Currently, My code does not properly detect if the string has a specific character.
for simple explanation. please refer to this:
Dim strSample as String = "Y1-K99"
and I tried this code if the string will detect if the strSample has a letter K.
If (strSample Like "##K*" Or strSample Like "###K*") Then
'Do Something
End if
But It does not trigger or go to the Inside Function of the If Else condition even if it has a K.
If you are after "strSample has a letter K"
, you will need to change your IF condition to be
If (strSample Like "*K*") Then
'Do Something
End if
Have a read at this
Like Operator
Related
Here is a code of my VBA script:
Function custom_if_formula(condition)
MsgBox(condition)
End Function
I am pasting the formula to any cell:
=custom_if_formula(B1="something")
The result in MsgBox is: TRUE or FALSE. Is it possible get in MsgBox as a result B1="something" as instead?
A pseudo-code of what I would like to achieve:
Function custom_if_formula(condition)
condition = condition.formula 'any method which take a literal string
MsgBox(condition)
End Function
P.S. My goal is to implement my own IFS function who behave identically like in Excel 2016. I am just curious if its possible. Tha'ts why I don't want to pass a string as an argument.
I think you're looking for:
Function custom_if_formula(condition)
If condition.Value = "something" Then MsgBox "something"
End Function
Where B1 is taken as the argument: =custom_if_formula(B1). Putting this in any cell (when B1 contains the string "something") will return:
You should really clarify what your intent is here, though. A UDF should return a value to its cell. Right now, it will just say 0 in the UDF's cell. Additionally, looking for "something" could be interpreted as looking for anything, and using this kind of verbiage leads to the "Who's on first, what's on second" ordeal...
Ok, I found a way how to do it:
Function custom_if_formula(condition)
cell_formula = Range(Application.Caller.Address).Formula 'I get an adress of a cell which call an UDF, and then take all string it contains
arg = Mid(cell_formula, 12, 100)
MsgBox(arg)
End Function
Is there a way to check if a string exists within another string in MATLAB. In python this is done easily with a in b. I do not want indexes or anything like that. I just want to check if its true or not. The answers that I find is "strcmp" or "strfind" and also regexp. regexp returns indexes. strcmp(a, b) does not seem to work. I have a string a = 'ac' and another string b = 'bc_gh_ac'. And want to check if a in b.
Best regards
The answer is indeed strfind. You have to be careful with the order of the parameters which at first seems unusual - the pattern is the second argument, not the first. The following code demonstrates:
a='ac';
b='bc_gh_ac';
strfind(b,a)
If you simply want to test whether the string is present or not then use the isempty function:
if ~isempty(strfind(b,a))
disp('String is present');
end
I'm trying to read a string in a specific format
RealSociedad
this is one example of string and what I want to extract is the name of the team.
I've tried something like this,
houseteam = sscanf(str, '%s');
but it does not work, why?
You can use regexprep like you did in your post above to do this for you. Even though your post says to use sscanf and from the comments in your post, you'd like to see this done using regexprep. You would have to do this using two nested regexprep calls, and you can retrieve the team name (i.e. RealSociedad) like so, given that str is in the format that you have provided:
str = 'RealSociedad';
houseteam = regexprep(regexprep(str, '^<a(.*)">', ''), '</a>$', '')
This looks very intimidating, but let's break this up. First, look at this statement:
regexprep(str, '^<a(.*)">', '')
How regexprep works is you specify the string you want to analyze, the pattern you are searching for, then what you want to replace this pattern with. The pattern we are looking for is:
^<a(.*)">
This says you are looking for patterns where the beginning of the string starts with a a<. After this, the (.*)"> is performing a greedy evaluation. This is saying that we want to find the longest sequence of characters until we reach the characters of ">. As such, what the regular expression will match is the following string:
<ahref="/teams/spain/real-sociedad-de-futbol/2028/">
We then replace this with a blank string. As such, the output of the first regexprep call will be this:
RealSociedad</a>
We want to get rid of the </a> string, and so we would make another regexprep call where we look for the </a> at the end of the string, then replace this with the blank string yet again. The pattern you are looking for is thus:
</a>$
The dollar sign ($) symbolizes that this pattern should appear at the end of the string. If we find such a pattern, we will replace it with the blank string. Therefore, what we get in the end is:
RealSociedad
Found a solution. So, %s stops when it finds a space.
str = regexprep(str, '<', ' <');
str = regexprep(str, '>', '> ');
houseteam = sscanf(str, '%*s %s %*s');
This will create a space between my desired string.
