标题: 【汉化资料】通用解包器制作工具QuickBMS图文教程2 [打印本页] 作者: shane007 时间: 2009-6-11 20:21 标题: 【汉化资料】通用解包器制作工具QuickBMS图文教程2 ok so set up our c:\temp directory bye extracting our file BoneObject.hsp + ?& W: n0 G3 u ^ f" S9 ?/ C' u1. to c:\temp& ?7 c6 }: k) z1 W( D' _* ?
2. create a new text document called astro.bms 6 g2 N9 L0 A! x4 P5 ~3. and place the newest version of quickbms in the folder also. $ X& t# h, }+ J$ U- \0 n- F H% t & L# u- W& O6 j P6 d7 ~Ok so open up BoneObject.hsp in your hex editor and lets take a look at it.6 f, g6 `# o* w8 I
[attach]13365[/attach] 4 E/ m1 e, D7 v5 T. M' qgood we have some plain text.& g- L' c8 ^; n7 J7 u% z% _
you will notice I highlighted the first 4 bytes 20 50 53 48 or " PSH" that is a space followed bye P S H. $ i$ R/ b1 a E" [# ?5 H( N9 thmm that seems familiar that is the file extension only backwards. this is know as the idstring# N# W4 n- l u4 ~
so up until now you would think to write in bms ! C! P Q' R5 [# lget IDSTRING long $ ?" w8 O0 S, x4 y( R9 [3 ^: dthere is nothing wrong with that but there is a better command$ ~; l/ K% u! _0 ?9 | c! l7 ?
idstring " PSH" ! s% {1 g8 o% `5 m. V+ D. T+ O# Tmake sure you include the quotes. : I! M0 G: ^4 V7 Z5 {so open your bms string and on the first line type 3 i* l2 q8 I; ]idstring " PSH" 3 H9 y& U4 k) L2 g6 Nthe reason this command is better is it will tell the program not to run if it does not find that string don't try to extract that file. % V7 ?1 o2 V' T( {4 h5 y# [7 y"aka noob proofing it"* ?* [3 ?, w5 l9 t2 B
1 K N9 a1 Z5 b5 \Ok so now lets look at what we can read I see ( Y3 `& \, I( q2 HDatas\Texture\BoneObject\npc_nagoya_octopus01_body.dds , Datas\Texture\BoneObject\Toon.bmp , Datas\Texture\BoneObject\Toon_a.bmp , Datas\Texture\BoneObject\Toon_zero.bmp$ T& M) x' d; c! r0 p
so I will assume there are 4 files in this archive./ H, e: |; `5 C
well lets look at the next 4 bytes and see what it is 01 00 00 00 hmm that is equal to 00 00 00 01 or 1 and we have more files in this archive than that so we do not know what this represents & l6 }- V: ^0 H. x% ^4 Z1 U& B kso lets write that in bms language / e/ L1 ^' c' c9 |" \ b" J& B( c4 j4 Hget UNK1 long! f9 D& P; P N, L$ e; @+ q
this saves those 4 bytes as the variable UNK1.) v, X% A9 D @" V0 b( h1 C) a8 `
+ o; u1 V% b6 j% g F; Yok the next 4 bytes are 04 00 00 00 hmm this translates into 00 00 00 04 or 4% o( i$ `8 \' n5 @* g) D- C, I
hey that is the number of files we counted so lets write that in bms 3 F- l0 O# H- Nget FILES long ; s: h& Q! |8 N2 M5 Vthis saves those 4 bytes as the variable FILES. , A1 U. F" L6 C4 M& q' I2 @/ F
the next 4 bytes are 00 00 00 00 well that is equal to zero so for now I will write that in bms1 {, W$ d) }" L) D( G$ o6 k
