I am trying to count the characters without the spaces.
Here is my code:
word = 'Ricardo Cadenas'
def string_lenght(word):
count = 0
for char in word:
count = count + 1
return count - word.count('')
print(string_lenght(word))
my out put is-1 ??? any ideas?
You have a typo. I think you meant to put word.count(' ') but you put word.count('') (no space.)
Also instead of looping to calculate count, you can initialize count = len(word).
Correct, changing “” to “ “ should do the trick!
Related
To enter the correct mob number and dob we have to check the code in progres
Use LENGTH(STRING(mobnum)). This will convert it to a character string and give you the length of it. You can check to see if it is 10 characters long.
Making minimal changes to your approach, something along these lines should work:
define var mobnum as INT64 label "Mobile No:" format "9999999999" .
update mobnum.
if mobnum >= 1000000000 and mobnum < 10000000000 then
do:
display "Correct number".
end.
else
do:
display "enter exactly 10 digits please".
update mobnum.
end.
You can use LENGTH(mobnum) to get the quantity of digits into the number... like:
This make possible to create separate message for each of the lengths of mobnum.
DEF VAR mobnum AS INT NO-UNDO FORMAT "9999999999" INITIAL "99999999".
DEF VAR ii AS INT NO-UNDO.
ii = LENGTH(mobnum).
CASE ii:
WHEN 9 THEN DO:
DISPLAY "Number OK".
END.
OTHERWISE DO:
DISPLAY "Number is not 9-digit long".
END.
END CASE.
But if you want to make it simple, just use:
DEF VAR mobnum AS INT NO-UNDO FORMAT "9999999999" INITIAL "99999999".
DEF VAR ii AS INT NO-UNDO.
ii = LENGTH(mobnum).
IF ii = 9 THEN DO:
DISPLAY "Number correct".
END. ELSE DO:
DISPLAY "Number incorrect".
END.
For example:
Regex: District ([0-9]{1,2})([^0-9]|$)
Input District 12 2021 returns 12
Input Southern District 3 returns 3
Input FooBar returns NULL
The function REGEXP_SUBSTR doesn't allow extracting a single capturing group.
You can use e.g. REGEXP_REPLACE(input, regex, '\\1') to replace occurrences of regex in input with the first capture group of regex.
The following stored function makes this easy to use:
DELIMITER $$
CREATE FUNCTION regexp_extract(inp TEXT, regex TEXT, capture INT) RETURNS TEXT DETERMINISTIC
BEGIN
DECLARE capstr VARCHAR(5);
DECLARE mregex TEXT;
IF inp IS NULL OR LENGTH(inp) = 0 OR inp NOT REGEXP regex THEN
RETURN NULL;
END IF;
SET capstr = CONCAT('\\', capture);
SET mregex = CONCAT('.*', regex, '.*'); -- Want to match the entire input string so it all gets replaced
RETURN REGEXP_REPLACE(inp, mregex, capstr);
END;
$$
DELIMITER ;
Used like so:
SELECT regexp_extract('District 12 2021', 'District ([0-9]{1,2})([^0-9]|$)', 1);
For those users who might be stuck with an earlier version of MySQL or MariaDB which does not have REGEXP_REPLACE available, we can also use SUBSTRING_INDEX here:
SELECT SUBSTRING_INDEX(
SUBSTRING_INDEX('Southern District 3', 'District ', -1), ' ', 1); -- 3
In SQLite I need to format a number to display it with the thousand separator and decimal separator. Example: The number 123456789 should be displayed as 1,234,567.89
What I did partially works because it does not display the thousand separator as I expected:
select *, printf ("U$%.2f", CAST(unit_val AS FLOAT) / 100) AS u_val FROM items;
u_val shows: U$1234567.89 but I need U$1,234,567.89
The following is one way that you could accomplish the result:-
select *, printf ("U$%.2f", CAST(unit_val AS FLOAT) / 100) AS u_val FROM items;
Could become :-
SELECT
*,
CASE
WHEN len < 9 THEN myfloat
WHEN len> 8 AND len < 12 THEN substr(myfloat,1,len - 6)||','||substr(myfloat,len - 5)
WHEN len > 11 AND len < 15 THEN substr(myfloat,1,len -9)||','||substr(myfloat,len-8,3)||','||substr(myfloat,len-5)
WHEN len > 14 AND len < 18 THEN substr(myfloat,1,len - 12)||','||substr(myfloat,len -11,3)||','||substr(myfloat,len-8,3)||','||substr(myfloat,len-5)
END AS u_val
FROM
(
SELECT *, length(myfloat) AS len
FROM
(
SELECT *,printf("U$%.2f",CAST(unit_val AS FLOAT) / 100)) AS myfloat
FROM Items
)
)
The innermost SELECT extracts the original data plus a new column as per your orginal SELECT.
