I'm using code below
ThisWorkbook.Sheets(1).Cells(1, 2).EntireColumn.Replace what:="&^%", replacement:="LOL"
Data(in cell(2,2)): Sometext&^%, just for example
Expected: SometextLOL, just for example
Actual Result: LOL, Just for example
Even the excel "Find and Replace" behaves the same way.
Is it considering the "what" text as regular expression? If so how to overcome this.
Related
I have had to look up hundreds (if not thousands) of free-text answers on google, making notes in Excel along the way and inserting SAS-code around the answers as a last step.
The output looks like this:
This output contains an unnecessary number of blank spaces, which seems to confuse SAS's search to the point where the observations can't be properly located.
It works if I manually erase superflous spaces, but that will probably take hours. Is there an automated fix for this, either in SAS or in excel?
I tried using the STRIP-function, to no avail:
else if R_res_ort_txt=strip(" arild ") and R_kom_lan=strip(" skåne ") then R_kommun=strip(" Höganäs " );
If you want to generate a string like:
if R_res_ort_txt="arild" and R_kom_lan="skåne" then R_kommun="Höganäs";
from three variables, let's call them A B C, then just use code like:
string=catx(' ','if R_res_ort_txt=',quote(trim(A))
,'and R_kom_lan=',quote(trim(B))
,'then R_kommun=',quote(trim(C)),';') ;
Or if you are just writing that string to a file just use this PUT statement syntax.
put 'if R_res_ort_txt=' A :$quote. 'and R_kom_lan=' B :$quote.
'then R_kommun=' C :$quote. ';' ;
A saner solution would be to continue using the free-text answers as data and perform your matching criteria for transformations with a left join.
proc import out=answers datafile='my-free-text-answers.xlsx';
data have;
attrib R_res_ort_txt R_kom_lan length=$100;
input R_res_ort_txt ...;
datalines4;
... whatever all those transforms will be performed on...
;;;;
proc sql;
create table want as
select
have.* ,
answers.R_kommun_answer as R_kommun
from
have
left join
answers
on
have.R_res_ort_txt = answers.res_ort_answer
& have.R_kom_lan = abswers.kom_lan_answer
;
I solved this by adding quotes in excel using the flash fill function:
https://www.youtube.com/watch?v=nE65QeDoepc
On SO, I was just given two answers that both work when called a single time. Now I want to put them in a loop and loop over several rows of data. However, I'm having a heck of time getting the code correct. I'm suspect it has to how I'm handling the double quotes.
The stand alone code lines are as follows.
Var = ActiveSheet.Evaluate("And(A1:F1)") and
Var = Application.WorksheetFunction.And(Range("A1:F1"))
for the first example I tried:
for i = 2 to 20
Var = ActiveSheet.Evaluate("And(A & i & :F & i)")
Next i
This produces "Error 2015"
for the second:
for i = 2 to 20
Var = Application.WorksheetFunction.And(Range("A" & i & ":F" & i))
Next i
This produces a line of red code
What am I doing wrong?
The Visual Basic Editor is making this harder than it should be, because its default syntax highlighting is making string literals the same color as identifiers:
You can change that under Tools/Options, and make Identifier Text a different color - here teal:
Now string literals are still black, but now identifiers look visually distinctive:
What you want to make sure, is that your variables are syntax-highlighted like identifiers - so they're teal, not black - like in your second example:
Contrast with your first attempt, where i doesn't get syntax-highlighted as the identifier it should be:
And since you know that i is a VBA variable and you want VBA to concatenate its value into this string, then i being syntax-highlighted as any other string literal (and not as an identifier) is your visual cue that something's off!
Compare to #JNevill's fixed version:
With Identifier Text having a different syntax highlighting than string literals in the editor, it becomes much easier to quickly locate a variable that's accidentally inside a string literal.
That first snippet isn't working, because ActiveSheet.Evaluate takes its parameter and gives it to Excel's expression evaluation engine, ...which has no idea what to do with this i. Variable i only exists in the execution context of the VBA code: only VBA code can evaluate its value.
Hi I would like to extract dynamically the numbers from string in Excel.
I have the following strings and I would like to have only the numbers before ". pdf". taken out of the string into the next column.
As you can see the number of characters varies from line to line.
I have invented something like this:
=MID(M20;SEARCH("_";M20);20)
But this takes out only the numbers after "_" and .pdf after this....
How to make it the way I like?
