mysql to append a number of stars based on the string length - mysql

I have a requirement that when a string length is less than 10, I need to append as many stars(*) at the end till length equals to 10.
ex:
ipString = "Stack"
opString = "Stack*****"
Is there any way to do it?

Use RPAD
SELECT RPAD('STACK',10,'*')
More accurately:
SET #str := 'Stack ';
SELECT
IF(LENGTH(#str) < 10, RPAD(#str,10,'*') ,#str)
WORKING DEMO If the length of the string is less than 10
WORKING DEMO If the length of the string is greater than or equal to 10
Note on RPAD:
Description
MySQL RPAD() function pads strings from right. The actual string which is to be padded, length of the string returned after padding and string which is used for padding - all these are passed as arguments.
Syntax
RPAD(str, len, padstr)
Arguments
Name Description
str The actual string which is to be padded.
len Length of the string returned after padding.
padstr String which will be used for padding.

Related

Separating character and numeric variables without delimiter in SAS

I have below data with 5 variables with 5 observations: Name Age Gender Weight country
data have;
length string $30.;
input string$;
datalines;
Naresh30Male70India
Venkey29Male50Kenya
Ravi30Male56Pak
Sai67Female40iran
Divya89Female78Dubai
;
run;
I want to Separate these 5 variables in 5 observations
Help me on it
This is fixed code that should work for any range of ages/weights. Not the most elegant solution, but it works. But I would still rather fix that at the input site, and put some delimiters into the original string.
data want;
set have;
mixed2=substr(Mixeddata, anydigit(Mixeddata), anydigit(Mixeddata,-50)-anydigit(Mixeddata)+1);
mixed3=substr(mixed2, notdigit(mixed2));
name=substr(Mixeddata,1, anydigit(Mixeddata)-1);
age=substr(mixed2,1,notdigit(mixed2)-1);
gender=substr(mixed2, notdigit(mixed2), anydigit(mixed2, 5)-notdigit(mixed2));
country=substr(Mixeddata, anydigit(Mixeddata, -50)+1);
weight=substr(mixed3, anydigit(mixed3));
drop Mixeddata mixed2 mixed3;
run;
Consider the makeup of a single string:
Naresh30Male70India
Each desired column is either a number or a character. If we could break this out into two strings, one with only numbers and one with only characters, we can easily pull the needed values:
string_num: 30 70
string char: Naresh Male India
We can do this with regular expressions by replacing letters with spaces and numbers with spaces.
data want;
set have;
string_num = compbl(prxchange('s/[0-9]/ /', -1, string) );
string_char = compbl(prxchange('s/[a-zA-Z]/ /', -1, string) );
name = scan(string_num, 1);
age = scan(string_char, 1);
gender = scan(string_num, 2);
weight = scan(string_char, 2);
country = scan(string_num, 3);
drop string_num string_char;
run;
Note that we use the compbl function to remove any extra spaces to make it easier to read for learning purposes, but this is an optional step.

