The VBA InStr function is one of the most used functions in VBA. It is used to find a string within a string and indeed it does a very fine job.
However, it is often used to help extract part of a string and for this task it performs badly.
If you have found string extraction in VBA to be a painful process, then read on. This post will show you a simpler and better way using three real world examples!
A Quick Guide to this Post
The following table provides a quick reference guide to what is covered in this post.
|1234ABC334||Fixed size||get left 4 chars||Left(s,4)|
|1234ABC334||Fixed size||get right 3 chars||Right(s,3)|
|1234ABC334||Fixed size||get chars 5,6,7||Mid(s,5,3)|
|"John Henry Smith"||Variable size||get first name||Split(s," ")(0)|
|"John Henry Smith"||Variable size||get second name||Split(s," ")(1)|
|"John Henry Smith"||Variable size||get third name||Split(s," ")(2)|
|"John Henry Smith"||Variable size||Get last name||Dim v As Variant
v = Split(s, " ")
Quick Reference Notes
To find out more about the items referenced in the post check out the following links
If you would like to know more about the InStr or InStrRev functions then please read Searching within a string.
If you would like to know more about Mid, Left or Right functions then check out Extracting Part of a String.
For more about the Split function check out String to Array using Split.
The Like operator is covered in Pattern Matching
I use Debug.Print in my examples. It prints values to the Immediate Window which you can view by pressing Ctrl and G (or select View->Immediate Window)
In this post, I’m going to show you a better way to extract values from a string than using then VBA InStr function with Left, Right or Mid.
This post is broken down as follows
- Section 1: How to extract from fixed sized strings.
- Section 2: How to extract from variable sized strings.
- Section 3: How to extract from variable sized string using the Split function.
- Sections 4 to 6: Some real world examples.
When VBA InStr, Left, Right and Mid are useful
If you want to check if a string contains a value then InStr is fine for the job. If you want to do a simple extraction then Left, Right and Mid also fine to use.
Using InStr to check if string contains text
In the following example, we check if the name contains “Henry”. If the return value of InStr is greater than zero then the string contains the value we are checking for.
' Check if string contains Henry If InStr("John Henry Smith", "Henry") > 0 Then Debug.Print "Found" End If
Extracting with Left, Right and Mid
The Left function is used to get characters from the left of a string.
The Right function is used to get characters from the right of a string.
The Mid function is used for the middle of the string. It is the same as Left except that you give it a starting position.
Sub ExtractString() Dim s As String: s = "ABCD-7789.WXYZ" Debug.Print Left(s, 2) ' Prints AB Debug.Print Left(s, 4) ' Prints ABCD Debug.Print Right(s, 2) ' Prints YZ Debug.Print Right(s, 4) ' Prints WXYZ Debug.Print Mid(s, 1, 2) ' Prints AB Debug.Print Mid(s, 6, 4) ' Prints 7789 End Sub
These three functions work fine if the text you require is always the same size and in the same place. For other scenarios, they require the use of InStr to find a particular position in the string. This makes using them complicated.
|Use Left, Right or Mid when the characters will always be in the same position.|
Dealing with Strings of Varying Lengths
Many of the strings you will deal with will be of different lengths. A simple example is when you are dealing with a list of names. The string length and part you require(e.g. the first name) may be of different each time. For example
(If you need random list of test names then try this random name generator)
Using the VBA InStr Function with Left
In the following example, we are going to get the first name from a string. In this string the first name is the name before the first space.
We use the VBA InStr function to get the position of the first space. We want to get all the characters before the space. We subtract one from the position as this gives us the position of the last letter of the name.
Sub GetFirstname() Dim s As String, lPosition As Long s = "John Henry Smith" ' Prints John lPosition = InStr(s, " ") - 1 Debug.Print Left(s, lPosition) s = "Lorraine Huggard" ' Prints Lorraine lPosition = InStr(s, " ") - 1 Debug.Print Left(s, lPosition) End Sub
Let’s look at the first example in the above code. The first space is at position 5. We substract 1 so which gives us position 4. This is the position of the last letter of John i.e.n.
