MID(stringvar, start [,length]) = string

Replaces a specified number of characters with characters from another string.


stringvarThe text string (String).
startThe character position to start (Long).
length(Optional) The number of characters to return (Long).
stringThe string expression that replaces part of stringvar (String).

REMARKS
* The number of characters replaced is always less than or equal to the number of characters in stringvar.
* Use the MidB statement with byte data contained in a string. In the MidB statement, start specifies the byte position within stringvar where replacement begins and length specifies the numbers of bytes to replace.
* This example uses the Mid statement to replace a specified number of characters in a string variable with characters from another string.
* You can use the MID - Function to replace a specified number of characters.
* For the Microsoft documentation refer to docs.microsoft.com

Dim sText As String 
sText = "The dog jumps" ' Initialize string.
Mid(sText, 5, 3) = "fox" ' sText = "The fox jumps".
Mid(sText, 5) = "cow" ' sText = "The cow jumps".
Mid(sText, 5) = "cat runs fast" ' sText = "The cat runs".
Mid(sText, 5, 3) = "rabbit" ' sText = "The rab runs".

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