Converting from base-32 to decimal without VBA

I've been searching and searching and my Google-fu has failed me. I'm trying to convert an encoded number from base-32 to decimal using either expressions or a macro, but I'm not finding anything. I know Excel has the "Decimal" function, I've been hoping that I could stumble onto something similar.
I'm reluctant to use VBA as I don't want to spend time re-learning the language right now and I'm worried that my organization will flag it as potentially dangerous (which could kill my attempts at making any databases).
With an input of "16O9E55"
I expect a result of 1300543653.
I should clarify that this is "base32hex" according to Wikipedia. It's 0-9, A-V. It's only 7 characters of base-32 that needs to convert to 10 digits of decimal. My use case is decoding a barcode into the data I need.
I doubt this can be accomplished without VBA. Consider code adapted from https://www.excelbanter.com/excel-worksheet-functions/150198-formulat-convert-base-32-decimal.html
Public Function Base32ToDec(Num As String) As Variant
Static Digits As String
Dim i As Integer
Dim myIndex As Integer
Dim myStr As String
Digits = "0123456789ABCDEFGHIJKLMNOPQRSTUV"
For i = Len(Num) To 1 Step -1
myStr = Mid(Num, i, 1)
myIndex = InStr(Digits, myStr) - 1
Base32ToDec = Base32ToDec + myIndex * 32 ^ (Len(Num) - i)
Next i
End Function
According to Wikipedia, Base32 uses a 32-character set comprising the twenty-six upper-case letters A–Z, and the digits 2–7. The variant base32hex starts with 0 - 9 and uses the letters A to V.
If only numbers with a limited range have been encoded, you can decode them easily with VBA, otherwise you would have to return an array of bytes and process it further.
You write, that you have up to 10 decimal digits. The question is, what the maximum number is. The Long type can store numbers up to 2,147,483,647. This are ten digits; however, with 10 digits you could store a number as big as 9,999,999,999.
Therefore, the following function returns the number as Double. If you know that your number will never exceed 2,147,483,647, then you can exchange the Double type by Long for the sum variable and the function return type.
Public Function DecodeBase32hex(ByVal encoded As String) As Double
Dim ch As String
Dim sum As Double
Dim d As Long, i As Long
For i = 1 To Len(encoded)
ch = Mid$(encoded, i, 1)
If ch >= "A" And ch <= "Z" Then
d = Asc(ch) - Asc("A") + 10
ElseIf ch >= "0" And ch <= "9" Then
d = Asc(ch) - Asc("0")
Else
Exit For 'E.g. padding charachters
End If
sum = 32 * sum + d
Next i
DecodeBase32hex = sum
End Function
Test in Access' immediate window:
?DecodeBase32hex("16O9E55")
1300543653

Remove All data in string after 10th entry

I have this data in a string 0871234567ThisPartOfTheStringIsRandom
How Do I update the string to just keep the first 10 Chars?
Please Keep in mind I have thousands of entries where 'ThisPartOfTheStringIsRandom' is different in every case
The LEFT function is a string function that returns the left part of a string with a specified length.
UPDATE TableA
SET YourColumn = LEFT(YourColumn,10)

How to convert binary floating point numbers to decimal numbers with verilog?

I am writing a program. there is a binary floating number like this format : XX.XXX. for example,binary floating number 01.101 convert to decimal number is 1.625. I tried it for a long time, but couldn't work it out.
I use [4:0]num to store the number. num[4:3] is the integer part, num[2:0] is the floating part. the integer part is easy, when num[2:0]=3'b101, it means that the binary floating part is 0.101, and convert to decimal number is 0.625. so how can I convert the sequence"101", get a sequence "625"?
The quickest way is probably just to use a LUT (lookup table). Since the fractional portion is only 3 bits, that leaves you with only 8 possibilities. And you could use an 12bit value where each nibble is a digit that could be sent to a display etc.
reg result[11:0] // each nibble represents a digit
always #(*) begin
case (num[2:0])
3'b000 : result = 12'h000;
3'b001 : result = 12'h125;
3'b010 : result = 12'h250;
3'b011 : result = 12'h375;
3'b100 : result = 12'h500;
3'b101 : result = 12'h625;
3'b110 : result = 12'h750;
default : result = 12'h875;
endcase
end

How to split this String in two parts?

I would like to split a string like this in Access 2000 (Visual Basic function):
"[Results]
[Comments]"
in two parts:
the results part
the comments part
As you can notice, these two parts are separated by an empty line (always, this is our separator).
[Results] and [Comments] are blocks of text. We don't care what's in it except:
the results part doesn't have any empty lines in it, so the first empty line we see is the separator one.
I want my function to extract the Comments part only.
Here is what i tried:
Public Function ExtractComm(txt As String) As String
Dim emptyLine As Integer
txt = Trim(txt)
'emptyLine = first empty line index ??
emptyLine = InStrRev(txt, (Chr(13) + Chr(10)) & (Chr(13) + Chr(10)))
'Comments part = all that is after the empty line ??
ExtractComm = Mid(txt, emptyLine + 4)
End Function
But it doesn't work well.
If I do:
ExtractComm(
"Res1
Res2
Comment1
Comment2"
)
I want to obtain:
"Comment1
Comment2"
but I only obtain Comment2. Any idea to extract the comment part ?
Thanks a lot !
Maybe you need to use InStr instead of InStrRev
InStrRev
Returns the position of the first occurrence of one string within another, starting from the right side of the string.
InStr
Returns an integer specifying the start position of the first occurrence of one string within another.