!(.-)!>"))

Note that you don't need to escape ! or < in a pattern. Of course escaping them is not an error.

To get the string without the date/time part, replace it with an empty string:

local result = string.gsub(s, "<!.-!>", "")
print(result)

You can also expand the pattern .- to validate the format of date/time more:

result = string.gsub(s, "<!%d%d%d%d%-%d%d%-%d%d%s+%d%d:%d%d:%d%d!>", "")

answered May 19, 2014 at 3:26

Yu Hao

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1 Comment

Amazing one ! =)