How to write a decryption function?

Hi! everyone ,


I see one encryption function in my database.


Select f_pwd_encrypt ('password') of double


-> 12412913141313139139130121


My question is


How to write a decryption function?


As:


Select f_pwd_decrypt ('12412913141313139139130121') of double


->password

CREATE OR REPLACE FUNCTION EPADM."F_PWD_ENCRYPT" ( vpwd in varchar2)
return varchar2
is
    vother_p   varchar2(9);
    vtr_pwd  varchar2(2048);
    i       number;
    j         number;
    vsubstr varchar2(9);
    vtemp1 varchar2(08);
    vvalue number := 0;
    vdb_pwd varchar2(100);


    function str_2_bit(vstring in varchar2)
    return varchar2
    is
       i number;
       vtemp number;
       v1 varchar(2048);


     function single_byte(vin in number)
       return varchar2
       is
         i number;
         vresult varchar2(08);
         vtemp number := vin;
       begin
         for i in 1..8 loop
           vresult := to_char(mod(vtemp,2))||vresult;
           vtemp := trunc(vtemp/2);
         end loop;
         return(vresult);
       end;
  --
    begin
      for i in 1..lengthb(vstring) loop
        select to_number(substrb(dump( vstring ,10,i,1),instr(dump( vstring ,10,i,1),' ',-1)+1))
          into vtemp
        from dual;
        v1 := v1 || single_byte(vtemp);
      end loop;


      return(v1);
    end;


begin
  vtr_pwd := str_2_bit(vpwd);
  vtr_pwd := substrb(vtr_pwd,4)||substrb(vtr_pwd,1,3);
  vvalue := 0;
  vdb_pwd := null;


   for i in  1..(lengthb(vtr_pwd)/4)  loop
     vtemp1 := substrb(vtr_pwd,(i-1)*4+1,4);
     for j in 1..4 loop
       vvalue :=  vvalue + to_number(substrb(vtemp1,j,1)) * power(2,j-1);
       dbms_output.put_line(j||' '||vvalue);
     end loop;
          vdb_pwd := to_char(vvalue) ||vdb_pwd;
          vvalue := 0;
   end loop;
  return(vdb_pwd);


END;
/

OK, after reviewing the, I don't think you'll be able to write a function of decryption for him.

The first thing he does is take the ascii value of each character in the password and converts them into a binary string.  The code it uses is far too complex and can be simplified, but which is not a problem here.

I've recreated the first step of SQL like this...

SQL > ed

A written file afiedt.buf

1 with chr_val like)

2. Select level l

3, dump('password',10,level,1) in the dmp

4, to_number (substrb (dump('password',10,level,1), instr (dump('password',10,level,1),' ', 1) + 1)) as chr_val

5, ascii (substr('password',level,1)) as chr_val - equivalent of extraction of useless dump

6 double

7. connect by level<=>

8        )

9, r (l, b, ch, chr_val, result, vtemp) as

10 (select l, 0 b, chr (chr_val), chr_val)

11, cast (null as varchar2 (8)) as a result

12, chr_val as vtemp

13 of chr_val

14 union of all the

15 select l, b + 1, b, ch, chr_val

16, to_char (mod(vtemp,2)) | result as a result

17, trunc(vtemp/2) as vtemp

18 r

where the 19 b + 1<=>

(20) depth search first by l, defined b seq

21, as)

22 select l, ch, chr_val, str_to_bit result

23 r

where the 24 b = 8

25 arrested by l, seq

26            )

27 select listagg (ch) within the Group (order) as password

28, listagg (chr_val, ',') within the Group (order) byte_vals

29, listagg (str_to_bit) within the Group (order) bit_vals

30 sec.

SQL > /.

PASSWORD BYTE_VALS BIT_VALS

--------------- ---------------------------------------- ----------------------------------------------------------------------------------------------------

password 0111000001100001011100110111001101110111011011110111001001100100 112,97,115,115,119,111,114,100

Then he takes this string binary ("bit_vals" in my example) and does the following:

1. take the first 3 bits of left and to transpose on the right end of the string.

2 chops the resultant bit string upward into sections of 4 bits (which is known as a 'nibble' inside)

3. for each bit in the nibble, he treats the bits in binary reverse to normal and gives them a value of 1,2,4 or 8 from left to right for each bit set to 1

4. for each nibble it adds the value of 1,2,4,8 bits to give a value from 0 to 15

To show that the use SQL...

