im having very big string on 1st row.so 1st row contains lots of items with comma like below
12345,54322,44444,222222222,444444,121,333,44444,........
I just need to split this till 50 items in every row. lets assume there are 700 items separated with comma and I want to keep till 50 items only in 1st row and then next 50 in 2nd row and so on.
I tried with the below code which splits till 50 for sure but im not sure if this will works going forward. so need help on this
OutData = Split(InpData, ",")(50)
MsgBox OutData
You can do this in many more ways, but one would be to replace every nth comma. For example through Regular Expressions:
Sub Test()
Dim s As String: s = "1,2,3,4,5,6,7,8,9,10,11"
Dim n As Long: n = 2
Dim arr() As String
With CreateObject("vbscript.regexp")
.Global = True
.Pattern = "([^,]*(?:,[^,]*){" & n - 1 & "}),"
arr = Split(.Replace(s, "$1|"), "|")
End With
End Sub
The pattern used means:
( - Open 1st capture group;
[^,]* - Match 0+ (Greedy) characters other than comma;
(?: - Open a nested non-capture group;
,[^,]* - Match a comma and again 0+ characters other than comma;
){1} - Close the non-capture group and match n-1 times (1 time in the given example);
), - Close the capture group and match a literal comma.
Replace every match with the content of the 1st capture group and a character you know is not in the full string so we can split on that character. See an online demo
I suppose you can do whatever you like with the resulting array. You probably want to transpose it into the worksheet.
I need help. I developed an Excel sheet that as I scan an employee barcode it will extract the base-32 code information so I can get the employee ID number, first name and last name using different formulas. Excel Sheet
The only problem is the formulas to extract this data is different based on how the code starts out as seen in the Excel Sheet. I can use the IFS formula in Excel on O365 but all of our agencies use the standard desktop version of Excel.
My question; is there a way to code out in VBA that when an ID is scanned, regardless of what the scanned code starts with, that it will perform the needed formula to extract the three items I need which is ID, first name and last name? Below are the formulas I use:
Scan starting with "M"
Base-32 =MID(A2,2,7)
First Name =PROPER(MID(A2,17,20))
Last Name =PROPER(MID(A2,38,20))
Scan Starting with "N"
Base-32 =MID(A3,9,7)
First Name =PROPER(MID(A3,16,20))
Last Name =PROPER(MID(A3,36,26))
Scan Starting with "C"
Base -32 =MID(A4,8,7)
First Name =PROPER(MID(A4,15,20))
Last Name =PROPER(MID(A4,35,20))
ID NUMBER
The ID number for each of them is calculated the same (based on the cell the scan goes in to) using:
=IF(C2="","0",SUMPRODUCT(POWER(32,LEN(C2)-ROW(INDIRECT("1:"&LEN(C2)))),(CODE(UPPER(MID(C2,ROW(INDIRECT("1:"&LEN(C2))),1)))-48*(CODE(MID(C2,ROW(INDIRECT("1:"&LEN(C2))),1))<58)-55*(CODE(MID(C2,ROW(INDIRECT("1:"&LEN(C2))),1))>64))))
Thank you in advance to anyone that can help.
Not sure if this is exactly in line with your requirements, but the following UDF could be used to retrieve your data:
Function GetData(inp As String, grp As Long) As String
With CreateObject("VBScript.RegExp")
.Pattern = "^(?:M|N.{7}|C.{6})(.{7})\S*([A-Z][a-z]+)\s*(\S+)"
If .Test(inp) Then
GetData = .Execute(inp)(0).Submatches(grp - 1)
Else
GetData = "No Data Found"
End If
End With
End Function
Here is an online demo how the pattern works. It would match:
^ - Start line anchor.
(?:M|N.{7}|C.{6}) - A non-capture group to either capture a literal 'M', or a literal 'N' followed by 7 characters other than newline, or a literal 'C' followed by 6 of those characters.
(.{7} - Then a 1st capture group of 7 characters to mimic the MID() functionality, capturing the Base-32 code.
