blocks|key|444500|text|最安全最简单的就是循环；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|444501|Dim+buff()+As+String
ReDim+buff(Len(my_string)+-+1)
For+i+=+1+To+Len(my_string)
++++buff(i+-+1)+=+Mid$(my_string,+i,+1)
Next|code-block|syntax|javascript|444502|如果你保证只使用ansi字符，你可以；|444503|Dim+buff()+As+String
buff+=+Split(StrConv(my_string,+vbUnicode),+Chr$(0))
ReDim+Preserve+buff(UBound(buff)+-+1)|444504|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Safest &amp; simplest is to just loop;

<pre><code>Dim buff() As String
ReDim buff(Len(my_string) - 1)
For i = 1 To Len(my_string)
 buff(i - 1) = Mid$(my_string, i, 1)
Next
</code></pre>

If your guaranteed to use ansi characters only you can;

<pre><code>Dim buff() As String
buff = Split(StrConv(my_string, vbUnicode), Chr$(0))
ReDim Preserve buff(UBound(buff) - 1)
</code></pre>

blocks|key|444527|text|您可以将字符串分配给一个字节数组(反之也是可能的)。结果是每个字符对应2个数字，因此Xmas转换为包含{88,0,109,0,97,0,115,0}的字节数组|type|unstyled|depth|inlineStyleRanges|entityRanges|data|444528|或者，您可以使用StrConv|444529|Dim+bytes()+as+Byte
bytes+=+StrConv("Xmas",+vbFromUnicode)|code-block|syntax|javascript|444530|这将给你{88,109,97,115}，但在这种情况下，你不能将字节数组分配回一个字符串。|444531|可以使用Chr()函数将字节数组中的数字转换回字符|444532|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You can just assign the string to a byte array (the reverse is also possible). The result is 2 numbers for each character, so Xmas converts to a byte array containing {88,0,109,0,97,0,115,0} or you can use StrConv

<pre><code>Dim bytes() as Byte
bytes = StrConv("Xmas", vbFromUnicode)
</code></pre>

which will give you {88,109,97,115} but in that case you cannot assign the byte array back to a string. You can convert the numbers in the byte array back to characters using the Chr() function

blocks|key|226388|text|这是在VBA中实现的另一种方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|226389|Function+ConvertToArray(ByVal+value+As+String)
++++value+=+StrConv(value,+vbUnicode)
++++ConvertToArray+=+Split(Left(value,+Len(value)+-+1),+vbNullChar)
End+Function
Sub+example()
++++Dim+originalString+As+String
++++originalString+=+"hi+there"
++++Dim+myArray()+As+String
++++myArray+=+ConvertToArray(originalString)
End+Sub|code-block|syntax|javascript|226390|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's another way to do it in VBA.

<pre><code>Function ConvertToArray(ByVal value As String)
 value = StrConv(value, vbUnicode)
 ConvertToArray = Split(Left(value, Len(value) - 1), vbNullChar)
End Function
Sub example()
 Dim originalString As String
 originalString = "hi there"
 Dim myArray() As String
 myArray = ConvertToArray(originalString)
End Sub
</code></pre>

blocks|key|444589|text|根据this+code+golfing+solution+by+Gaffi的说法，以下是可行的：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|444590|a+=+Split(StrConv(s,+64),+Chr(0))|code-block|syntax|javascript|444591|entityMap|0|LINK|mutability|MUTABLE|url|http://gaffiprog.blogspot.com/2013/04/golfing-tip-vba-split-string-into.html^0|2|Z|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

According to <a href="http://gaffiprog.blogspot.com/2013/04/golfing-tip-vba-split-string-into.html" rel="nofollow noreferrer">this code golfing solution by Gaffi</a>, the following works:

<pre><code>a = Split(StrConv(s, 64), Chr(0))
</code></pre>

blocks|key|444563|text|问题是在vb中没有内置的方法(或者至少我们都找不到)来做到这一点。但是，有一个可以在空格上拆分字符串，所以我只是重新构建字符串并添加到空格中……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|444564|Private+Function+characterArray(ByVal+my_string+As+String)+As+String()
++'create+a+temporary+string+to+store+a+new+string+of+the+same+characters+with+spaces
++Dim+tempString+As+String+=+""
++'cycle+through+the+characters+and+rebuild+my_string+as+a+string+with+spaces+
++'and+assign+the+result+to+tempString.++
++For+Each+c+In+my_string
+++++tempString+&=+c+&+"+"
++Next
++'return+return+tempString+as+a+character+array.++
++Return+tempString.Split()
End+Function|code-block|syntax|javascript|444565|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

the problem is that there is no built in method (or at least none of us could find one) to do this in vb. However, there is one to split a string on the spaces, so I just rebuild the string and added in spaces.... 

