我想用它们比较Excel中的两个列表（列）来查找匹配项。 由于这是一个相当复杂的操作，我过去使用Excel中的几个不同的功能（非VBA）来执行它，但事实证明它充其量是笨拙的，因此我想尝试一体化如果可能的话，VBA解决方案。
第一列的名称有不规则性（例如引用的昵称，后缀如'jr'或'sr'，以及括号'首选'的名字'）。此外，当存在中间名时，它们可以是名称或初始名称。
第一栏中的顺序是：
<first name or initial>
<space>
<any parenthetical 'preferred' names - if they exist>
<space>
<middle name or initial - if it exists>
<space>
<quoted nickname or initial - if it exists>
<space>
<last name>
<comma - if necessary><space - if necessary><suffix - if it exists>
第二栏中的顺序是：
`<lastname><space><suffix>,<firstname><space><middle name, if it exists>`
，没有第一列的'不规则'。
我的主要目标是按照以下顺序“清理”第一列：
`lastname-space-suffix,firstname-space-preferred name-space-
middle name-space-nickname`
虽然我在这里保留了“违规行为”，但我可以在比较代码中使用某种“标志”来逐个提醒我。
我一直在尝试几种模式，这是我最近的模式：
["]?([A-Za-z]?)[.]?["]?[.]?[\s]?[,]?[\s]?
但是，我想允许使用姓氏和后缀（如果存在）。我用'global'测试了它，但是我无法弄清楚如何通过反向引用来分隔姓氏和后缀，例如。
然后，我想比较两个列表之间的最后一个，第一个，中间的首字母（因为大多数名称只是第一个列表中的首字母）。
An example would be:
(1st list)
John (Johnny) B. "Abe" Smith, Jr.
turned into:
Smith Jr,John (Johnny) B "Abe"
or
Smith Jr,John B
and
(2nd list)
Smith Jr,John Bertrand
turned into:
Smith Jr,John B
Then run a comparison between the two columns.
这个列表比较有什么好的开始或持续点？
2012年4月10日附录：
作为旁注，我需要从首选名称中删除昵称和括号中的引号。我可以将分组引用进一步分解为子组（在下面的示例中）吗？
(?: ([ ] \( [^)]* \)))? # (2) parenthetical 'preferred' name (optional)
(?: ([ ] (["'] ) .*?) \6 )? # (5,6) quoted nickname or initial (optional)
我可以这样分组：
(?:(([ ])(\()([^)]*)(\))))? # (2) parenthetical 'preferred' name (optional)
not sure how to do this one - # (5,6) quoted nickname or initial (optional)
我在“正则表达式教练”和“RegExr”中尝试过它们，但它们运行良好，但在VBA中，当我希望后向引用返回时 \ 11，\ 5 所有返回的是名字，数字1和逗号（例如“Carl1”）。我要回去查看任何错别字了。谢谢你的帮助。
2012年4月17日附录：
我忽略了一个名称'情况'，这是由2个或更多单词组成的姓氏，例如： 'St Cyr'或'Von Wilhelm' 是否会增加以下内容
`((St|Von)[ ])?
在这个正则表达式中工作，你提供了什么？
`((St|Von)[ ])?([^\,()"']+)
我在Regex Coach和RegExr中的测试并没有完全奏效，因为替换返回前面有空格的'St'。
答案 0 :(得分：2)
这是一个可能有用的正则表达式，它将按以下顺序为您提供6个捕获组：名字，前缀名称，中间名，昵称，姓氏，后缀。
([a-z]+)\.?\s(?:(\([a-z]+\))\s)?(?:([a-z]+)\.?\s)?(?:("[a-z]+")\s)?([a-z]+)(?:,\s([a-z]+))?
以下是解释：
([a-z]+)\.?\s # First name, followed by optional '.' (required)
(?:(\([a-z]+\))\s)? # Preferred name, optional
(?:([a-z]+)\.?\s)? # Middle name, optional
(?:("[a-z]+")\s)? # Nickname, optional
([a-z]+) # Last name, required
(?:,\s([a-z]+))? # Suffix, optional
例如，您可以将John (Johnny) B. "Abe" Smith, Jr.组合成Smith Jr,John (Johnny) B "Abe"组合，或者使用\5 \6,\1 \2 \3 \4将其转换为Smith Jr,John B
答案 1 :(得分：2)
重做 -
这是不同的方法。它可能适用于您的VBA，只是一个例子。
我在Perl中对它进行了测试，效果很好。但是，我不会显示perl代码，
只是正则表达式和一些解释。
这是一个两步过程。
. - 在全球范围内搜索\.，无需替换'' \s+，替换为单个空格[ ] （请注意，如果无法规范化，无论尝试什么，我都没有太大的成功机会）
规范化列值（对两列都做）后，通过这些正则表达式运行它。
第1列正则表达式
^
[ ]?
