解析文本中的子字符串

本文介绍了解析文本中的子字符串的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在编写一个宏，该宏将LDAP格式的名称列表转换为第一，最后(区域)".

对于那些不知道LDAP外观的用户，请参见以下内容:

CN = John Smith(地区)，OU =法律，DC =示例，DC = comand

在Excel VBA中，我似乎无法使用string.substring(开始，结束).在Google上进行的搜索似乎表明，最好使用Mid(字符串，开始，结束).

问题是这样的:在Mid中，end的整数是距起点的距离，而不是字符的实际索引位置.这意味着不同的名称大小将具有不同的结束位置，并且我不能使用索引)"来查找区域的结尾.由于所有名称均以CN =开头，因此我可以正确找到第一个子字符串的末尾，但由于名称长度不同，因此无法正确找到)".

我下面有一些代码:

mgrSub1 = Mid(mgrVal, InStr(1, mgrVal, "=") + 1, InStr(1, mgrVal, "\") - 4) mgrSub2 = Mid(mgrVal, InStr(1, mgrVal, ","), InStr(1, mgrVal, ")") - 10) manager = mgrSub1 & mgrSub2

是否有一种方法可以使用设置的终点，而不是距离起点有那么多值的终点?

解决方案

这是vba .. no string.substring;)

这更像是VB 6(或下面的任何一个)..因此，您陷入了mid，instr，len(获取字符串总len)的问题.我想您错过了len以获取字符总数一串?如果您需要澄清，请发表评论.

另一个快速hack ..

Dim t As String t = "CN=Smith, John (region),OU=Legal,DC=example,DC=comand" Dim s1 As String Dim textstart As Integer Dim textend As Integer textstart = InStr(1, t, "CN=", vbTextCompare) + 3 textend = InStr(1, t, "(", vbTextCompare) s1 = Mid(t, textstart, textend - textstart) MsgBox s1 textstart = InStr(1, t, "(", vbTextCompare) + 1 textend = InStr(1, t, ")", vbTextCompare) s2 = Mid(t, textstart, textend - textstart) MsgBox s2

很明显，您的问题是，由于您需要对第二个参数进行区分，因此应始终对其进行一些数学运算.

I am writing a macro that converts a list of names that are in an LDAP format to First, Last (region).

For those who do not know what LDAP looks like, it is below:

CN=John Smith (region),OU=Legal,DC=example,DC=comand

In Excel VBA, I do not appear to be able to use string.substring(start, end). A search on Google seems to reveal that Mid(string, start, end) is the best option.

The problem is this: In Mid, the integer for end is the distance from start, not the actual index location of the character. This means that different name sizes will have different ending locations and I cannot use index of ")" to find the end of the region. Since all of the names start with CN=, I can find the end of the first substring properly, but I cannot find ")" properly because names are different lengths.

I have some code below:

mgrSub1 = Mid(mgrVal, InStr(1, mgrVal, "=") + 1, InStr(1, mgrVal, "\") - 4) mgrSub2 = Mid(mgrVal, InStr(1, mgrVal, ","), InStr(1, mgrVal, ")") - 10) manager = mgrSub1 & mgrSub2

Is there a way to use a set end point instead of an end point that is so many values away from the start?

解决方案

This is vba.. no string.substring ;)

this is more like VB 6 (or any one below).. so you are stuck with mid, instr, len (to get the total len of a string).. I think you missed len to get the total of chars in a string? If you need some clarification just post a comment.

edit:

Another quick hack..

Dim t As String t = "CN=Smith, John (region),OU=Legal,DC=example,DC=comand" Dim s1 As String Dim textstart As Integer Dim textend As Integer textstart = InStr(1, t, "CN=", vbTextCompare) + 3 textend = InStr(1, t, "(", vbTextCompare) s1 = Mid(t, textstart, textend - textstart) MsgBox s1 textstart = InStr(1, t, "(", vbTextCompare) + 1 textend = InStr(1, t, ")", vbTextCompare) s2 = Mid(t, textstart, textend - textstart) MsgBox s2

Clearly your problem is that since you need a diference for the second parameter, you should always do some math for it...