# How to write a unique number in a single Excel cell

I'm using Labview 2013 SP1 with the Microsoft Report Builder and I can't understand how to write a number of DP to a specific cell in an Excel spreadsheet. I have no problem writing tables of numbers to specific cell ranges, formatting of cells, the writing of titles, using models, everything I need, except the possibility to write a number to a specific cell. All vi who write the numbers seems to require tables 2D for the entries. I must be missing something, but I think I crossed every vi in the palette of MRG and found nothing.

Any help/examples were greatly appreciated.

... and you probably know that the best way to do it is with build table, twice.

Note the lines is 'thick' goes from left to right.

Bob Schor

Tags: NI Software

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2
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;

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Hi! everyone ,

I see one encryption function in my database.

-> 12412913141313139139130121

My question is

As:

# Select f_pwd_decrypt ('12412913141313139139130121') of double

```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
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vdb_pwd := null;

for i in  1..(lengthb(vtr_pwd)/4)  loop
vtemp1 := substrb(vtr_pwd,(i-1)*4+1,4);
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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

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 > /.

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

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.

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Argument1,

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Hello

< pre >

SELECT DISTINCT A.LKP_ID AS ID
D.PART_CNDTN_CD as a CONDITION
D.PART_DISPOSTN_CD AS PROVISION
C.LKP_NM AS DESCRIPTION
OF FWOWNER. CLM_LKP_VALUE HAS
FWOWNER. CLM_LKP_USAGE B
FWOWNER. C CLM_LKP_VALUE_LANG
IWOWNER. WC_CLM_SRVC_PART D
WHERE B.LKP_USED_AS IN ('RR_CAUSAL_PART_CNDTN_CODE', 'rr_parts_disposition')
AND A.LKP_REF BETWEEN B.LKP_REF AND B.LKP_REF + '.' 99999
and A.LKP_ID = C.LKP_ID
AND C.LANG_CD = 'ENG_USA. '
AND A.LKP_LVL = 3
and A.LKP_VALUE = BOX
WHEN B.LKP_USED_AS IN ('RR_CAUSAL_PART_CNDTN_CODE')
THEN D.PART_CNDTN_CD
WHEN B.LKP_USED_AS IN ('rr_parts_disposition')
THEN D.PART_DISPOSTN_CD
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< / pre >

top query let me out like,

< pre >
AVAILABLE FOR THE ID CODE DESCRIPTION
----- ------- ---------- ---------------------------------
25789 CP22 leak (null)
25795 CP28 scars / covered
25796 CP29 (null) Warped/distorted/curly
25768 CP01 PROCESSED Bent/pleated/Fold/bent/blister
CP10 25777 chipped
Contamination CP12 25779
25786 CP19 grooved/crosses (null)
25793 pitted CP26 (null)
25768 CP01 (null) Bent/Creased/Fold/twisted/Crinkled
25777 chipped CP10 (null)
25783 dented CP16 (null)
25790 obsolete CP23 / supplant (null)
25797 CP30 (null) worn too
* 25799 CP32 (null) other *.
* Other CP32 25799 *.
28225 CP07 REVIEW REVIEW
25771 CP04 Broken/fractured (null)
25779 CP12 Contamination (null)
25780 CP13 corroded/eroded (null)
25781 CP14 Cracked (null)
25785 CP18 Galled/frettees (null)
25788 CP21 (null) failure to point
25795 CP28 Scarred/Spalled (null)
CP30 25797 door too
28225 CP01 REVIEW REVIEW

28 response (s) selected [extract metadata: 0/ms] [extract data: 0/ms]

< / pre >

Please see the part "BOLD", I get two duplicates with a vertical drop of NULL and empty space I want to generate a unique number on the CONDITION column and available how to do this.

MERGE TWO COLUMNS?
)

This is not Oracle...

In any case:

``````SELECT  DISTINCT A.LKP_ID          AS ID
,       D.PART_CNDTN_CD          as CONDITION
,       CASE WHEN LENGTH(D.PART_DISPOSTN_CD)=0 or D.PART_DISPOSTN_CD IS NULL THEN null ELSE D.PART_DISPOSTN_CD END  AS DISPOSITION
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FROM     FWOWNER.CLM_LKP_VALUE          A
,     FWOWNER.CLM_LKP_USAGE          B
,     FWOWNER.CLM_LKP_VALUE_LANG      C
,     IWOWNER.WC_CLM_SRVC_PART     D
WHERE     B.LKP_USED_AS          IN ( 'RR_CAUSAL_PART_CNDTN_CODE', 'rr_parts_disposition')
AND     A.LKP_REF          BETWEEN     B.LKP_REF     AND     B.LKP_REF + '.99999'     -- No quotes
and     A.LKP_ID          = C.LKP_ID
AND     C.LANG_CD          = 'ENG_USA'
AND     A.LKP_LVL          = 3
and      A.LKP_VALUE          = CASE
WHEN  B.LKP_USED_AS     IN ( 'RR_CAUSAL_PART_CNDTN_CODE')
THEN  D.PART_CNDTN_CD
WHEN  B.LKP_USED_AS     IN ('rr_parts_disposition')
THEN  D.PART_DISPOSTN_CD
END```
```

Max
[My Italian blog Oracle | http://oracleitalia.wordpress.com/2010/02/07/aggiornare-una-tabella-con-listruzione-merge/]