get NULL1 long + j2 r! H: \8 f+ B' mthis saves those 4 bytes as the variable NULL1 # K6 e4 y. Q' E8 N- D; k$ X0 F) Z4 P Q) W
ok now we have reached the first file name Datas\Texture\BoneObject\npc_nagoya_octopus01_body.dds 7 x9 `2 a& K" J5 N- y$ Ithis is 0x36 bytes long but wait there was no indicator like the last file that told us how long the name is how do we write this? ) G+ N; m: B% z6 _ i: Wwell lets look for a pattern8 z9 x- @% ?; ^! r- U: J, o
Datas\Texture\BoneObject\npc_nagoya_octopus01_body.dds is 0x36, m8 h/ Z" N1 }1 o8 b9 l8 Y9 w
Datas\Texture\BoneObject\Toon.bmp is 0x210 b3 z$ O6 c+ B: m; A
Datas\Texture\BoneObject\Toon_a.bmp is 0x23 4 z s+ ?/ F5 h# b# S1 E% \Datas\Texture\BoneObject\Toon_zero.bmp is 0x26* j2 Q1 ~9 h! I7 ^' V e
hmm I don't see anything that makes that a pattern. ! k3 X/ X! O7 r* a. \but I do see all the names are followed bye a lot of zero's. how long is the name + the zeros of each file?& F1 n3 f- M$ a8 a
Datas\Texture\BoneObject\npc_nagoya_octopus01_body.dds + 0's is 0x80; {0 `' d" Q" c- X
Datas\Texture\BoneObject\Toon.bmp + 0's is 0x80 . r0 e/ o, B0 f7 _- \4 |Datas\Texture\BoneObject\Toon_a.bmp + 0's is 0x80$ D$ E% Q2 P: s- f& \4 Z
Datas\Texture\BoneObject\Toon_zero.bmp + 0's is 0x802 }) K% D% @) ]6 o0 Z' t) H# V
hey they are all the same size when I include the 0's 1 v5 C6 B5 r, M( F- Rso in bms I would write this as & L2 N# U$ R7 r2 F0 f, Z/ Mgetdstring NAME 0x809 W0 ~4 ]' ]/ L7 b% o2 s
this tells it to grab 0x80 bytes and store the text value of it% W* p; V" m+ F! d+ B
and as an added feature it will automatically remove trailing 0's 8 D% \* J0 V" |* E* b# } 0 p- T' t; {# [7 Z) Xok so now we have 0xC bytes before I see the next file name5 o2 i( o: b# h. ]5 r
which is 3 long values + l4 _; \4 Y. Q* gso lets write those in and we will figure out what they represent later. " Q# k: r: D! `( B0 d' Zget UNK2 long9 R6 p' l, x5 q0 r v8 ~
get UNK3 long $ |$ J7 k( b# B1 C1 t2 Y3 c& Yget UNK4 long : w' J, |* |* y, \2 E* |/ d6 p {( q- E5 U1 y& R4 ?9 i1 @1 J) w
ok so now we see the name again 5 u' d) K, q% `* `8 y$ {7 y; `7 J2 Ewe have our pattern so lets write our script based on what we learned9 ~( i8 ~. ]6 Z5 Q
so it would look like this up until now" P/ h" Z! u4 O( A
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Code:0 D) A0 C' N0 {3 z7 i( b0 Q5 I
idstring " PSH" / ]; v, l* R3 X7 Wget UNK1 long % c# f: S0 V: n6 @get FILES long " Z* V; O2 F0 l- a2 kget NULL1 long , ^* `: w$ K+ f% L; a2 Qfor i = 0 < FILES2 u7 x7 w+ \' r$ Y; N
getdstring NAME 0x80 % o" Q6 G0 B$ x. _" R( y* L& Bget UNK2 long _" ?( x" k. A6 Nget UNK3 long ' l/ X' C) M3 Z+ k; c: fget UNK4 long % K/ ?7 y* _# U' bclog NAME OFFSET ZSIZE SIZE * y; r1 I( N2 |$ M' ?1 H6 Vnext i % u g6 b4 e6 [! k# A # v4 }' Z, ^& v4 \- G" b- y% }" H; h* H6 w3 |3 r4 V+ }: ?7 {