<pre><code>Private Function characterArray(ByVal my_string As String) As String()
 'create a temporary string to store a new string of the same characters with spaces
 Dim tempString As String = ""
 'cycle through the characters and rebuild my_string as a string with spaces 
 'and assign the result to tempString. 
 For Each c In my_string
 tempString &amp;= c &amp; " "
 Next
 'return return tempString as a character array. 
 Return tempString.Split()
End Function
</code></pre>

blocks|key|226559|text|要将字符串拆分为任意所需长度的子字符串数组，请执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|226560|Function+charSplitMulti(s+As+Variant,+splitLen+As+Long)+As+Variant
++++
++++++++Dim+padding+As+Long:+padding+=+0
++++++++Dim+l+As+Long:+l+=+0
++++++++Dim+v+As+Variant
++++++++
++++++++'Pad+the+string+so+it+divides+evenly+by
++++++++'+the+length+of+the+desired+sub-strings
++++++++Do+While+Len(s)+Mod+splitLen+>+0
++++++++++++s+=+s+&+"x"
++++++++++++padding+=+padding+%2B+1
++++++++Loop
++++++++
++++++++'Create+an+array+with+sufficient
++++++++'+elements+to+hold+all+the+sub-strings
++++++++Do+Until+Len(v)+=+(Len(s)+/+splitLen)+-+1
++++++++++++v+=+v+&+","
++++++++Loop
++++++++v+=+Split(v,+",")
++++++++
++++++++'Populate+the+array+by+repeatedly
++++++++'+adding+in+the+first+[splitLen]
++++++++'+characters+of+the+string,+then
++++++++'+removing+them+from+the+string
++++++++Do+While+Not+s+Like+""
++++++++++++v(l)+=+Mid(s,+1,+splitLen)
++++++++++++s+=+Right(s,+Len(s)+-+splitLen)
++++++++++++l+=+l+%2B+1
++++++++Loop
++++++++
++++++++'Remove+any+padding+characters+added+at+step+one
++++++++v(UBound(v))+=+Left(v(UBound(v)),+Len(v(UBound(v)))+-+padding)
++++++++
++++++++'Output+the+array
++++++++charSplitMulti+=+v
++++
++++End+Function|code-block|syntax|javascript|226561|您可以将字符串作为字符串传递给它：|226562|Sub+test_charSplitMulti_stringInput()

++++Dim+s+As+String:+s+=+"123456789abc"
++++Dim+subStrLen+As+Long:+subStrLen+=+4
++++Dim+myArray+As+Variant
++++
++++myArray+=+charSplitMulti(s,+subStrLen)
++++
++++For+i+=+0+To+UBound(myArray)
++++++++MsgBox+myArray(i)
++++Next

End+Sub|226563|…或者已经解密为变体：|226564|Sub+test_charSplitMulti_variantInput()

++++Dim+s+As+Variant:+s+=+"123456789abc"
++++Dim+subStrLen+As+Long:+subStrLen+=+5
++++
++++s+=+charSplitMulti(s,+subStrLen)
++++
++++For+i+=+0+To+UBound(s)
++++++++MsgBox+s(i)
++++Next

End+Sub|226565|如果所需的子字符串的长度不能平均划分为字符串的长度，则数组的最上面的元素将更短。(它将等于strLength+Mod+subStrLength。这可能是显而易见的。)|226566|我发现最经常使用它来将字符串拆分成单个字符，所以我添加了另一个函数，这样我就可以懒惰了，在这种情况下不必传递两个变量：|226567|Function+charSplit(s+As+Variant)+As+Variant

++++charSplit+=+charSplitMulti(s,+1)

End+Function

Sub+test_charSplit()

++++Dim+s+As+String:+s+=+"123456789abc"
++++Dim+myArray+As+Variant
++++
++++myArray+=+charSplit(s)
++++
++++For+i+=+0+To+UBound(myArray)
++++++++MsgBox+myArray(i)
++++Next

End+Sub|226568|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

To split a string into an array of sub-strings of any desired length:
<pre><code>Function charSplitMulti(s As Variant, splitLen As Long) As Variant
 
 Dim padding As Long: padding = 0
 Dim l As Long: l = 0
 Dim v As Variant
 
 'Pad the string so it divides evenly by
 ' the length of the desired sub-strings
 Do While Len(s) Mod splitLen &gt; 0
 s = s &amp; &quot;x&quot;
 padding = padding + 1
 Loop
 