\S* - 0+ (Greedy) non-whitespace characters, upto:
([A-Z][a-z]+) - A 2nd capture group to capture the lastname through a single uppercase alphachar and 1+ (Greedy) lowercase ones.
\s* - 0+ (Greedy) whitespace characters upto:
(\S+) - A 3rd capture group to catch the first name through 1+ (Greedy) non-whitespace characters.
You'd call this function in your sheet through =GetData(A1,1) to get the 'Base-32' code and use integer 2 to get the last name and a 3 to get the first name. I hope that helped a bit.
Given a sentence "hi I stay at 4th cross street and my ssn number is 56tyuh". I want to remove words such as alphanumeric ( 4th and 56tyuh ). Does isalpha() is used only to check if there are alphanumerics in sentences? If not how do I use it to remove alphanumerics
You'll need to use regex for this. Regex can be confusing but in this case, it's quite straight forward.
import re
s = 'hi I stay at 4th cross street and my ssn number is 56tyuh'
r = r'\S*\d+\S*'
cut_string = re.sub(r, '', s)
Let's break this down:
r is a regex variable, which detects character sequences of 0-n leading non-whitespace characters, followed by 1-n numeric charcters and again 0-n trailing non-whitespace characters.
re.sub replaces the matches of our regex with the second parameter, in our case an empty string. Thus it removes all matches of our regex from the string.
Edit:
This will also remove numbers. If you only want to remove alphanumeric words, make the follwing change:
r = r'([a-zA-Z]*\d+[a-zA-Z]+|[a-zA-Z]+\d+[a-zA-Z]*)'
Note the | in the center of the variable. This means either match the first part within the parentheses or the second. The first would match 4th but not ep95, the opposite is true for the second.
Hello I have a column with strings (names of products) in it.
Now these are formatted as Name LenghtxWidth, example Green box 20x30. Now I need to change the 20 with the 30 in this example so I get Green box 30x20, any ideas how I can achieve this?
Thanks
Here is both a formula solution, as well as a VBA solution using Regular Expressions:
Formula
=LEFT(A1,FIND(TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)),A1)-1)&
MID(TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)),SEARCH("x",TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)))+1,99)&
"x"&
LEFT(TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)),SEARCH("x",TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)))-1)
UDF
Option Explicit
Function RevWL(S As String)
Dim RE As Object
Const sPat As String = "(\d+.?\d*)x(\d+.?\d*)"
'If L or W might start with a decimal point, and not a digit,
'Then change sPat to: (\d*.?\d+)x(\d*.?\d+)
Set RE = CreateObject("vbscript.regexp")
With RE
.Global = True
.ignorecase = True
.Pattern = sPat
RevWL = .Replace(S, "$2x$1")
End With
End Function
Here is an example of the kinds of data I tested with:
The Formula works by looking at the last space-separated substring which would be LxW, then reversing the portion after and before the x, then concatenating everything back together.
The regex pattern captures the two numbers (could be integers or decimals, so long as the start with an integer -- although that could be changed if needed), and reversing them.
Here is a more detailed explanation of the regex (and the replacement string) with links to a tutorial:
(\d+.?\d*)x(\d+.?\d*)
(\d+.?\d*)x(\d+.?\d*)
Options: Case insensitive; ^$ don’t match at line breaks
Match the regex below and capture its match into backreference number 1 (\d+.?\d*)
Match a single character that is a “digit” \d+
Between one and unlimited times, as many times as possible, giving back as needed (greedy) +
Match any single character that is NOT a line break character .?
Between zero and one times, as many times as possible, giving back as needed (greedy) ?
Match a single character that is a “digit” \d*
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) *
Match the character “x” literally x
Match the regex below and capture its match into backreference number 2 (\d+.?\d*)
Match a single character that is a “digit” \d+
Between one and unlimited times, as many times as possible, giving back as needed (greedy) +
Match any single character that is NOT a line break character .?