ok this may look complex but it is almost identical to the first tutorial file except we added 1 more variable : I; a6 p0 U8 b4 J* q+ g7 L$ FZSIZE this represents the compressed file size while SIZE represents the decompressed file size ! [2 D/ v! f, nand we also changed the log command to clog to represent it is a compressed file. ' w4 H! I4 @9 D2 R$ p: M 7 `4 a. [" `4 l% A6 Y! Z# `ok so now we have our loop and the commands to extract our files but we still need to fill in the variables/ M8 c6 R3 m X2 W4 K+ X7 s
OFFSET ZSIZE SIZE / \3 H; Q# c) A+ f7 ~so that means our 3 unknown values must represent that but how do we know what order they are in? 6 i3 l9 z% W' v1 \3 \& m, K" {; b: {% h) I7 Y
Well ill let you in on a cool trick follow the file loop to the end start at Datas\Texture\BoneObject\npc_nagoya_octopus01_body.dds and highlight the whole 0x80 length9 P e* u4 |. E2 D m
then add our 3 unknown variables so that means we are highlighting 0x8C for our length. the first file is from 0x10 - 0x9B 2 V& _/ A' s% k$ q" Q# ~& @: zso now do this for the rest of the files and you end up at highlighting 0x1B4 - 0x23F* b2 w- V- b) l
[attach]13366[/attach] 0 k* z& A* ^4 m, iok so we reached the end of our loop now what? 4 r5 D: H9 {& s$ C, N' wwell the next 2 bytes are 78 9C and this is an archive extractors best friend when you see this at the start of a file. + b8 w! G! w9 n& V: |3 M+ `78 9C represents the standard zlib compression header ; y* a- l" H3 Q' sok so this means our first file starts there which is at offset 0x240 1 T: t% x6 X5 D7 H. v5 }4 N. j" v7 c1 n# ^
well lets go back to our first file in the list and look at those unknown variables.' g- ^& c* |* k0 f' n
24 72 00 00 is = 00 00 72 24 = 0x7224 ! x( u1 U- C6 }+ V7 @* P* z' L( w80 00 02 00 is = 00 02 00 80 = 0x20080. G( y+ J/ X- |% ^% K
40 02 00 00 is = 00 00 02 40 = 0x240' ^) J2 Y9 z6 W( O, |2 m% W
I think we have a winner so the third variable is 0x240 aka the offset 9 J1 {, v% e X: B5 g. cso lets update our script6 r/ R7 z8 R$ e
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$ R# h9 Q9 L' \% f. cCode: , z" k6 E7 I/ @" N- O# V$ _2 }idstring " PSH" - ?" x# ~. x5 Z! O5 O9 cget UNK1 long& c2 c! c% k0 c3 x7 V
get FILES long 6 Q% ^2 q, ]" o$ n. t" fget NULL1 long 2 u" f# p0 s* a" z. tfor i = 0 < FILES8 E+ q6 H0 i# O1 ^# C8 x2 @
getdstring NAME 0x803 A# H1 i- E9 y
get UNK2 long8 \# z4 { i$ @
get UNK3 long* q2 d5 V; L: V$ }8 _8 m
get OFFSET long ' k' N- c( J- p8 ~clog NAME OFFSET ZSIZE SIZE 8 g( a& [" C* K4 Dnext i1 J( L! s4 Q5 e) w6 a) I! y% M
