.net 如何将字符串拆分为固定长度的字符串数组？
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How to split a string into a fixed length string array?
提问by Yoga Fire
I have a long string like this
我有一个像这样的长字符串
dim LongString as String = "123abc456def789ghi"
And I want to split it into a string array. Each element of the array should be in 3 characters length
我想把它拆分成一个字符串数组。数组的每个元素的长度应为 3 个字符
For example,
例如，
Dim LongArray(5) As String
LongArray(0) = "123"
LongArray(1) = "abc"
LongArray(2) = "456"
LongArray(3) = "def"
LongArray(4) = "789"
LongArray(5) = "ghi"
How do I split it using VB.net code?
如何使用 VB.net 代码拆分它？
采纳答案by chrissie1
This could work.
这可以工作。
Module Module1
Sub Main()
Dim LongString As String = "123abc456def789ghi"
Dim longlist As New List(Of String)
For i As Integer = 0 To Convert.ToInt32(LongString.Length / 3) - 1
longlist.Add(LongString.Substring(i * 3, 3))
Next
For Each s As String In longlist
Console.WriteLine(s)
Next
Console.ReadLine()
End Sub
End Module
And this should work in .Net 1.1
这应该适用于 .Net 1.1
Module Module1
Sub Main()
Dim LongString As String = "123abc456def789ghi"
Dim longlist(Convert.ToInt32(LongString.Length / 3) - 1) As String
For i As Integer = 0 To Convert.ToInt32(LongString.Length / 3) - 1
longlist(i) = (LongString.Substring(i * 3, 3))
Next
For i As Integer = 0 To Convert.ToInt32(LongString.Length / 3) - 1
Console.WriteLine(longlist(i))
Next
Console.ReadLine()
End Sub
End Module
回答by Aaron
You could use LINQ like so:
你可以像这样使用 LINQ：
' VB.NET
Dim str = "123abc456def789ghij"
Dim len = 3
Dim arr = Enumerable.Range(0, str.Length / len).Select (Function(x) str.Substring(x * len, len)).ToArray()
// C#
var str = "123abc456def789ghij";
var len = 3;
var arr = Enumerable.Range(0, str.Length / len).Select (x => str.Substring(x * len, len)).ToArray();
Note this will only take completeoccurrences of length (i.e. 3 sets in a string 10 characters long).
请注意，这将只需要完整出现的长度（即 10 个字符长的字符串中的 3 个集合）。
回答by Jon Skeet
This C# code should work:
这个 C# 代码应该可以工作：
public static string[] SplitByLength(string text, int length)
{
// According to your comments these checks aren't necessary, but
// I think they're good practice...
if (text == null)
{
throw new ArgumentNullException("text");
}
if (length <= 0)
{
throw new ArgumentOutOfRangeException("length");
}
if (text.Length % length != 0)
{
throw new ArgumentException
("Text length is not a multiple of the split length");
}
string[] ret = new string[text.Length / length];
for (int i = 0; i < ret.Length; i++)
{
ret[i] = text.Substring(i * length, length);
}
return ret;
}
Reflector converts that to VB as:
反射器将其转换为 VB 为：
Public Shared Function SplitByLength(ByVal [text] As String, _
ByVal length As Integer) As String()
' Argument validation elided
Dim strArray As String() = New String(([text].Length \ length) - 1) {}
Dim i As Integer
For i = 0 To ret.Length - 1
strArray(i) = [text].Substring((i * length), length)
Next i
Return strArray
End Function
It's possible that that isn't idiomatic VB, which is why I've included the C# as well.
这可能不是惯用的 VB，这就是我也包含 C# 的原因。
回答by vangli
I'm splitting the string by 35.
我将字符串拆分为 35。
var tempstore ="12345678901234567890123456789012345";
for (int k = 0; k < tempstore.Length; k += 35) {
PMSIMTRequest.Append(tempstore.Substring(k,
tempstore.Length - k > 35 ? 35 : tempstore.Length - k));
PMSIMTRequest.Append(System.Environment.NewLine);
}
messagebox.Show(PMSIMTRequest.tostring());
回答by Kan
Last array is missing:
缺少最后一个数组：
Public Function SplitByLength(ByVal text As String, _
ByVal length As Integer) As String()
' Argument validation elided
Dim strArray As String() = New String((text.Length \ length) - 1) {}
Dim i As Integer
For i = 0 To text.Length - 1
strArray(i) = text.Substring((i * length), length)
Next i
' Get last array:
ReDim Preserve strArray(i)
strArray(i) = text.Substring(i * length)
Return strArray
End Function
回答by Amir
I have added some more logic to @jon code. This will work perfectly for the string which has length less than length passed.
我在@jon 代码中添加了更多逻辑。这对于长度小于传递的长度的字符串非常有效。
Public Shared Function SplitByLength(ByVal [text] As String, ByVal length As Integer) As String()
Dim stringLength = text.Length
Dim arrLength As Integer = ([text].Length \ length) - 1 + IIf(([text].Length
Mod length) > 0, 1, 0)
Dim strArray As String() = New String(arrLength) {}
Dim returnString As String = ""
Dim i As Integer
Dim remLength As Integer = 0
For i = 0 To strArray.Length - 1
remLength = stringLength - i * length
If remLength < length Then
strArray(i) = [text].Substring((i * length), remLength)
Else
strArray(i) = [text].Substring((i * length), length)
End If
Next i
Return strArray
END FUNCTION
回答by xanatos
Dim LongString As String = "1234567"
Dim LongArray((LongString.Length + 2) \ 3 - 1) As String
For i As Integer = 0 To LongString.Length - 1 Step 3
LongArray(i \ 3) = IF (i + 3 < LongString.Length, LongString.Substring(i, 3), LongString.Substring(i, LongString.Length - i))
Next
For Each s As String In LongArray
Console.WriteLine(s)
Next
There are some interesting parts, the use of the \integer division (that is always rounded down), the fact that in VB.NET you have to tell to DIM the maximum element of the array (so the length of the array is +1) (this is funny only for C# programmers) (and it's solved by the -1 in the dim), the "+ 2" addition (I need to round UP the division by 3, so I simply add 2 to the dividend, I could have used a ternary operator and the modulus, and in the first test I did it), and the use of the ternary operator IF() in getting the substring.
有一些有趣的部分，\整数除法的使用（总是向下取整），事实上在 VB.NET 中你必须告诉 DIM 数组的最大元素（所以数组的长度是 +1） （这仅对 C# 程序员来说很有趣）（并且它由昏暗中的 -1 解决），“+ 2”加法（我需要将除法四舍五入为 3，所以我只需将 2 添加到股息中，我可以已经使用了三元运算符和模数，并且在第一个测试中我做到了），以及使用三元运算符 IF() 来获取子字符串。