Between zero and one times, as many times as possible, giving back as needed (greedy) ?
Match a single character that is a “digit” \d*
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) *
$2x$1
Insert the text that was last matched by capturing group number 2 $2
Insert the character “x” literally x
Insert the text that was last matched by capturing group number 1 $1
Created with RegexBuddy
Here is a VBA solution that will work for you:
Option Explicit
Function Switch(r As Range) As String
Dim measurement As String
Dim firstPart As String
Dim secondPart As String
measurement = Right(r, Len(r) - InStrRev(r, " "))
secondPart = Right(measurement, Len(measurement) - InStr(1, measurement, "x"))
firstPart = Left(measurement, InStr(1, measurement, "x") - 1)
Switch = Left(r, InStrRev(r, " ") - 1) & " " & secondPart & "x" & firstPart
End Function
You can paste this in a regular module in the VBE (Visual Basic Editor) and use it as a regular function/formula. If your value is in cell A1 then type =Switch(A1) in cell B1. Hope it helps!
Ok, so it is really easier to use VBA, but if you want only some formulas you can use some columns to split your text and then concatenate your cells.
Here is a little example:
Of course B1-4 are optional. It is here only to have something more readable, but you can do use only one formula
=CONCATENATE(LEFT(A1, SEARCH(" ",A1,1)-1)," ",RIGHT(RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)),LEN(RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)))-SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)),1)),"x",LEFT(RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)), SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)),1)-1))
If you have several spaces in your names, you can use this formula that will search the last space in the text
=CONCATENATE(LEFT(A1, SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))-1)," ",RIGHT(RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))),LEN(RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))))-SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))),1)),"x",LEFT(RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))), SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))),1)-1))
How do I concatenate together two strings, of unknown length, in COBOL? So for example:
WORKING-STORAGE.
FIRST-NAME PIC X(15) VALUE SPACES.
LAST-NAME PIC X(15) VALUE SPACES.
FULL-NAME PIC X(31) VALUE SPACES.
If FIRST-NAME = 'JOHN ' and LAST-NAME = 'DOE ', how can I get:
FULL-NAME = 'JOHN DOE '
as opposed to:
FULL-NAME = 'JOHN DOE '
I believe the following will give you what you desire.
STRING
FIRST-NAME DELIMITED BY " ",
" ",
LAST-NAME DELIMITED BY SIZE
INTO FULL-NAME.
At first glance, the solution is to use reference modification to STRING together the two strings, including the space. The problem is that you must know how many trailing spaces are present in FIRST-NAME, otherwise you'll produce something like 'JOHNbbbbbbbbbbbbDOE', where b is a space.
There's no intrinsic COBOL function to determine the number of trailing spaces in a string, but there is one to determine the number of leading spaces in a string. Therefore, the fastest way, as far as I can tell, is to reverse the first name, find the number of leading spaces, and use reference modification to string together the first and last names.
You'll have to add these fields to working storage:
WORK-FIELD PIC X(15) VALUE SPACES.
TRAILING-SPACES PIC 9(3) VALUE ZERO.
FIELD-LENGTH PIC 9(3) VALUE ZERO.
Reverse the FIRST-NAME
MOVE FUNCTION REVERSE (FIRST-NAME) TO WORK-FIELD.
WORK-FIELD now contains leading spaces, instead of trailing spaces.
Find the number of trailing spaces in FIRST-NAME
INSPECT WORK-FIELD TALLYING TRAILING-SPACES FOR LEADING SPACES.
TRAILING-SPACE now contains the number of trailing spaces in FIRST-NAME.
Find the length of the FIRST-NAME field
COMPUTE FIELD-LENGTH = FUNCTION LENGTH (FIRST-NAME).
Concatenate the two strings together.
STRING FIRST-NAME (1:FIELD-LENGTH – TRAILING-SPACES) “ “ LAST-NAME DELIMITED BY SIZE, INTO FULL-NAME.
You could try making a loop for to get the real length.