Find first instance of a pattern in Lua and remove it from string

Question

I get strings in the following format:

abc:321,cba:doodoo,hello:world,eat:mysh0rts

I'd like to grab one instance of the data pairing from the string and remove it from the string, so for example if I wanted to grab the value following hello:world I'd like this to happen:

local helloValue, remainingString = GetValue("hello")

Which will return world for hellovalue and abc:321,cba:doodoo,eat:mysh0rts for remainingString.

I'm doing this rather cumbersomely with loops, what would a better way of doing it be?

It's better to show what you are doing, so that we know how to improve. — Yu Hao
– Yu Hao, Commented Dec 3, 2014 at 6:06
What's your expected output, if the input is abc:321,cba:doodoo,eat:mysh0rts,hello:world ? — Avinash Raj
– Avinash Raj, Commented Dec 3, 2014 at 6:11

Yu Hao · Accepted Answer · 2014-12-03 08:43:15Z

2

This is one way:

local str = 'abc:321,cba:doodoo,hello:world,eat:mysh0rts'

local t = {}
for k, v in str:gmatch('(%w+):(%w+)') do
    if k ~= 'hello' then
        table.insert(t, k .. ':' .. v)
    else
        helloValue = v
    end
end

remainingString = table.concat(t, ',')
print(helloValue, remainingString)

You can turn this to a more general GetValue function yourself.

edited Dec 3, 2014 at 8:43

answered Dec 3, 2014 at 6:14

Yu Hao

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Comments

lhf · Accepted Answer · 2014-12-03 10:38:36Z

1

Try also this:

local str = 'abc:321,cba:doodoo,hello:world,eat:mysh0rts'

function GetValue(s,k)
    local p=k..":([^,]+),?"
    local a=s:match(p)
    local b=s:gsub(p,"")
    return a,b
end

print(GetValue(str,"hello"))
print(GetValue(str,"eat"))

If you want to parse the whole string into key-value pairs, try this:

for k,v in str:gmatch("(.-):([^,]+),?") do
    print(k,v)
end

edited Dec 3, 2014 at 10:38

answered Dec 3, 2014 at 10:28

lhf

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Comments

vks · Accepted Answer · 2014-12-03 05:51:31Z

0

(hello:[^,]+,)

Just do a replace with empty string.The replace data and $1 are the things you want.See demo.

http://regex101.com/r/yR3mM3/24

answered Dec 3, 2014 at 5:51

vks

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1 Comment

Yu Hao Over a year ago

The question was mistakenly tagged with regex, Lua pattern isn't regular expression, so this won't work.

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