小编Lt_*_*ade的帖子

Is it possible to change a value inside a Lua bytecode? How? Any idea?

I got a script that is no longer supported and I'm looking for a way to change the value of a variable in it... The script is encrypted (loadstring/bytecode/something like that) e.g.: loadstring('\27\76\117\97\81\0\1\4\4\4\8\0\')

我可以找到我想要更改的内容(在编译脚本后通过记事本),但是如果我尝试更改值,脚本将无法工作,如果我更改并尝试重新编译它仍然无法工作: "luac: Testing09.lua: unexpected end in precompiled chunk" ...

有任何想法吗?我用一个程序做了类似的事情,使用ollydbg,但我不能用lua脚本...我有点迷失在这里,做一些谷歌搜索相当一段时间找不到方法......任何想法?

lua bytecode recompile decompiler

4
推荐指数
1
解决办法
885
查看次数

hibernate左边连接多个where条件

------我正在写sql查询它的工作正常------------------

SELECT SUM (lb.QUANTITY*mp.DB_PRICE) 
FROM ledger_balance lb 
LEFT JOIN material_price mp ON (mp.MATERIAL_ID = lb.MATERIAL_ID) 
WHERE lb.QTY_TYPE ='G' 
AND lb.BALANCE_TYPE='CUMU' 
AND lb.CUSTOMER_ID = 13
AND mp.CUSTOMER_ID = 13 AND  lb.MATERIAL_ID IN (226, 230, 362, 365)
AND mp.MATERIAL_ID IN (226, 230, 362, 365);
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--- hibernate查询它不工作------

SELECT SUM (B.quantity*P.dbPrice) 
FROM com.efl.efms.data.domain.common.LedgerBalance as B 
LEFT JOIN com.efl.efms.data.domain.material.MaterialPrice as P   
with (B.materialId = P.materialCatalog.materialId)  
WHERE   B.balanceType=:balanceType 
AND B.qtyType=:qtyType and B.customerId in (:customerId) AND P.customer.customerId IN (:customerId)  
AND B.materialId in (:materialPriceId)  
AND P.materialCatalog.materialId in (:materialPriceId) …
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java sql hibernate hql

2
推荐指数
1
解决办法
5035
查看次数

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bytecode ×1

decompiler ×1

hibernate ×1

hql ×1

java ×1

lua ×1

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