**Please Note: **
This article is written for users of the following Microsoft Excel versions: 97, 2000, 2002, and 2003. If you are using a later version (Excel 2007 or later), *this tip may not work for you*. For a version of this tip written specifically for later versions of Excel, click here: Pulling Initial Letters from a String.

Rajeev needs a formula that will extract the first letters of a series of words. For instance, if a cell contains the text "Rajeev Kumar Pandey" he would like to extract, into another cell, the letters "RKP". The number of words in series can vary from cell to cell.

There are a couple of ways that this task can be approached. It is assumed, to begin with, that you don't want to modify the structure of your worksheet by adding intermediate columns. This assumption precludes, as well, the use of the Text to Columns feature to split the original string into individual words.

The key to the problem is making sure that your formula can determine where the spaces are in the original string. You might think that a formula such as the following will do the job:

=LEFT(A1,1)&MID(A1,FIND(" ",A1,1)+1,1)&MID(A1, FIND(" ",A1,FIND(" ",A1,1)+1)+1,1)

This formula works partially. It works just fine if the original string has two spaces separating three words. If there are any fewer words then the formula returns an error. If there are any more words, then it returns only the first letters of the first three words (it ignores anything after the third word).

This means that the formula needs to not only check for spaces, but handle errors if there are no spaces or if there are too few spaces. The error checking means that the formula becomes much longer:

=IF(ISERR(LEFT(A1,1)&MID(A1,SEARCH(" ",A1)+1,1) &MID(A1,SEARCH(" ",A1,SEARCH(" ",A1)+1)+1,1) &MID(A1,SEARCH(" ",A1,SEARCH(" ",A1,SEARCH(" ",A1)+1)+1)+1,1)), IF(ISERR(LEFT(A1,1)&MID(A1,SEARCH(" ",A1)+1,1) &MID(A1,SEARCH(" ",A1,SEARCH(" ",A1)+1)+1,1)), IF(ISERR(LEFT(A1,1)&MID(A1,SEARCH(" ",A1)+1,1)), IF(ISERR(LEFT(A1,1)),"",LEFT(A1,1)),LEFT(A1,1)&MID(A1,SEARCH(" ",A1)+1,1)), LEFT(A1,1)&MID(A1,SEARCH(" ",A1)+1,1)&MID(A1,SEARCH(" ",A1, SEARCH(" ",A1)+1)+1,1)),LEFT(A1,1)&MID(A1,SEARCH(" ",A1)+1,1) &MID(A1,SEARCH(" ",A1,SEARCH(" ",A1)+1)+1,1) &MID(A1,SEARCH(" ",A1,SEARCH(" ",A1,SEARCH(" ",A1)+1)+1)+1,1))

This formula will handle, properly, anything from 0 to 4 words in a string. It also assumes that the string doesn't start or end with a space and that it doesn't contain multiple numbers of spaces between words. If you want to handle a larger number of words or other potential complications (such as the number of spaces between words), then it is best to use a user-defined function.

There are any number of ways that a user-defined function could pull the leading characters from the words of a string. In fact, I received quite a few variations that accomplish the same thing. The following example, however, is perhaps the most concise code that I ran across:

Function Initials1(Raw As String) As String Dim Temp As Variant Dim J As Integer Application.Volatile Temp = Split(Trim(Raw)) For J = 0 To UBound(Temp) Initials1 = Initials1 & Left(Temp(J), 1) Next J End Function

This code will work in any version of VBA starting with Excel 2000. The Split function "tears apart" a string based on where spaces occur within it. The individual words in the string are placed into an array (in this case, Temp) where you can then access individual words. To use the function in your worksheet, simply use something like this:

=Initials1(A1)

If you are using a version of Excel that doesn't support the Split function, then the following code will work just fine, as well:

Function Initials2(Raw As String) Dim p As Integer Application.Volatile Initials2 = "" p = 0 Do Initials2 = Trim(Initials2) & Mid(Raw, p + 1, 1) p = InStr(p + 1, Raw, " ") Loop Until p = 0 End Function

*Note:*

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This tip (8661) applies to Microsoft Excel 97, 2000, 2002, and 2003. You can find a version of this tip for the ribbon interface of Excel (Excel 2007 and later) here: **Pulling Initial Letters from a String**.

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