D:\Users\xxxx\Desktop\1610\ts25b_4462.pdf
D:\Users\xxx\Desktop\1610\ts02b_39522.pdf
D:\Users\xxxxx\Desktop\1610\ts02b_except_39511.pdf
D:\Users\xxxx\Desktop\1610\ts02b_except_39555.pdf
D:\Users\xxxx\Desktop\1610\ts22b_6118.pdf
So that I have just :
4462
39522
39511
39555
6118
and so on...
Thank you!!!
With VBA, try to do it like this:
Public Function splitThings(strInput As String) As String
splitThings = Split(Split(strInput, "_")(1), ".")(0)
End Function
Concerning your formula, try to use =LEFT(MID(M20;SEARCH("_";M20);20),K), where K is the difference of the length of ts22b_6118.pdf and 4 (.pdf). 4 is the length of .pdf.
Something like this should do the work:
=LEFT(MID(I3,SEARCH("_",I3)+1,LEN(I3)),LEN(MID(I3,SEARCH("_",I3),LEN(I3)))-5)
You should do it using Excel formula. For example:
=SUBSTITUTE(LEFT(A1,FIND(".pdf",A1)-1),LEFT(A1,FIND("_",A1)),"")
Using the first line as an example, with LEFT(A1,FIND(".pdf",A1)-1) you will have D:\Users\xxxx\Desktop\1610\ts25b_4462 and with the LEFT(A1,FIND("_",A1)) D:\Users\xxxx\Desktop\1610\ts25b_, if you SUBSTITUTE the first part by "" you will have 4462.
Hope this can help.
With this formula, you should be able to get the numbers you want:
=MID(A1,FIND("|",SUBSTITUTE(A1,"_","|",LEN(A1)-LEN(SUBSTITUTE(A1,"_",""))))+1,FIND(".",A1)-FIND("|",SUBSTITUTE(A1,"_","|",LEN(A1)-LEN(SUBSTITUTE(A1,"_",""))))-1)
Basically, this is the initial fomula:
=MID(A1,FIND("_",A1)+1,FIND(".",A1)-FIND("_",A1)-1)
But since there may be two _ in the string so this is the one to find the 2nd _:
=SUBSTITUTE(A1,"_","|",LEN(A1)-LEN(SUBSTITUTE(A1,"_","")))
Now just replace this SUBSTITUTE with A1 above and you get that long formula. Hope this helps.
This will return the number you want regardless of extension (could be .pdf, could be .xlsx, etc) and regardless of the number of underscores present in the filename and/or filepath:
=TRIM(LEFT(RIGHT(SUBSTITUTE(SUBSTITUTE(M20,".",REPT(" ",LEN(M20))),"_",REPT(" ",LEN(M20))),LEN(M20)*2),LEN(M20)))
In a spreadsheet formula, =VALUE("$100") will evaluate to the numeric value of 100. I then tried to access this function in VBA via WorksheetFunction object, however it is missing.
In VBA I tried the conversion function Val("$100"), however that returns 0. So how can I accomplish this via VBA?
Val() only really works if the string is all numbers I'm afraid - currency signs cause it a problem.
If you're always going to have the same currency sign in the string, it might be worth using something like
StringName = replace(StringName, "$", "")
to take out the $ by replacing it with "" - otherwise if your strings aren't always going to be this predictable the below question might help:
How to find numbers from a string?
see https://learn.microsoft.com/en-us/office/vba/api/excel.worksheetfunction.numbervalue
example of using above, which will return a value of -1234.56:
MsgBox WorksheetFunction.NumberValue("-$1,234.56", ".", ",")
Note that if the result is non-numeric, it throws an error. For example (swapping the comma grouping and decimal character params which is invalid in this case):
MsgBox WorksheetFunction.NumberValue("-$1,234.56", ".", ",")
I don't understand why the above link doesn't have any version info. It is currently dated 2019-05-23 - no idea if that's because it is new or if it was recently updated.
how can i resovlve the follwing issue in VBA
if string contains "SeachText" replace whole string.
ex.
input: "Test_1_The_Text_SeachString_Sth"
input: "Test_2_The_Text_SeachString_Sth"
look for: "SeachrString"
replace with "New_Text"
so after execute the code the
"Test_1_Of_The_Text_SeachString_Sth" will become "New_Text"
as well as:
"Test_2_Of_The_Text_SeachString_Sth" will become "New_Text"
Use Instr function like this:
ip1= "Test_1_The_Text_SeachString_Sth"
lookfor="SeachString"
If instr(ip1,lookfor)>0 then
ip1= "newtext"
End if