We then give 4 to the Left function and it returns the first four characters e.g. “John”
We can perform the same task in one line by passing the return value from InStr to the Left function.
Dim s As String s = "John Henry Smith" ' Prints John Debug.Print Left(s, InStr(s, " ") - 1)
Using the VBA InStr Function with Right
In this example, we will get the last word in the string i.e. Smith. We can use the InStrRev function to help us. This is the same as InStr except it searches from the end of the string.
It’s important to note that InStrRev gives us the position from the start of the string . Therefore, we need to use it slightly differently than we used InStr and Left.
Sub GetLastName() Dim s As String: s = "John,Henry,Smith" Dim Position As Long, Length As Long Position = InStrRev(s, ",") Length = Len(s) ' Prints Smith Debug.Print Right(s, Length - Position) ' Alternative method. Prints Smith - do in one line Debug.Print Right(s, Len(s) - InStrRev(s, ",")) End Sub
How this the above example works
- We get the position of the last space using InStrRev: 11
- We get the length of the string: 16.
- We subtract the position from the length: 16-11=5
- We give 5 to the Right function and get back Smith
Using the VBA InStr Function with Mid
In the next example, we will get “Henry” from the string. The word we are looking for is between the first and second space.
We will use the Mid function here.
Sub GetSecondName() Dim s As String: s = "John Henry Smith" Dim firstChar As Long, secondChar As Long Dim count As Long ' Find space position plus 1. Result is 6 firstChar = InStr(s, " ") + 1 ' find 2nd space position. Result is 11 secondChar = InStr(firstChar, s, " ") ' Get numbers of characters. Result is 5 count = secondChar - firstChar ' Prints Henry Debug.Print Mid(s, firstChar, count) End Sub
You can see this is tricky to do and requires a bit of effort to figure out. We need to find the first space. Then we need to find the second space. Then we have to substract one from the other to give us the number of characters to take.
If have a string with a lot of words then this can get very tricky indeed. Luckily for us there is a much easier was to extract characters from a string. It’s called the Split function.
The Split Function
We can use the Split function to perform the above examples. The Split function splits a string into an array. Then we can easily access each individual item.
Let’s try the same three examples again and this time we will use Split.
Dim s As String: s = "John Henry Smith" Debug.Print Split(s, " ")(0) ' John Debug.Print Split(s, " ")(1) ' Henry Debug.Print Split(s, " ")(2) ' Smith
Boom! What a difference using Split makes. The way it works is as follows
- The Split function splits the string wherever there is a space.
- Each item goes into an array location starting at location zero.
- Using the number of a location we can access an array item.
The following table shows what the array might look like after Split has been used.
Note: the first position in the array is zero. Having zero based arrays is standard in programming languages.
In the above code we split the string each time we used it. We could also split the string once and store it in an array variable. Then we can access it when we want.
Sub SplitName() Dim s As String: s = "John Henry Smith" Dim arr() As String arr = Split(s, " ") Debug.Print arr(0) ' John Debug.Print arr(1) ' Henry Debug.Print arr(2) ' Smith End Sub
If you would like to know more about arrays then I wrote an entire post about them called The Complete Guide to Using Arrays in Excel VBA.
In the next sections, we will look at some real world examples. You will see the benefit of using Split instead of the InStr function.
Please feel free to try these yourself first. It is a great way to learn and you may have fun trying to figure them out(or maybe that’s just me!)
Example 1: Getting part of a file name
Imagine we want to extract the numbers from the following filenames
This is similar to the example about where we get the second item. To get the values here we use the underscore(i.e. “_”) to split the string. See the code example below
Sub GetNumber() ' Prints 23476 Debug.Print Split("VB_23476_Val.xls", "_")(1) ' Prints 987 Debug.Print Split("VV_987_Val.txt", "_")(1) ' Prints 12223 Debug.Print Split("ABBZA_12223_Val.doc", "_")(1) End Sub
In the real world you would normally read strings like these from a range of cells. So let’s say these filenames are stored in cells A1 to A3. We will adjust the code above slightly to give us:
Sub ReadNumber() Dim c As Range For Each c In Range("A1:A3") ' Split each item as you read it Debug.Print Split(c, "_")(1) Next c End Sub
Example 2: IP Address Range
The example here is taken from a question on the StackOverflow website.