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now that just leaves ZSIZE and SIZE - x4 f* U/ H1 p7 I( l$ J7 t5 U) ]) E& @well bye process of elimination the decompressed file must be bigger than the compressed file so we compare the 2 variables , V# a5 W9 `0 B( ~' h% ]+ T- t24 72 00 00 is = 00 00 72 24 = 0x7224 8 q; U) z z" _0 ~6 k80 00 02 00 is = 00 02 00 80 = 0x20080 & i" j( P t% X6 {well 0x20080 is definitely bigger so we now know the last 2 variables' ^7 C' Z8 i0 n7 ~. M( j
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Code:& [4 v- C6 ^$ V( x1 ]
idstring " PSH" : @5 {/ Y: i1 ?; }! s' E; \6 ~get UNK1 long 1 o) S, K6 J- Rget FILES long ' ]6 T t: j/ |& h: s. p4 xget NULL1 long& c* l4 }& R' C+ q1 K
for i = 0 < FILES! B. B8 i8 o; Q( x# b
getdstring NAME 0x80' c ^) b: @. T) \: Z( P4 A
get ZSIZE long # l8 X" `* d( D3 u& xget SIZE long2 f$ |- }- _, \0 Y! m$ ~
get OFFSET long ! `8 _8 c B+ }# g2 mclog NAME OFFSET ZSIZE SIZE 9 j4 Q9 A3 u/ M: S( p) o5 onext i 9 y7 C* K7 G/ Y, X/ t( {* I! u( S: I( G/ P/ ^5 B+ [* X$ B. x
" ?# D& [' @# P& D" f; vnow try our code out on the file 9 u. f6 [* M$ r& [6 d* q1 @/ ]open the command prompt and change to the directory ; @3 l8 B8 E! }* E6 \! Wc:\temp & ~6 ?$ D, o( n+ N" Fnow type 3 v) ? |9 `; ]3 k% [quickbms.exe -l astro.bms BoneObject.hsp .% r7 N5 r# n7 I+ _5 j7 s z& b! M
yay it listed our files without any errors now lets try extracting them 2 [# U3 ^; D0 n) Z" s& Screate a folder called extract( Q5 O+ w! B l6 g0 {' O1 j9 j
and type ( o1 d$ l5 g) m- t: j5 i Pquickbms.exe astro.bms BoneObject.hsp extract 6 r# c5 Y! t3 P2 N0 R' y% E6 Aif we look in there we now have folders and in those folders are 4 pictures ' o1 r. _% S) |7 z9 Fwe did it. . Z4 A# d" g" C% X . `' C3 e7 a; k1 ?+ q6 e# H# VLet me know if you want more pictures or any way I can improve the tutorials.# \7 k) a* P& }5 K. X3 K& w
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Last edited by chrrox on Tue Jun 09, 2009 2:33 pm, edited 1 time in total.作者: herojimmy 时间: 2009-9-9 21:53
cool turotials ,thx!作者: 深绿 时间: 2009-10-7 21:48
应群里朋友之请，翻译了一下这篇教程，不是完全照翻的，后面差不多是自己rewrite了，希望会有帮助 . T# }3 s w' \, i* R7 Y—————————————————————————————— ; j, P$ {- r" n: s8 Y6 P建立一个目录C:\\temp来放我们解包出的文件 ; J9 e& f; X/ p* z8 O! z, V1 O" Y" W$ x; D; _
1，进入C:\\temp文件夹 7 ^$ ~ G2 g1 i7 L2，建立一个新文件 astro.bms（QuickBMS解包脚本） 0 x% [4 ?, O* z# j, y: e3，把最新版的quickbms也放到这个文件夹; R+ p% b4 f0 L5 t0 h. _
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现在，用你的十六进制编辑器打开BoneObject.hsp，来好好观察一下6 @8 [# Z+ S, p. X2 l. M' l) j
（图） 7 r2 V& M; z* Q, J : Y$ k6 p9 j6 m3 t) u7 I' a很好，我们看到了一些清楚的文字& Z! s* _! a1 X9 _2 i
8 s% _/ g$ e. e. z- Q' Y: ~你会注意到最开始4个字节20 50 53 48，是空格跟上PSH% j. |2 G. |# F
看起来就是文件后缀名的反向排列 , B( a/ \+ H- } D z - ]0 L2 G& p; k6 g6 I0 |这被称为idstring（标识字串） @. T3 i& I1 a' k% l5 T% k
所以，现在在脚本里写上一句5 Z8 E7 \: T2 x" O& o. V
