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Print result as string (SOLVED)

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AylinCE
Grandmaster Cheater Supreme

Reputation: 37

Joined: 16 Feb 2017
Posts: 1550

Posted: Sat Aug 03, 2019 10:12 am Post subject: Print result as string (SOLVED)

Why the question is name-specific?
Because @ DaSpamer has original ideas for such complex encodings. Smile

If you have any ideas of course, please help.

The following code is an @Corroder publication.
He performs an Aobs scan and prints all the results.

I tried, but I couldn't find a solution Smile

I need two additions.

Code:

1) I need to get the result as "String"
Sample:

Code:

aobResult("22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 ?? ?? ?? ?? ?? ?? ?? ?? ?? 7D")

Code:

(Print) Result: "22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 32 31 35 30 32 31 34 31 34 7D"

What I want is that it returns the result as a string as follows.

Code:

New String Format Result: "", "rank": 0, "score": 215021414}"

2) Here's a question specific to @DaSpamer:
What I need are the numbers in the code.
I just have to print those numbers as a string.

Sample:

Code:

aobResult("22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 ?? ?? ?? ?? ?? ?? ?? ?? ?? 7D", 23, 9)

Code:

New String Format Result: "215021414"

Of course, if you have any relevant solution ideas, please write.
Thanks ..
_________________

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Last edited by AylinCE on Mon Aug 05, 2019 10:27 pm; edited 1 time in total

Lynxz Gaming
Expert Cheater

Reputation: 4

Joined: 01 Jul 2017
Posts: 208
Location: help

Posted: Sat Aug 03, 2019 7:54 pm Post subject: Re: Print result as string (! DaSpamer)

if im not misunderstanding
my way is i add address manually in table and make type string and set the length and i make the description as "to string"

Code:

table = getAddressList().getMemoryRecordByDescription
scan = AOBScan("22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 ?? ?? ?? ?? ?? ?? ?? ?? ?? 7D")

table("to string").address = scan[0]
print(table("to string").value)

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Corroder
Grandmaster Cheater Supreme

Reputation: 75

Joined: 10 Apr 2015
Posts: 1668

Posted: Sat Aug 03, 2019 9:43 pm Post subject:

No. 1 :

Code:

function string.fromhex(str)
str = str:gsub("%s+", "")
str = string.gsub(str, "%s+", "")
return (str:gsub('..', function (cc)
return string.char(tonumber(cc, 16)) end))
end

Result = "22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 32 31 35 30 32 31 34 31 34 7D"
new_str = string.fromhex(str)
print(new_str) --> ", "rank": 0, "score": 215021414}

No. 2

Code:

local new_str = '"rank": 0, "score": 215021414}'
local extracted = string.match(new_str, '"score":(.*)')
local repl = string.gsub(extracted,'}','')
local text = 'Result : '..repl
print(text) --> Result : 215021414

_________________

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AylinCE
Grandmaster Cheater Supreme

Reputation: 37

Joined: 16 Feb 2017
Posts: 1550

Posted: Sun Aug 04, 2019 5:26 am Post subject:

@Lynxz Gaming and @Corroder ,
Thanks for your ideas.
But the codes don't match.
Here's the code to use:

Code:

function getByteString(address, bytecount)
local bytes = readBytes(address, bytecount, true)
if bytes then
local result = ""
for i = 1, #bytes do
if #result > 0 then result = result .. " " end
result = result .. string.format("%02X", bytes[i]) end
return result end
end

function aobResult(aobStr)
res = AOBScan(aobStr)
j = 0
cnt = 1
if (res~=nil) then
while j < res.Count do
a = getByteString(res[j], 51)

--UDF1.CEMemo1.append("Address "..cnt.." = "..res[j].." --> AOB : "..a)
UDF1.CEMemo1.append(a)
j=j+1
cnt = cnt+1
end
res.destroy()
res=nil
else
UDF1.CEMemo1.append("No results found")
end
end

function num2hex(num)
local hexstr = '0123456789ABCDEF'
local s = ' '
while num > 0 do
local mod = math.fmod(num, 16)
s = string.sub(hexstr, mod+1, mod+1) .. s
num = math.floor(num / 16)
end
if s == '' then s = '0' end
return s
end

function str2hex(str1)
local hex = ''
while #str1 > 0 do
local hb = num2hex(string.byte(str1, 1, 1))
if #hb < 2 then hb = '1' .. hb end
hex = hex .. hb
str1 = string.sub(str1, 2)
end
return hex
end;
-----------------------------------
function CEButton1Click(sender)
UDF1.CEEdit2.Text=""
UDF1.CEMemo1.Clear()

a = UDF1.CEEdit1.Text -->The zID number given by the game to the user to challenge his / her own score.
b = str2hex(a)
UDF1.CEEdit2.Text=("22 69 64 22 3A 20 22 "..b.." 22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ??") --> Search Aob
end