SQL > byte (select ' 0111000001100001011100110111001101110111011011110111001001100100' as pieces of double)

2, swap3bit as (-take the binary string and put the first bits (high) 3 as a (low) bit of the right hand side)

3. Select bytes.bits

4, substr (bit 4) | substr (bits, 1, 3) as init_substr

5 bytes

6                   )

7, split4 as (-chop the string of bits nibbles (half bytes - 4 bits))

8. Select level l

9, substr (init_substr, ((level-1) * 4) + 1, 4) as a nibble

swap3bit 10

11. connect by level<=>

12                 )

13, bitpowers (select l

14, snack

15, to_number (substr(nibble,1,1)) * power (2, 1-1) as bitval1

16, to_number (substr(nibble,2,1)) * power (2, 2-1) as bitval2

17, to_number (substr(nibble,3,1)) * power (2, 3-1) as bitval3

18, to_number (substr(nibble,4,1)) * power (2, 4-1) as bitval4

19, to_number (substr(nibble,1,1)) * power (2, 1-1) +.

20 to_number (substr (nibble, 2, 1)) * power (2, 2-1) +.

21 to_number (substr (nibble, 3, 1)) * power (2, 3-1) +.

22 to_number (substr (nibble, 4, 1)) * power (2, 4-1) as total_val

23 of split4

24                   )

25 select * from bitpowers

26.

L NIBB BITVAL1 BITVAL2 BITVAL3 BITVAL4 TOTAL_VAL

---------- ---- ---------- ---------- ---------- ---------- ----------

1 1000          1          0          0          0          1

2 0011          0          0          4          8         12

3 0000          0          0          0          0          0

4 1011          1          0          4          8         13

5 1001          1          0          0          8          9

6 1011          1          0          4          8         13

7 1001          1          0          0          8          9

8 1011          1          0          4          8         13

9 1011          1          0          4          8         13

10 1011          1          0          4          8         13

11 0111          0          2          4          8         14

12 1011          1          0          4          8         13

13 1001          1          0          0          8          9

14 0011          0          0          4          8         12

15 0010          0          0          4          0          4

16 0011          0          0          4          8         12

16 selected lines.

These final values are then re-combination as strings in reverse order so that you get then:

'12' |' 4'||' 12' |' 9'||' 13'... and so on.

In SQL...

SQL > byte (select ' 0111000001100001011100110111001101110111011011110111001001100100' as pieces of double)

2, swap3bit as (-take the binary string and put the first bits (high) 3 as a (low) bit of the right hand side)

3. Select bytes.bits

4, substr (bit 4) | substr (bits, 1, 3) as init_substr

5 bytes

6                   )

7, split4 as (-chop the string of bits nibbles (half bytes - 4 bits))

8. Select level l

9, substr (init_substr, ((level-1) * 4) + 1, 4) as a nibble

swap3bit 10

11. connect by level<=>

12                 )

13, bitpowers (select l

14, snack

15, to_number (substr(nibble,1,1)) * power (2, 1-1) as bitval1

16, to_number (substr(nibble,2,1)) * power (2, 2-1) as bitval2

17, to_number (substr(nibble,3,1)) * power (2, 3-1) as bitval3

18, to_number (substr(nibble,4,1)) * power (2, 4-1) as bitval4

19, to_number (substr(nibble,1,1)) * power (2, 1-1) +.

20 to_number (substr (nibble, 2, 1)) * power (2, 2-1) +.

21 to_number (substr (nibble, 3, 1)) * power (2, 3-1) +.

22 to_number (substr (nibble, 4, 1)) * power (2, 4-1) as total_val

23 of split4

24                   )

25 select listagg (to_char (total_val)) the Group (order of the desc) as pwd

26 of bitpowers

27.

PWD

-------------------------------------------------------------------------------------------------------------------

12412913141313139139130121

Now, the problem of decryption is that these numbers are concatenated without padding for a fixed number of digits by value, so you don't know if it was

'12' |' 4'||' 12' |' 9'||' 13'... and so forth as we did it, or whether he was

'1'||' 2'||' 4'||' 12' |' 9'||' 1'||' 3'... and so on, or any other combination of values from 0 to 15

There is essentially no information to allow you to divide the string upwards in the correct components to allow the whole process be reversed.

So, you are out of luck... no chance of decrypting it.

Tags: Database

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    Thanks and greetings

    Srikkanth.M

    To submit a competing request from the back - end:

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    Summary

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    Description IN varchar2 default NULL,

    start_time IN varchar2 default NULL,

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    Argument1,

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    Related essays: overview of the Message dictionary (see page)

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    After a call to the FND_REQUEST. SUBMIT_REQUEST function, installation of all parameters are reset to their default values.

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    short name of the application associated with the concurrent request for enforcement.
    short simultaneous program (not the executable) name of the program for which the application must be made.
    Description Description of the application that appears in the form of concurrent requests (optional).
    start_time time during which demand is expected to start running in the (optional) HH24 or HH24:MI:SS format.
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