I would like to get a working code to simply remove from a text line a specific part that always begins with "(" and finish with ")".
Sample text : Hello, how are you (it is a question)
I want to remove this part: "(it is a question)" to only keep this message "Hello, how are you"
Lost...
Thanks
One way using Regular Expressions;
input = "Hello, how are you (it is a question)"
dim re: set re = new regexp
with re
.pattern = "\(.*\)\s?" '//anything between () and if present 1 following whitespace
.global = true
input = re.Replace(input, "")
end with
msgbox input
If the part to be removed is always at the end of the string, string operations would work as well:
msg = "Hello, how are you (it is a question)"
pos = InStr(msg, "(")
If pos > 0 Then WScript.Echo Trim(Left(msg, pos-1))
If the sentence always ends with the ( ) section, use the split function:
line = "Hello, how are you (it is a question)"
splitter = split(line,"(") 'splitting the line into 2 sections, using ( as the divider
endStr = splitter(0) 'first section is index 0
MsgBox endStr 'Hello, how are you
If it is in the middle of the sentence, use the split function twice:
line = "Hello, how are you (it is a question) and further on"
splitter = split(line,"(")
strFirst = splitter(0) 'Hello, how are you
splitter1 = split(line,")")
strSecond = splitter1(UBound(Splitter1)) 'and further on
MsgBox strFirst & strSecond 'Hello, how are you and further on
If there is only one instance of "( )" then you could use a '1' in place of the UBound.
Multiple instances I would split the sentence and then break down each section containing the "( )" and concatenate the final sentence.
Related
I have a file called worker.dat, and in that file, a list of information is stored as variables "Income","Promotion", "Age" etc.
And I want to read that information from the file and print on the screen.
So I used
Open App.Path & "worker.DAT" For Input As #1
and using the Print method, printed the information.
However for the sake of emphasis, I want to print some information in a bigger size and in different font etc.
So I wrote this.
Printer.FontSize = 16
Printer.Print "Income = "; Income
However this didn't work. Does anyone how to fix this problem?
Set the size on the font object:
Dim pt As Long
With Printer.Font
pt = .Size
Printer.Print "default text"
.Size = 16
Printer.Print "larger text"
.Size = pt
.Bold = True
Printer.Print "bold ";
.Bold = False
Printer.Print "in default size"
Printer.EndDoc
End With
Looking at my server, i see filenames, that theur characters have been messed up.
Example: ôøùú-ô÷åãé-1.mp3
should should be Hebrew letters.
While the files can be read in a browser and FTP, they cannot be read always - for example, in an online audio player, or VBScript FSO (File system object)
is there something that can be done to change the charset or something, so that i can read and then rename these files?
Thanks!
solved, using this link: http://pastebin.com/yXRWDggY
function convertChar(letter)
lat = array("à","á","â","ã","ä","å","æ","ç","è","é","ê","ë","ì","í","î","ï","ð","ñ","ò","ó","ô","õ","ö","÷","ø","ù","ú", " " )
heb = array("א","ב","ג","ד","ה","ו","ז","ח","ט","י","ך","כ","ל","ם","מ","ן","נ","ס","ע","ף","פ","ץ","צ","ק","ר","ש","ת", " ")
inarray = false
for ii = 0 to ubound(heb)
if letter = heb(ii) then
convertChar = lat(ii)
inarray = true
exit for
end if
next
if inarray = false then convertChar = letter
end function
I'm trying to grab the first characters before a space.
I know it can be done this way
str = "3 Hello World"
str = Mid(str, 1,2)
But how would i do this after a space?
Edit: Looks like you changed your question to get characters BEFORE the first space instead of AFTER. I've updated my examples.
Here's one way:
strTextBeforeFirstSpace = Split(str, " ")(0)
Assuming a space exists in your string, this would return everything up until the first space.
Another way would be:
strTextBeforeFirstSpace = Left(str, InStr(str, " ") - 1)
You can get the index of the first space with the InStr function
InStr(str, " ")
And use this as a parameter in your Mid function
Dim str, index
str = "3 Hello World"
index = InStr(str," ")
'only neccessary if there is a space
If index > 0 Then
str = Mid(str,1,index - 1)
End If
I have not used Visual FoxPro for a while. Today, my ex-colleague asks me how to remove non character from name field, i.e. only a-z and A-Z are allowed. I remember I used a function called strstran to do it. I needed to define a variable contains a-z and A-Z. But I do not remember now. Does someone knows how to handle this problem. Thanks in advance.
Use the CHRTRAN() function.
FUNCTION GetAlphaCharacters
LPARAMETERS tcExpressionSearched
LOCAL lcAllowedCharacters
m.lcAllowedCharacters = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
RETURN CHRTRAN(m.tcExpressionSearched, CHRTRAN(m.tcExpressionSearched, m.lcAllowedCharacters, ""), "")
ENDFUNC
Another option is to use ISALPHA(). This only looks at the left most position in the string but it's not case sensitive.
***This should work, but I haven't tested it.
myresults = ""
myvar = "MyText12"
FOR(i = 1 TO LEN(myvar))
IF ISALPHA( SUBSTR(myvar, i, 1) )
myresults = myresults + SUBSTR(myvar, i, 1)
ENDIF
ENDFOR
RETURN myresults
I know I'm a bit late to the party, but here is a function I wrote to clean out all non-printable ASCII characters from a character string.
CLEAR
* Contains ASCII characters 1 (SOH) and 2 (STX)
cTest = "Garbage Data "
? cTest
cTest = RemoveNonPrintableCharacters(cTest)
? cTest
FUNCTION RemoveNonPrintableCharacters
LPARAMETERS tcExpressionSearched
cCleanExpression = tcExpressionSearched
* Cleans out the first 32 ASCII characters, which are not printable
FOR decCount = 0 TO 31
cCleanExpression = CHRTRAN(m.cCleanExpression, CHR(decCount), "")
ENDFOR
* Also cleans out the non-printable DEL character (ASCII 127)
cCleanExpression = CHRTRAN(m.cCleanExpression, CHR(127), "")
* Return the clean string
RETURN cCleanExpression
ENDFUNC
I want to check if an address starts with http://www.youtube.com.
If I have something like this
if rs("mainVideoName")="http://www.youtube.com*" then
This doesn't work, so how can I do it?
Try this:
Function UrlStartsWith(string1, string2)
UrlStartsWith = InStr(1, string1, string2, 1) = 1
End Function
If UrlStartsWith(rs("mainVideoName"), "http://www.youtube.com") Then
End If
Starts with is tested by using IntStr and ensuring that it returns 1 as the starting position that the search string is found. Since you are testing a URL the code above uses a TextCompare to make insensitive to case.
You can use the InStr() function for this:
Dim positionOfMatchedString
positionOfMatchedString= InStr(rs("mainVideoName"),"http://www.youtube.com")
If positionOfMatchedString > 0 Then
// Do your stuff here
End If
As Anthony points out, this tells you that string2 is contained in string1. You could write it as:
If positionOfMatchedString = 1 Then
to check for it beginning with.
How about...
Dim s: s = "http://www.youtube.com"
Dim l: l = Len(s)
If Left(rs("mainVideoName"), l) = s Then
' String begins with URL part '
End If