WebHelp! Separate string CO SM F PLUS S30 JYSN 08002000 if i use =mid(str1;7;(len(str1)-16) then result F PLUS S30 JYSN Constant number is first 6 and last 9 character Now how separate the last string part (blue) whose length is variable. second question count spaces in string?! (herein example is 6 space) ?! WebApr 6, 2015 · Pos = InStrRev (Str, ValueCrnt) If Pos > 0 Then Str = Mid (Str, 1, Pos - 1) & Replace (Str, ValueCrnt, ValueNew, Pos) End If Wscript.Echo Str Share Improve this …
Access VBA find the last occurrent of a string? - Stack Overflow
WebVBA InStrRev Second to Last Space. VBA InStrRev function can find the position of the second to last space in a string. We can use VBA InStrRev function to find the last … Webfrom your original string, I returned everything from the last "\" up to the third "-" following. Worksheet Excel function: =LEFT (SUBSTITUTE (TRIM (RIGHT (SUBSTITUTE (A1,"\",REPT (" ",99)),99)), "-",CHAR (1),3),FIND (CHAR (1),SUBSTITUTE (TRIM (RIGHT (SUBSTITUTE ( A1,"\",REPT (" ",99)),99)),"-",CHAR (1),3))-1) User Defined Function: black friday woche 2022 amazon
How to Locate the Last Space in the Text String - ExcelNotes
WebMay 18, 2024 · You could try Home>Editing>Find & Select (Ctrl+F) and in the Find what box put in " * " for leading spaces and " * " for trailing. 'Find All' then Ctrl+A with focus in the results to select them all. Then have an outer loop for the areas and an inner loop for the cells in each area. Is a possibility? – IvenBach WebJul 13, 2024 · Access VBA has Instr to return the position of the first occurrence of a string in another string. Instr ( [start], string_being_searched, string2, [compare] ) Is there any method to return the position of the last occurrence of a string in another string? ms-access vba ms-access-2007 Share Follow edited Jul 13, 2024 at 11:34 Erik A 31.4k 12 … WebJul 5, 2024 · In short: Use =LEFT (), =RIGHT () and =MID () to get parts of your string and concatenate the parts and your spaces. Edit: In VBA: Public Function StringWithSpaces (inpStr As String) As String StringWithSpaces = Left (inpStr, 5) & " " & Mid (inpStr, 6, 2) & " " & Right (inpStr, Len (inpStr) - 7) End Function Share Improve this answer Follow game shops in kingston