Strings (OpenOffice.org BASIC Runtime Library)
Working with Sets of Characters
When administering strings, OpenOffice.org Basic uses the set of Unicode characters. The Asc and Chr functions allow the Unicode value belonging to a character to be established and/or the corresponding character to be found for a Unicode value. The following expressions assign the various Unicode values to the code variable:
Code = Asc("A") ' Latin letter A (Unicode-value 65) Code = Asc("€") ' Euro character (Unicode-value 8364) Code = Asc("Л") ' Cyrillic letter Л (Unicode-value 1083)
Conversely, the expression
MyString = Chr(13)
ensures that the MyString string is initialized with the value of the number 13, which stands for a hard line break.
The Chr command is often used in Basic languages to insert control characters in a string. The assignment
MyString = Chr(9) + "This is a test" + Chr(13)
therefore ensures that the text is preceded by a tab character (Unicode-value 9) and that a hard line break (Unicode-value 13) is added after the text.
Accessing Parts of a String
OpenOffice.org Basic provides three functions that return partial strings, plus a length function:
- Left(MyString, Length)
- returns the first Length characters of MyString.
- Right(MyString, Length)
- returns the last Length characters of MyString.
- Mid(MyString, Start, Length)
- returns first Length characters of MyString as of the Start position.
- returns the number of characters in MyString.
Unlike array subscripts, character positions in a string start with 1. Here are a few example calls for the named functions:
Dim MyString As String Dim MyResult As String Dim MyLen As Integer MyString = "This is a small test" MyResult = Left(MyString,5) ' Provides the string "This " MyResult = Right(MyString, 5) ' Provides the string " test" MyResult = Mid(MyString, 8, 5) ' Provides the string " a sm" MyLen = Len(MyString) ' Provides the value 20
Search and Replace
OpenOffice.org Basic provides the InStr function for searching for a partial string within another string:
PositionOfMatch = InStr(MyString, StringToFind)
The StringToFind parameter specifies the string to be searched for within MyString. The function returns a number that contains the position at which the StringToFind first appears within MyString; a return value of zero indicates no match. If you want to find other matches for the string, the function also provides the opportunity to specify an optional start position from which OpenOffice.org Basic begins the search. In this case, the syntax of the function is:
PositionOfMatch = InStr(StartPosition, MyString, StringToFind)
In the previous examples, InStr ignores uppercase and lowercase characters. To change the search so that InStr is case sensitive, add the parameter 0, as shown in the following example:
PositionOfMatch = InStr(MyString, StringToFind, 0)
Using the previous functions for editing strings, programmers can search for and replace one string in another string:
Function Replace(Source As String, Search As String, NewPart As String) Dim Result As String Dim StartPos As Long Dim CurrentPos As Long Result = "" StartPos = 1 CurrentPos = 1 If Search = "" Then Result = Source Else Do While CurrentPos <> 0 CurrentPos = InStr(StartPos, Source, Search) If CurrentPos <> 0 Then Result = Result + Mid(Source, StartPos, _ CurrentPos - StartPos) Result = Result + NewPart StartPos = CurrentPos + Len(Search) Else Result = Result + Mid(Source, StartPos, Len(Source)) End If ' Position <> 0 Loop End If Replace = Result End Function
The function searches through the transferred Search string in a loop by means of InStr in the original term Source. If it finds the search term, it takes the part before the expression and writes it to the Result return buffer. It adds the NewPart section at the point of the search term Search. If no more matches are found for the search term, the function establishes the part of the string still remaining and adds this to the return buffer. It returns the string produced in this way as the result of the replacement process.
Since replacing parts of character sequences is one of the most frequently used functions, the Mid function in OpenOffice.org Basic has been extended so that this task is performed automatically. The following example replaces three characters with the string is from the sixth position of the MyString string.
Dim MyString As String MyString = "This was my text" Mid(MyString, 6, 3, "is")
Or this much simpler function can be used. First it splits "Source" dividing it at all "Search", and joins the parts back with "NewPart". So Replace("This is a test", " ", "_") yields "This_is_a_test" :D
Function Replace(Source As String, Search As String, NewPart As String) Dim Result As String Result = join(split(Source, Search), NewPart) Replace = Result End Function
|When it is used with 4 arguments, to replace a sub-string in a string, Mid is an instruction, not a function : it does not return any value !|
The Format function formats numbers as a string. To do this, the function expects a Format expression to be specified, which is then used as the template for formatting the numbers. Each place holder within the template ensures that this item is formatted correspondingly in the output value. The five most important place holders within a template are the zero (0), pound sign (#), period (.), comma (,) and dollar sign ($) characters.
The 0 character within the template ensures that a number is always placed at the corresponding point. If a number is not provided, 0 is displayed in its place.
A . stands for the decimal point symbol defined by the operating system in the country-specific settings.
The example below shows how the 0 and . characters can define the digits after the decimal point in an expression:
MyFormat = "0.00" MyString = Format(-1579.8, MyFormat) ' Provides "-1579,80" MyString = Format(1579.8, MyFormat) ' Provides "1579,80" MyString = Format(0.4, MyFormat) ' Provides "0,40" MyString = Format(0.434, MyFormat) ' Provides "0,43"
In the same way, zeros can be added in front of a number to achieve the desired length:
MyFormat = "0000.00" MyString = Format(-1579.8, MyFormat) ' Provides "-1579,80" MyString = Format(1579.8, MyFormat) ' Provides "1579,80" MyString = Format(0.4, MyFormat) ' Provides "0000,40" MyString = Format(0.434, MyFormat) ' Provides "0000,43"
A , represents the character that the operating system uses for a thousands separator, and the # stands for a digit or place that is only displayed if it is required by the input string.
MyFormat = "#,##0.00" MyString = Format(-1579.8, MyFormat) ' Provides "-1.579,80" MyString = Format(1579.8, MyFormat) ' Provides "1.579,80" MyString = Format(0.4, MyFormat) ' Provides "0,40" MyString = Format(0.434, MyFormat) ' Provides "0,43"
In place of the $ place holder, the Format function displays the relevant currency symbol defined by the system (this example assumes a European locale has been defined):
MyFormat = "#,##0.00 $" MyString = Format(-1579.8, MyFormat) ' Provides "-1.579,80 €" MyString = Format(1579.8, MyFormat) ' Provides "1.579,80 €" MyString = Format(0.4, MyFormat) ' Provides "0,40 €" MyString = Format(0.434, MyFormat) ' Provides "0,43 €"
The format instructions used in VBA for formatting date and time details can also be used:
sub main dim myDate as date myDate = "01/06/98" TestStr = Format(myDate, "mm-dd-yyyy") ' 01-06-1998 MsgBox TestStr end sub
|Content on this page is licensed under the Public Documentation License (PDL).|