The intermediate SELECT adds another column for the length of the new column as derived from unit_val via the printf (this could have been done in the first/innermost SELECT, getting this value simplifies (in my opinion) the outermost SELECT, or you could use multiple length(myfloats) in the outermost SELECT).
RESULT - Example
The following is the result from a testing (source column is myfloat) :-
The highlighted columns being the original columns.
The circled data being the result.
The other 2 columns are intermediate columns.
Edit
As you've clarified that the input is an integer, then :-
SELECT *,'U$'||printf('%,d',(unit_val/100))||'.'||CAST((unit_val % 100) AS INTEGER) AS u_val FROM Items
would work assuming that you are using at least version 3.18 of SQLite.
Correction
Using the SQL immediately above if the value of the last part (the cents) is less than 10 then the leading 0 is dropped. So the correct SQL is. Note for simplicity the CAST has also been dropped and rather than concatening the . it has been added to the printf formatter string so :-
SELECT
'U$' ||
printf('%,d', (unit_val / 100)) ||
printf('.%02d',unit_val % 100)
AS u_val
FROM Items
Or as a single line
SELECT 'U$' || printf('%,d', (unit_val / 100)) || printf('.%02d',unit_val % 100) AS u_val FROM Items
Here is a suggestion:
WITH cte AS (SELECT 123456789 AS unit_val)
SELECT printf('%,d.%02d', unit_val/100, unit_val%100) FROM cte;
The Common Table Expression is just there to supply a dummy value, in the absence of variables.
The %,d format adds thousands separators, but, as many have pointed out, only for integers. Because of that, you will need to use the unit_val twice, once for the integer part, and again to calculate the decimal part.
SQLite truncates integer division, so unit_val/100 gives you your dollar part. The % operator is a remainder operator (not strictly the same as “mod”), so unit_val%100 gives the cents part, as another integer. The %02d format ensures that this is always 2 digits, padding with zeroes if necessary.
I have raw finance text files that I'm importing into Access 2010 and exporting in Excel format. These files contain several 14 character length fields which represent dollar values. I'm having issues converting these fields into currency because of the 14th character. The 14th character is a number represented by a bracket or letter. It also dictates whether the unique field is a positive or negative value.
Positive numbers 0 to 9 start with open bracket { being zero, A being one, B being two,...I being nine.
Negative numbers -0 to -9 (I know, -0 is a mathematical faux pas but stay with me. I don't know how else to explain it.) start with close bracket } being -0, J being -1,K being -2,...R being -9.
Example data (all belonging to the same field/column):
0000000003422{ converted is $342.20
0000000006245} converted is -$624.50
0000000000210N converted is -$21.05
0000000011468D converted is $1,146.84
Here's the query that I'm working with. Each time I execute it, the entire field is deleted though. I would prefer to stick to a SQL query if possible but I'm open to all methods of resolution.
SET FIELD_1 = Format(Left([FIELD_1],12) & "." & Mid([FIELD_1],13,1) & IIf(Right([FIELD_1],1)="{",0,IIf(Right([FIELD_1],1)="A",1,IIf(Right([FIELD_1],1)="B",2,IIf(Right([FIELD_1],1)="C",3,IIf(Right([FIELD_1],1)="D",4,IIf(Right([FIELD_1],1)="E",5,IIf(Right([FIELD_1],1)="F",6,IIf(Right([FIELD_1],1)="G",7,IIf(Right([FIELD_1],1)="H",8,IIf(Right([FIELD_1],1)="I",9,"")))))))))),"$##0.00"), IIf(Right([FIELD_1],1)="}",0,IIf(Right([FIELD_1],1)="J",1,IIf(Right([FIELD_1],1)="K",2,IIf(Right([FIELD_1],1)="L",3,IIf(Right([FIELD_1],1)="M",4,IIf(Right([FIELD_1],1)="N",5,IIf(Right([FIELD_1],1)="O",6,IIf(Right([FIELD_1],1)="P",7,IIf(Right([FIELD_1],1)="Q",8,IIf(Right([FIELD_1],1)="R",9,"")))))))))),"-$##0.00")
here is a function that you can call to convert an input string like the ones in your example into a string formatted as you desire.