The user has a string with an IP address in the format “BE-ABCDDD-DDS 172.16.23.3”.
He wants an IP of the range 172.16 to 172.31 to be valid. So for example
“BE-ABCDDD-DDS 172.16.23.3″ is valid
“BE-ABCDDD-DDS 172.25.23.3″ is valid
“BE-ABCDDED-DDS 18.104.22.168″ is not valid
“BE-ABCDDDZZ-DDS 22.214.171.124″ is not valid
This is how I would do this. First I split the string by the periods. The number we are looking for is between the first and second period. Therefore, it is the second item. When we split the string it is placed at position one in the array (remember that the array starts at position zero).
The resulting array will look like this
The code below shows how to do this
Sub IPAdd() ' Check the number to test different ip addresses Dim s1 As String: s1 = "BE-ABCDDD-DDS 172.31.23.3" ' Split the string using the period symbol Dim num As Long num = Split(s1, ".")(1) ' Check the number is valid Debug.Print num >= 16 And num <= 31 End Sub
Example 3: Check if a filename is valid
In this final example, we want to check that a file name is valid. There are three rules
- It must end with .pdf
- It must contain AA
- It must contain 1234 after AA
The following tables shows some valid and invalid items
|AA1234.pdf1||Not valid - doesn't end with .pdf|
|1234 AA.pdf||Not valid - AA does not come before 1234|
First we will do this using the InStr and Right functions.
Sub UseInstr() Dim f As String: f = "AA_1234_(5).pdf" ' Find AA first as the 1234 must come after this Dim lPos As Long: lPos = InStr(f, "AA") ' Search for 1234 and ensure last four chars are .pdf Debug.Print InStr(lPos, f, "1234") > 0 And Right(f, 4) = ".pdf" End Sub
This code is very messy. Luckily for us, VBA has Pattern Matching. We can check the pattern of a string without having to search for items and positions etc. We use the Like operator in VBA for pattern matching. The example below shows how to do it.
Sub UsePattern() Dim f As String: f = "AA_1234_(5).pdf" ' Define the pattern Dim pattern As String: pattern = "*AA*1234*.pdf" ' Check each item against the pattern Debug.Print f Like pattern ' False End Sub
In the above example, the asterisk in the pattern refers to any number of characters.
Let’s break down this pattern *AA*1234*.pdf
* – any group of characters
AA – the exact characters AA
* – any group of characters
1234 – the exact characters 1234
* – any group of characters
.pdf – the exact characters .pdf
To show this works correctly, let’s try it on all the example names in the table
Sub UsePatternTest() ' Create a collection of file names Dim coll As New Collection coll.Add "AA1234.pdf" coll.Add "AA_ljgslf_1234.pdf" coll.Add "AA1234.pdf1" coll.Add "1234 AA.pdf" coll.Add "12_AA_1234_NM.pdf" ' Define the pattern Dim pattern As String: pattern = "*AA*1234*.pdf" ' Check each item against the pattern Dim f As Variant For Each f In coll Debug.Print f Like pattern Next f End Sub
The output is
To find out more about Pattern Matching and the Like keyword please check out this post.
InStr and InStrRev are really only useful for simple tasks like checking if text exists in a string.
Left, Right and Mid are useful when the position of the text is always the same.
The Split function is the best way to extract from a variable string.
When trying to check the format of a string that is not fixed in size, the Like keyword(i.e. Pattern Matching) will generally provide an easier solution.
If you want to read about more VBA topics you can view a complete list of my posts here. I also have a free eBook(see below) which you will find useful if you are new to VBA.
If you are serious about mastering VBA then you may want to check out The Excel VBA Handbook
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