UDF1.Show()

Result:

Code:

22 69 64 22 3A 20 22 33 31 31 30 39 38 38 35 30 36 35 22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 32 31 35 30 32 31 34 31 34 7D
22 69 64 22 3A 20 22 33 31 31 30 39 38 38 35 30 36 35 22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 32 31 35 30 32 31 34 31 34 7D
22 69 64 22 3A 20 22 33 31 31 30 39 38 38 35 30 36 35 22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 32 31 35 30 32 31 34 31 34 7D

I have to fill Memo.Lines.Text with the number given for String and "Score". Wink

_________________

Corroder
Grandmaster Cheater Supreme

Reputation: 75

Joined: 10 Apr 2015
Posts: 1668

Posted: Sun Aug 04, 2019 5:55 am Post subject: Re: Print result as string (! DaSpamer)

Aylin wrote:

1) I need to get the result as "String"
Sample:

Code:

aobResult("22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 ?? ?? ?? ?? ?? ?? ?? ?? ?? 7D")

Code:

(Print) Result: "22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 32 31 35 30 32 31 34 31 34 7D"

What I want is that it returns the result as a string as follows.

Code:

New String Format Result: "", "rank": 0, "score": 215021414}"

2) Here's a question specific to @DaSpamer:
What I need are the numbers in the code.
I just have to print those numbers as a string.

Sample:

Code:

aobResult("22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 ?? ?? ?? ?? ?? ?? ?? ?? ?? 7D", 23, 9)

Code:

New String Format Result: "215021414"

Those above the first what you want. Of course, if you have any relevant solution ideas, please write.
Thanks ..

Of course, the code I wrote which tried to give you solutions as you wish, doesn't match. Because what you want on the first topic is different from what you want on your last post. I didn't surprise about it. Has become a habit, that I knew.
_________________

Stealing Code From Stolen Code...
And Admit It.. Hmmm....Typically LOL

AylinCE
Grandmaster Cheater Supreme

Reputation: 37

Joined: 16 Feb 2017
Posts: 1550

Posted: Mon Aug 05, 2019 10:26 pm Post subject: Re: Print result as string (! DaSpamer)

Corroder wrote:

@Corroder, I was hasty again.
I apologize and thank you for the Code.
I made some arrangements with your Uncle Lua's support,

http://lua-users.org/wiki/StringLibraryTutorial

The code you provide is now compatible with the main code
And it works fine.

with codes and regulations that you gave, compatible version:

Code:

function CEButton1Click(sender)
UDF1.CEEdit2.Text=""
UDF1.CEEdit3.Text=""
a = UDF1.CEEdit1.Text
b = str2hex(a)
c = ("22 69 64 22 3A 20 22 "..b.." 22 2C 20 22 72 61 6E 6B 22 3A 20 30 2C 20 22 73 63 6F 72 65 22 3A 20 ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ??")

aobResult(c)
UDF1.CETimer1.Enabled = true
end

function CETimer1Timer(sender)
function string.fromhex(str1)
local str1 = UDF1.CEEdit3.Text
str1 = str1:gsub("%s+", "")
str1 = string.gsub(str1, "%s+", "")
return (str1:gsub('..', function (cc)
return string.char(tonumber(cc, 16)) end))
end

UDF1.CEEdit2.Text = string.fromhex(str1)

local new_str = UDF1.CEEdit2.Text
--print(new_str) (result ("id": "31109885065", "rank": 0, "score": 215022254}, {)
local re = string.find(new_str, 'e":') +3
local rep = string.find(new_str, '}') -1
local repl = string.sub(new_str, re, rep)
local text = (cnt..'_ Result : '..repl)
UDF1.CEMemo1.append(text)
UDF1.CETimer1.Enabled = false
end

UDF1.CEEdit2.visible = false
UDF1.CEEdit3.visible = false
UDF1.CETimer1.Enabled = false
UDF1.Show()

-- Result: "4_ Result : 215022254"

Thanks again @Corroder, you always deserve +1. Smile

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