在Lua中格式化整数 [英] Format integer in Lua

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本文介绍了在Lua中格式化整数的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我想将数字格式化为类似于 1,234 1,234,432 123,456,789 的数字,您就明白了.我尝试这样做,如下所示:

I'd like to format a number to look like 1,234 or 1,234,432 or 123,456,789, you get the idea. I tried doing this as follows:

function reformatint(i)
    local length = string.len(i)
    for v = 1, math.floor(length/3) do
        for k = 1, 3 do
            newint = string.sub(mystring, -k*v)
        end
        newint = ','..newint
    end
    return newint
end

如您所见,一次失败的尝试,我的问题是我无法弄清楚错误是什么,因为我正在其中运行该程序的程序拒绝向我报告错误.

As you can see, a failed attempt, my problem is that I can't figure out what the error is because the program I am running this in refuses to report an error back to me.

推荐答案

这是一个将负数和小数部分都考虑在内的函数:

Here's a function that takes negative numbers, and fractional parts into account:

function format_int(number)

  local i, j, minus, int, fraction = tostring(number):find('([-]?)(%d+)([.]?%d*)')

  -- reverse the int-string and append a comma to all blocks of 3 digits
  int = int:reverse():gsub("(%d%d%d)", "%1,")

  -- reverse the int-string back remove an optional comma and put the 
  -- optional minus and fractional part back
  return minus .. int:reverse():gsub("^,", "") .. fraction
end

assert(format_int(1234)              == '1,234')
assert(format_int(1234567)           == '1,234,567')
assert(format_int(123456789)         == '123,456,789')
assert(format_int(123456789.1234)    == '123,456,789.1234')
assert(format_int(-123456789.)       == '-123,456,789')
assert(format_int(-123456789.1234)   == '-123,456,789.1234')
assert(format_int('-123456789.1234') == '-123,456,789.1234')

print('All tests passed!')

这篇关于在Lua中格式化整数的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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