This is my first question on the forum, but reading previous questions has been enormously helpful in the project I'm working on, so already my thanks. I couldn't find an answer to this, but apologies if I overlooked something.
I am writing an excel macro in vba, and am trying to create a select case... statement in which the expression has a variable boolean and numeric component. For example, the macro can pull "> 3" or "< 3" from another worksheet.
My hope had been that I could assign to a string all of these parameters, i.e.:
test1 = "is " & BoolOperator1 & " " & NumericValue1
and then
Select case ValuetoCompare
Case test1
'Do something
Case test2
'...
Is there a way to do this? I suppose the alternative would be to nest the case with the numeric variable inside a select function that determines the operator, but I thought this would be more elegant.
Thanks in advance for your guidance--
Josh
Assuming that you'll get a string BoolOperator1 that is a valid operator, e.g. >=, =, and a numeric value NumericValue1, the easiest way to execute this comparison on another numeric value ValueToCompare is to use the Evaluate function. This will execute a string as VBA and return it's result.
In your case, you could simply use:
If Evaluate(ValueToCompare&BoolOperator1&NumericValue1) Then ...
If you want to use this in a Select Case statement, you'd either need to use a simple If ... ElseIf ... statement - or use this trick:
Select Case True
Case Evaluate(ValueToCompare&BoolOperator1&NumericValue1): ...
Case Evaluate(ValueToCompare&BoolOperator2&NumericValue2): ...
Case Else ...
End Select
I recently inherited a VBA macro that needs to have validation logic added to it. I need to be able to determine if any characters in a text based cell are non ASCII characters (i.e. have a binary value > 0x7F). The cells may contain some carriage control values (particularly linefeeds) that need to be retained (so the CLEAN function does not work for this validation). I have tried the IsText function, but found that it will interpret UTF-8 character sequences as valid text (which I don't want).
I don't need to actually manipulate the string, I just want to display an error to the user that runs the macro to tell him that there are invalid (non-ASCII) characters in a specific cell.
If you want a technically pure approach you might try a regular expression. Add a reference in VBA to the Microsoft Scripting library, and try this code. It looks a little complex, but you will be blown away by what regular expressions can do, and you will have a valuable tool for future use.
Function IsTooHigh(c As String) As Boolean
Dim RegEx As Object
Set RegEx = CreateObject("vbscript.regexp")
With RegEx
.Global = True
.MultiLine = True
.Pattern = "[^\x00-\x7F]"
End With
IsTooHigh = RegEx.Test(c)
End Function
This function gives TRUE if any character in string c is not (^) in the range 0 (x00) to 127 (x7F).
You can Google for "regular expression" and whatever you need it to do, and take the answer from almost any language, because like SQL, regular expression patterns seem to be language agnostic.
The asc(character) command will convert a character to it's ASCII value.
hex(asc(character)) will convert the character to it's HEX value.
Once you've done that you can easily do some comparisons to determine if the data is bad and toss the errors if required.
Here's some sample code:
http://www.freevbcode.com/ShowCode.asp?ID=4486
Function IsGoodAscii(aString as String) as Boolean
Dim i as Long
Dim iLim as Long
i=1
iLim=Len(aString)
While i<=iLim
If Asc(Mid(aString,i,1))>127 then
IsGoodAscii=False
Exit Function
EndIf
i=i+1
Wend
IsGoodAscii=True
End Function