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get IDSTRING long (将四个字节（long）存为IDSTRING) ' l2 r# E7 S7 k. C4 M5 `) ^8 j% M3 e
这没什么错误，不过我们有一条更好的指令/ K2 P6 G/ W o8 X2 M& H. e
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idstring " PSH" 7 N' q. i8 `! x4 J' s
4 F" @9 _; ]3 l# ` g6 b j* A确保你没漏掉引号。) C4 Y' m: ?% \, o( d
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这条指令更好是因为你可以告诉程序，如果没有在开头找到这个标识符，那么就不要解包这个文件。 ; T5 c' O# t. L2 R8 o6 d. V) R6 c& h! n+ c
之后继续观察文件，我们可以看到" N* V: A3 y/ D2 A# [
Datas\\Texture\\BoneObject\\npc_nagoya_octopus01_body.dds , Datas\\Texture\\BoneObject\\Toon.bmp , Datas\\Texture\\BoneObject\\Toon_a.bmp , Datas\\Texture\\BoneObject\\Toon_zero.bmp $ r0 _$ J3 v8 ~+ A9 Y/ y所以我假设有4个文件在这个包里。: L/ x$ {& b5 Y4 U, h
. M9 H6 b1 ]+ yok，回到开头标识符，接下来看之后的四个字节，是01 00 00 00，那等于00 00 00 01或者1，文件数量比这要多，所以我们不明白这代表什么& b, x# H' `' ?8 N, Q. r% q
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那么我们在脚本里写这么一句 1 r* @6 x4 u! h: Y8 K2 C ( {$ A; G" p8 l# k T0 Hget UNK1 long , T6 N7 R6 D) k% D这句指令把4个字节存为变量UNK10 W# g4 A% y9 y: E. P, i
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ok，之后四个字节是04 00 00 00，就是00 00 00 04或者4 # d/ @. L; X' T7 u3 m! \这就是包里的文件数量，所以我们在脚本里写这么一句： / _' x/ i( W P& I: I ( H( H' I. {1 j: I/ Xget FILES long( H' n s" z3 {* I
这一句把4个字节存为变量FILES $ p% z s/ Q+ l, @4 z$ Q! y% H4 n, P K( r0 I, |& U7 ^
之后四个字节是00 00 00 00，嗯，那就代表0 ( k: G ?+ R2 W. Z e6 c于是我们这么写" j' \2 [, v9 o- W! M
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get NULL1 long) t5 c8 ?) M" \, w8 S
把这四个字节存为变量 NULL1 $ g1 ~% p3 q' x: e4 N8 s" b* F- ], w5 ^
好了，现在我们到达了第一个文件的文件名部分 7 G# u" p6 U2 }1 z: gDatas\\Texture\\BoneObject\\npc_nagoya_octopus01_body.dds4 y, G' V; M/ K( @9 c9 G
这个字串的长度是0x36，不过等等，这儿没有一个标示符告诉我们文件名的长度，那么我们该怎么写脚本呢？/ X3 x/ ?) f$ Z: |! w' s3 c8 W
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well，我们来找找规律 * E6 w# z# y& Y& j0 ZDatas\\Texture\\BoneObject\\npc_nagoya_octopus01_body.dds is 0x36 6 h) N$ v9 y, o4 L7 H* ^, Y
Datas\\Texture\\BoneObject\\Toon.bmp is 0x21% b8 M3 e& N" C0 x/ i2 t J
Datas\\Texture\\BoneObject\\Toon_a.bmp is 0x23 ) C$ c$ N4 b" s7 y9 b# J* ^2 X. lDatas\\Texture\\BoneObject\\Toon_zero.bmp is 0x26, n+ q( V& A d- m* @" A
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看起来没什么规律，呵呵% W- _% m, Y" E& {; u- C. X- t
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不过我注意到，文件名之后都跟着一大堆的00，那么把文件名加上那些0，长度是多少呢？9 I' i; d$ w0 B! U
. p- t1 n1 ]7 Y# H: n- f3 gDatas\\Texture\\BoneObject\\npc_nagoya_octopus01_body.dds + 0's is 0x80 $ s' u; [' ~; S, U8 BDatas\\Texture\\BoneObject\\Toon.bmp + 0's is 0x80 - W& O0 P5 ?' q1 X" MDatas\\Texture\\BoneObject\\Toon_a.bmp + 0's is 0x80! j, D4 U* g4 c$ O# Q+ O$ i
Datas\\Texture\\BoneObject\\Toon_zero.bmp + 0's is 0x80, w# q. E* Z& Y* L. Q