Lesson learnt today!
The old application designed for 16 bit or 32-bit may not run properly in new terminal server setting. Believe me we had a big one!! The application was originally designed in/or 16-bit architecture and we tried to run it in new MS Terminal Server will all bell and whistle but it did not work and gave us memory access violation.
To solve this following steps were followed in the MS Terminal Server:
1. Right click My Computer
2. Select Properties
3. Click on Advance System Settings
4. Click on Settings under Advance > Performance
5. Select Data Execution Prevention tab
6. Select Turn on DEP for all programs and services except those I select: option
7. Click Add
8. Browse to the exe of the file, it will show up in the "selected" box
9. Click OK
After this the application ran fine.
astu...
Wednesday, July 11, 2012
Friday, July 6, 2012
Padding zeros (0) in front of number or numerical string in T-SQL
To skip story start from the START:
I had to create a Alternate ID from the Employee ID in the following format
ABCD00000
4 characters followed by 5 digits. Employee ID was 4 or 5 digit so would have to pad 0 if 4 digits. I used STR combined with REPLACE function to do this.
START:
Syntax
STR (expression [, length [, decimal] ] )
Return type: varchar
expression: numerical vaue (numerical string would work too)
length: Length of resulting string including - + sign, decimal and digits
decimal: digits after decimal
REPLACE(string, pattern, replacement)
Return type: nvarchar or varchar
string: where you want to replace
pattern: what you want to replace
replacement: string to replace with
Example:
SELECT REPLACE(STR(123.45,10,4), ' ', '0')
SELECT REPLACE(STR('123.45',10,4), ' ', '0')
Result:
00123.4500
SELECT REPLACE(STR(123.45,10,0), ' ', '0')
SELECT REPLACE(STR('123.45',10,0), ' ', '0')
Result:
0000000123
SELECT REPLACE(STR(12345,10,0), ' ', '0')
SELECT REPLACE(STR('12345',10,0), ' ', '0')
Result:
0000012345
Astu...
I had to create a Alternate ID from the Employee ID in the following format
ABCD00000
4 characters followed by 5 digits. Employee ID was 4 or 5 digit so would have to pad 0 if 4 digits. I used STR combined with REPLACE function to do this.
START:
Syntax
STR (expression [, length [, decimal] ] )
Return type: varchar
expression: numerical vaue (numerical string would work too)
length: Length of resulting string including - + sign, decimal and digits
decimal: digits after decimal
REPLACE(string, pattern, replacement)
Return type: nvarchar or varchar
string: where you want to replace
pattern: what you want to replace
replacement: string to replace with
Example:
SELECT REPLACE(STR(123.45,10,4), ' ', '0')
SELECT REPLACE(STR('123.45',10,4), ' ', '0')
Result:
00123.4500
SELECT REPLACE(STR(123.45,10,0), ' ', '0')
SELECT REPLACE(STR('123.45',10,0), ' ', '0')
Result:
0000000123
SELECT REPLACE(STR(12345,10,0), ' ', '0')
SELECT REPLACE(STR('12345',10,0), ' ', '0')
Result:
0000012345
Astu...
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