Returns a Variant (String) containing a specified number of characters from a string.
Mid(string, start[, length])
The Mid function syntax has these named arguments:
Required; Long. Character position in string at which the part to be taken begins. If start is greater than the number of characters in string, Mid returns a zero-length string ("").
Optional; Variant (Long). Number of characters to return. If omitted or if there are fewer than length characters in the text (including the character at start), all characters from the start position to the end of the string are returned.
To determine the number of characters in string, use the Len function.
Use the MidB function with byte data contained in a string, as in double-byte character set languages. Instead of specifying the number of characters, the arguments specify numbers of bytes. For sample code that uses MidB, see the second example in the example topic.
The first example uses the Mid function to return a specified number of characters from a string.
Dim MyString, FirstWord, LastWord, MidWords MyString = "Mid Function Demo" ' Create text string. FirstWord = Mid(MyString, 1, 3) ' Returns "Mid". LastWord = Mid(MyString, 14, 4) ' Returns "Demo". MidWords = Mid(MyString, 5) ' Returns "Function Demo".
The second example use MidB and a user-defined function (MidMbcs) to also return characters from string. The difference here is that the input string is ANSI and the length is in bytes.
Function MidMbcs(ByVal str as String, start, length) MidMbcs = StrConv(MidB(StrConv(str, vbFromUnicode), start, length), vbUnicode) End Function Dim MyString MyString = "AbCdEfG" ' Where "A", "C", "E", and "G" are DBCS and "b", "d", ' and "f" are SBCS. MyNewString = Mid(MyString, 3, 4) ' Returns ""CdEf" MyNewString = MidB(MyString, 3, 4) ' Returns ""bC" MyNewString = MidMbcs(MyString, 3, 4) ' Returns "bCd"