MS Access 2003 VBA 字符串在换行符处拆分
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MS Access 2003 VBA String Split on Line Break
提问by jim
I have a textbox on a MS Access form that users are going to copy a column of numbers into from an excel spreadsheet. I need to take this input and use it as parameters to build a query. I have code that looks like this
我在 MS Access 表单上有一个文本框，用户将从 Excel 电子表格中复制一列数字。我需要获取此输入并将其用作参数来构建查询。我有看起来像这样的代码
Dim data as variant
Dim input as String
data = Split(input,vbLf)
I want to be able to build a list of the input from the users but I can't figure out how to split it on the line break. I've tried "\n\r", "\n". "\r", vbCrLf, vbLf. The input looks like "12345[][]23456" with the box characters between each number
我希望能够构建用户输入的列表，但我不知道如何在换行符处拆分它。我试过“\n\r”、“\n”。"\r", vbCrLf, vbLf。输入看起来像“12345[][]23456”，每个数字之间有方框字符
Thanks
谢谢
回答by Jay Riggs
I got Split to work for me using vbCrLf. I also wrote the result of Split to a String array.
我让 Split 使用 vbCrLf 对我来说有效。我还将 Split 的结果写入了一个 String 数组。
Here's my code:
这是我的代码：
Dim data() As String
Dim yourInput As String
data = Split(yourInput, vbCrLf)
回答by manji
vbCRLF worked for me, try: Strings.Chr(13) & Strings.Chr(10) (which is vbCRLF)
vbCRLF 对我来说有效，尝试： Strings.Chr(13) & Strings.Chr(10) (which is vbCRLF)
try to see what is the ASCII code of those 2 boxes:
试着看看这两个盒子的 ASCII 码是什么：
//ex for input = "12345[][]23456"
Strings.Asc(Strings.Mid(input, 6, 1))
Strings.Asc(Strings.Mid(input, 7, 1))
