MID - Statement |
MID(stringvar, start [,length]) = string |
Replaces a specified number of characters with characters from another string. |
stringvar | The text string (String). |
start | The character position to start (Long). |
length | (Optional) The number of characters to return (Long). |
string | The string expression that replaces part of stringvar (String). |
REMARKS |
* You cannot prefix this statement with "VBA." * This statement will replace a specified number of characters in a string variable. * VBA.Mid always calls the MID Function which means you will probably see a Run-time error 424 Object required. * The number of characters replaced is always less than or equal to the number of characters in stringvar. * The first character position is 1. * You can use the LEFT function to return a substring from the left of a string. * You can use the MID Function to return a substring from the middle, left or right of a string. * You can use the RIGHT function to return a substring from the right of a string. * For more information, refer to the Replacing Strings page. * For the Microsoft documentation refer to learn.microsoft.com |
Dim sText As String
sText = "hello world"
Mid(sText, 1, 3) = "XXX"
Debug.Print sText '= "XXXlo world"
sText = "hello world"
Mid(sText, 7, 3) = "XXX"
Debug.Print sText '= "hello XXXld"
sText = "The dog jumps"
Mid(sText, 5, 3) = "fox"
Debug.Print sText '= "The fox jumps"
Mid(sText, 5) = "cow"
Debug.Print sText '= "The cow jumps"
Mid(sText, 5) = "cat runs fast"
Debug.Print sText '= "The cat runs"
Mid(sText, 5, 3) = "rabbit"
Debug.Print sText '= "The rab runs"
VBA.Mid(sText, 5, 3) = "rabbit" 'Runtime error 424 Object required
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