([^\ ,()"']+) # (1) first name or initial (required)
(?: ([ ] \( [^)]* \)) )? # (2) parenthetical 'preferred' name (optional)
(?:
([ ] [^\ ,()"'] ) # (3,4) middle initial OR name (optional)
([^\ ,()"']*) # name and initial are both captured
)?
(?: ([ ] (["'] ) .*?) \6 )? # (5,6) quoted nickname or initial (optional)
[ ] ([^\ ,()"']+) # (7) last name (required)
(?:
[, ]* ([ ].+?) [ ]? # (8) suffix (optional)
| .*?
)?
$
更换取决于您的需要。
定义了三种类型（根据需要将$替换为\）：
$7$8,$1$2$3$4$5$6 $7$8,$1$2$3$5$6 $7$8,$1$3 转换示例：
Input (raw) = 'John (Johnny) Bertrand "Abe" Smith, Jr. '
Out type 1 full middle = 'Smith Jr,John (Johnny) Bertrand "Abe"'
Out type 1 middle initial = 'Smith Jr,John (Johnny) B "Abe"'
Out type 2 middle initial = 'Smith Jr,John B'
第2列正则表达式
^
[ ]?
([^\ ,()"']+) # (1) last name (required)
(?: ([ ] [^\ ,()"']+) )? # (2) suffix (optional)
,
([^\ ,()"']+) # (3) first name or initial (required)
(?:
([ ] [^\ ,()"']) # (4,5) middle initial OR name (optional)
([^\ ,()"']*)
)?
.*
$
更换取决于您的需要。
定义了两种类型（根据需要将$替换为\）：
$1$2,$3$4$5 $1$2,$3$4 转换示例：
Input = 'Smith Jr.,John Bertrand '
Out type 1 full middle = 'Smith Jr,John Bertrand'
Out type 1 middle initial = 'Smith Jr,John B'
VBA替换帮助
这适用于Excel的一个非常旧的副本，创建一个VBA项目 这些是为了展示一个例子而创建的两个模块 他们都做同样的事情。
第一个是可能的所有替换类型的详细示例 第二个是仅使用类型2比较的修剪版本。
我之前没有做过VB，但是它应该很简单
为了你收集更换的工作原理，以及如何配合excel
列。
如果你只做一个扁平的比较，你可能想做一个col 1 val 一次，然后检查第2列中的每个值，然后转到下一个val 第1列，然后重复。
要以最快的方式执行此操作，请创建2个额外的列，转换为尊重的
列vals为type-2（变量strC1_2和strC2_2，见例），然后复制它们
到新栏目
之后，您不需要正则表达式，只需比较列，找到匹配的行，
然后删除type-2列。
详细 -
Sub RegexColumnValueComparison()
' Column 1 and 2 , Sample values
' These should probably be passed in values
' ============================================
strC1 = "John (Johnny) Bertrand ""Abe"" Smith, Jr. "
strC2 = "Smith Jr.,John Bertrand "
' Normalization Regexs for whitespace's and period's
' (use for both column values)
' =============================================
Set rxDot = CreateObject("vbscript.regexp")
rxDot.Global = True
rxDot.Pattern = "\."