 'Create an array with sufficient
 ' elements to hold all the sub-strings
 Do Until Len(v) = (Len(s) / splitLen) - 1
 v = v &amp; &quot;,&quot;
 Loop
 v = Split(v, &quot;,&quot;)
 
 'Populate the array by repeatedly
 ' adding in the first [splitLen]
 ' characters of the string, then
 ' removing them from the string
 Do While Not s Like &quot;&quot;
 v(l) = Mid(s, 1, splitLen)
 s = Right(s, Len(s) - splitLen)
 l = l + 1
 Loop
 
 'Remove any padding characters added at step one
 v(UBound(v)) = Left(v(UBound(v)), Len(v(UBound(v))) - padding)
 
 'Output the array
 charSplitMulti = v
 
 End Function
</code></pre>
You can pass the string into it either as a string:
<pre><code>Sub test_charSplitMulti_stringInput()

 Dim s As String: s = &quot;123456789abc&quot;
 Dim subStrLen As Long: subStrLen = 4
 Dim myArray As Variant
 
 myArray = charSplitMulti(s, subStrLen)
 
 For i = 0 To UBound(myArray)
 MsgBox myArray(i)
 Next

End Sub
</code></pre>
…or already declard as a variant:
<pre><code>Sub test_charSplitMulti_variantInput()

 Dim s As Variant: s = &quot;123456789abc&quot;
 Dim subStrLen As Long: subStrLen = 5
 
 s = charSplitMulti(s, subStrLen)
 
 For i = 0 To UBound(s)
 MsgBox s(i)
 Next

End Sub
</code></pre>
If the length of the desired sub-string doesn't divide equally into the length of the string, the uppermost element of the array will be shorter. (It'll be equal to strLength Mod subStrLength. Which is probably obvious.)
I found that most-often I use it to split a string into single characters, so I added another function, so I can be lazy and not have to pass two variables in that case:
<pre><code>Function charSplit(s As Variant) As Variant

 charSplit = charSplitMulti(s, 1)

End Function

Sub test_charSplit()

 Dim s As String: s = &quot;123456789abc&quot;
 Dim myArray As Variant
 
 myArray = charSplit(s)
 
 For i = 0 To UBound(myArray)
 MsgBox myArray(i)
 Next

End Sub
</code></pre>

blocks|key|226599|text|尝试来自Rara的这个迷你代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|226600|Function+charSplitMulti(TheString+As+Variant,+SplitLen+As+Long)+As+Variant
++++'Defining+a+temporary+array.
++++Dim+TmpArray()+As+String
++++'Checking+if+the+SplitLen+is+not+less+than+one.+if+so+the+function+returns+the+whole+string+without+any+changing.
++++SplitLen+=+IIf(SplitLen+>=+1,+SplitLen,+Len(TheString))
++++'Redefining+the+temporary+array+as+needed.
++++ReDim+TmpArray(Len(TheString)+\+SplitLen+%2B+IIf(Len(TheString)+Mod+SplitLen+<>+0,+1,+0))
++++'Splitting+the+input+string.
++++For+i+=+1+To+UBound(TmpArray)
++++++++TmpArray(i)+=+Mid(TheString,+(i+-+1)+*+SplitLen+%2B+1,+SplitLen)
++++Next
++++'Outputing+the+result.+
++++charSplitMulti+=+TmpArray
End+Function|code-block|syntax|javascript|226601|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this minicode From Rara:
<pre><code>Function charSplitMulti(TheString As Variant, SplitLen As Long) As Variant
 'Defining a temporary array.
 Dim TmpArray() As String
 'Checking if the SplitLen is not less than one. if so the function returns the whole string without any changing.
 SplitLen = IIf(SplitLen &gt;= 1, SplitLen, Len(TheString))
 'Redefining the temporary array as needed.
 ReDim TmpArray(Len(TheString) \ SplitLen + IIf(Len(TheString) Mod SplitLen &lt;&gt; 0, 1, 0))
 'Splitting the input string.
 For i = 1 To UBound(TmpArray)
 TmpArray(i) = Mid(TheString, (i - 1) * SplitLen + 1, SplitLen)
 Next
 'Outputing the result. 
 charSplitMulti = TmpArray
End Function
</code></pre>

How is it possible to split a VBA string into an array of characters? 

I tried <code>Split(my_string, "")</code> but this didn't work.

Split string into array of characters?

如何将VBA字符串拆分成字符数组？我尝试过Split(my_string, "")，但这不起作用。

问是否将字符串拆分成字符数组？
EN

回答 7

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问是否将字符串拆分成字符数组？EN

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问是否将字符串拆分成字符数组？
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