Private Function ConvertCurrency(strCur As String) As String
Const DIGITS = "{ABCDEFGHI}JKLMNOPQR"
Dim strAlphaDgt As String
Dim intDgt As Integer, intSign As Integer
Dim f As Integer
Dim curConverted As Currency
strAlphaDgt = Right(strCur, 1) ' Extract 1st char from right
f = InStr(DIGITS, strAlphaDgt) ' Search char in DIGITS. Its position is related to digit value
intDgt = (f - 1) Mod 10 ' Converts position into value of the digit
intSign = 1 - 2 * Int((f - 1) / 10) ' If it's in the 1st half is positive, if in the 2nd half of DIGITS it's negative
curConverted = intSign * _
CCur(Left(strCur, Len(strCur) - 1) & _
Chr(intDgt + 48)) / 100 ' Rebuild a currency value with 2 decimal digits
ConvertCurrency = Format(curConverted, _
"$#,###.00") ' Format output
End Function
If you need to have a Currency as returned value, you can change the type returned from String to Currency and return the content of curConverted variable.
Bye.
I'm extracting rows from a txt file.
This row contains values like this:
DESCRIPTION 1 1.234,00 15.980,00 [etc.]
I would like to extract these values (I mean only numeric values).
So I thought to find first comma, execute a for cycle backwards until first White space and execute a For cycle forward for decimals digits.
The I should go to the second comma and perform these cycles again.
Can you suggest some code that could be useful for my solution?
From your description, if you just need the decimal number before the comma, then you can do this with a pretty simple regex:
Dim s = "DESCRIPTION 1 1.234,00 15.980,00"
Dim pattern = "\d+(\.\d+)?,\d+"
Dim matches = System.Text.RegularExpressions.Regex.Matches(s, pattern)
For Each match in matches
Console.WriteLine(match.Value)
Next
'Outputs:
'
'1.234,00
'15.980,00
Here's a quick breakdown of the regex:
\d+ - \d is shorthand for [0-9], which just means "any numeric character". The + just indicates "one or more"
\. - this just matches a period character.
, - this just matches a comma.
( ... ) - parentheses just creates a group (think of it as a sub-regex)
? - question marks mean that the previous item is optional. In this case, that means that the group matching (\.\d+)? is optional, which allows you to match both 0.000,00 and 0,00
In that regex, if the comma and period are optional, then you can add a ? after them.
My Visual Basic knowledge is pretty limited, but can't you utilize the IsNumeric function available in VB.NET?
Someting like this:
' initial string/row/etc
Dim s As String = "DESCRIPTION 1 1.234,00 15.980,00"
' Split string based on spaces
Dim words As String() = s.Split(New Char() {" "c})
' Use For Each loop over split and display them
Dim word As String
For Each word In words
If IsNumeric(word) Then
Console.WriteLine(word & " is numeric")
Else
Console.WriteLine(word & " is not numeric")
End If
Next
I think you'll be needing to look at System.Text.Regex.
Match m = Regex.Match("DESCRIPTION 1 1.234,00 15.980,00", ".*?( [0-9]*?.(?'n1'[0-9]+),(?'n2'[0-9]+)))
While m.Success
System.Diagnostics.Debug.WriteLine(m.Groups["n1"].Value + " "+m.Groups["n2"].Value);
m = m.NextMatch()
End While
If the columns are fixed width, you can get the values like this:
Dim input As String = "DESCRIPTION 1 1.234,00 15.980,00"
Dim col1 As String = input.SubString(17, 12).Trim()
Dim col2 As String = input.SubString(29).Trim()