Lua replace string with function

Question

I want to replace matched string with some function.

I've used '%1' to find strings, but I couldn't use the matched strings.

print(text) showed me %1, not matched string.

original_text = "Replace ${test01} and ${test02}"

function replace_function(text)
    -- Matched texts are "test01" and "test02"
    -- But 'text' was "%1", not "test01" and "test02"
    local result_text = ""

    if(text == "test01") then
        result_text = "a"
    elseif(text == "test02") then
        result_text = "b"
    end

    return result_text
end

replaced_text = original_text:gsub("${(.-)}", replace_function("%1"))

-- Replace result was "Replace  and"
-- But I want to replace "Replace ${test01} and ${test02}" to "Replace a and b"
print(replaced_text)

How can I use matched string in gsub?

luther · Accepted Answer · 2021-07-20 00:25:43Z

2

The problem is that replace_function gets called before gsub can start running. replace_function doesn't know what %1 means, and it doesn't return a string that has any special meaning to gsub.

However, the following information from the gsub doc tells us that you can pass replace_function directly to gsub:

If repl is a function, then this function is called every time a match occurs, with all captured substrings passed as arguments, in order.

original_text:gsub("${(.-)}", replace_function)

answered Jul 20, 2021 at 0:25

luther

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