能力值:
( LV2,RANK:10 )
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2 楼
double pin = 3.14
string ls_no
string ls_name
string ls_key
double ll_no
double ll_name
double ll_system
double ll_enddate
double ll_diskid
double ll_key
integer i
string ls_zipdate
ls_key = string((rand(1000) * rand(1000)) * 3.14)
ls_key = left(ls_key,8)
if isnull(as_serial_no) then
return ls_key
end if
if trim(as_serial_no) = "" then
return ls_key
end if
if isnull(as_company_name) then
return ls_key
end if
if trim(as_company_name) = "" then
return ls_key
end if
ls_zipdate = f_zip_date(as_enddate)
as_serial_no = as_serial_no + "3.0"
for i = 1 to len(as_serial_no)
ll_no = ll_no + logten(asc(mid(as_serial_no,i,1)) / i + i)
next
for i = 1 to len(as_company_name)
ll_name = ll_name + logten(asc(mid(as_company_name,i,1)) / i + i)
next
for i = 1 to len(as_sys_name)
ll_system = ll_system + logten(asc(mid(as_sys_name,i,1)) / i + i)
next
for i = 1 to len(as_enddate)
ll_enddate = ll_enddate + logten(asc(mid(as_enddate,i,1)) / i + i)
next
for i = 1 to len(as_diskid)
ll_diskid = ll_diskid + logten(asc(mid(as_diskid,i,1)) / i + i)
next
ll_key = logten((ll_no + ll_name + ll_system + ll_enddate + ll_diskid) * 3.14)
ls_key = string(ll_key)
i = pos(ls_key,".")
if i > 0 then
ls_key = replace(ls_key,i,1,"")
end if
ls_key = left(ls_key,8)
return (ls_zipdate + ls_key)
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能力值:
( LV2,RANK:10 )
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3 楼
谢谢大家,给点提示。我写个注册机
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能力值:
( LV12,RANK:570 )
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4 楼
占个位置!其实看过去好象很简单的样子!不过现在比较忙.不好意思了!
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能力值:
( LV2,RANK:10 )
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-
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5 楼
trim(as_userid)
rand(1000)
mid(as_sys_name,i,1)
....
这些都是pb的函数,完成的功能从函数名就能推测到。
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