Question

I have a string that contains something like this:

##### abc 'foo'
/path/to/filename:1
##### abc 'bar'
/path/to/filename:1

The string can potentially be very long (say, 50 lines) and doesn’t change often.

I would like to fetch the last occurrence of text in between the single-quotes (bar in this example). This is similar to someone else’s Python problem (except the answer there doesn’t work for me in Lua, as seen far below).

I could parse each line, and put the results into an array, and then just take the last element of the array, but that doesn’t seem elegant to me:

local text = [[
    ##### abc 'foo'
    /path/to/filename:1
    ##### abc 'bar'
    /path/to/filename:1
]]

local arr = {}
local pattern = "abc '([^']+)'"
for s in text:gmatch(pattern) do
  table.insert(arr, s)
end
print('last:', arr[#arr])

I’m interested in using Lua string patterns to search the string from the end. The pattern I tried below starts from the beginning instead of the end:

local text = [[
    ##### abc 'foo'
    /path/to/filename:1
    ##### abc 'bar'
    /path/to/filename:1
]]

-- FIXME: pattern searches from beginning
local pattern = "abc '([^']+)'.*$"

local s = text:gmatch(pattern)()
assert(s == 'bar', 'expected "bar" but saw "'..s..'"')
print('last:', s)

This yields:

input:12: expected "bar" but saw "foo"

What string pattern specifies the “reverse search” I’m looking for?

Answer 1

You could use

local pattern = ".*abc '([^']+)'"

The .* is greedy so it chews up as much as it can before it matches (in this case, it chews up all the earlier matches and gives you the last).

Or if you really wanted, you could reverse your string and (sort of) your pattern too, but I think it’s better to rely on the greedy .* 😛

pattern = "'([^']+)' cba"
print(text:reverse():gmatch(pattern)())           -- rab
print(text:reverse():gmatch(pattern)():reverse()) -- bar