Set rxWSp = CreateObject("vbscript.regexp")
rxWSp.Global = True
rxWSp.Pattern = "\s+"
' Column 1 Regex
' ==================
Set rxC1 = CreateObject("vbscript.regexp")
rxC1.Global = False
rxC1.Pattern = "^[ ]?([^ ,()""']+)(?:([ ]\([^)]*\)))?(?:([ ][^ ,()""'])([^ ,()""']*))?(?:([ ]([""']).*?)\6)?[ ]([^ ,()""']+)(?:[, ]*([ ].+?)[ ]?|.*?)?$"
' Column 2 Regex
' ==================
Set rxC2 = CreateObject("vbscript.regexp")
rxC2.Global = False
rxC2.Pattern = "^[ ]?([^ ,()""']+)(?:([ ][^ ,()""']+))?,([^ ,()""']+)(?:([ ][^ ,()""'])([^ ,()""']*))?.*$"
' Normalize column 1 and 2, Copy to new var
' ============================================
strC1_Normal = rxDot.Replace(rxWSp.Replace(strC1, " "), "")
strC2_Normal = rxDot.Replace(rxWSp.Replace(strC2, " "), "")
' ------------------------------------------------------
' This section is informational
' Shows some sample replacements before comparison
' Just pick 1 replacement from each column, discard the rest
' ------------------------------------------------------
' Create Some Replacement Types for Column 1
' =====================================================
strC1_1a = rxC1.Replace(strC1_Normal, "$7$8,$1$2$3$4$5$6")
strC1_1b = rxC1.Replace(strC1_Normal, "$7$8,$1$2$3$5$6")
strC1_2 = rxC1.Replace(strC1_Normal, "$7$8,$1$3")
' Create Some Replacement Types for Column 2
' =====================================================
strC2_1b = rxC2.Replace(strC2_Normal, "$1$2,$3$4$5")
strC2_2 = rxC2.Replace(strC2_Normal, "$1$2,$3$4")
' Show Types in Message Box
' =====================================================
c1_t1a = "Column1 Types:" & Chr(13) & "type 1a full middle - " & strC1_1a
c1_t1b = "type 1b middle initial - " & strC1_1b
c1_t2 = "type 2 middle initial - " & strC1_2
c2_t1b = "Column2 Types:" & Chr(13) & "type 1b middle initial - " & strC2_1b
c2_t2 = "type 2 middle initial - " & strC2_2
MsgBox (c1_t1a & Chr(13) & c1_t1b & Chr(13) & c1_t2 & Chr(13) & Chr(13) & c2_t1b & Chr(13) & c2_t2)
' ------------------------------------------------------
' Compare a Value from Column 1 vs Column 2
' For this we will compare Type 2 values
' ------------------------------------------------------
If strC1_2 = strC2_2 Then
MsgBox ("Type 2 values are EQUAL: " & Chr(13) & strC1_2)
Else
MsgBox ("Type 2 values are NOT Equal:" & Chr(13) & strC1_2 & " != " & strC1_2)
End If
' ------------------------------------------------------
' Same comparison (Type 2) of Normalized column 1,2 values
' In esscense, this is all you need
' ------------------------------------------------------
If rxC1.Replace(strC1_Normal, "$7$8,$1$3") = rxC2.Replace(strC2_Normal, "$1$2,$3$4") Then
MsgBox ("Type 2 values are EQUAL")
Else
MsgBox ("Type 2 values are NOT Equal")
End If
End Sub
仅限类型2 -
Sub RegexColumnValueComparison()
' Column 1 and 2 , Sample values
' These should probably be passed in values
' ============================================
strC1 = "John (Johnny) Bertrand ""Abe"" Smith, Jr. "
strC2 = "Smith Jr.,John Bertrand "
' Normalization Regexes for whitespace's and period's
' (use for both column values)
' =============================================
Set rxDot = CreateObject("vbscript.regexp")
rxDot.Global = True
rxDot.Pattern = "\."
Set rxWSp = CreateObject("vbscript.regexp")
rxWSp.Global = True
rxWSp.Pattern = "\s+"
' Column 1 Regex
' ==================
Set rxC1 = CreateObject("vbscript.regexp")
rxC1.Global = False
rxC1.Pattern = "^[ ]?([^ ,()""']+)(?:([ ]\([^)]*\)))?(?:([ ][^ ,()""'])([^ ,()""']*))?(?:([ ]([""']).*?)\6)?[ ]([^ ,()""']+)(?:[, ]*([ ].+?)[ ]?|.*?)?$"
' Column 2 Regex
' ==================
Set rxC2 = CreateObject("vbscript.regexp")
rxC2.Global = False
rxC2.Pattern = "^[ ]?([^ ,()""']+)(?:([ ][^ ,()""']+))?,([^ ,()""']+)(?:([ ][^ ,()""'])([^ ,()""']*))?.*$"
' Normalize column 1 and 2, Copy to new var
' ============================================
strC1_Normal = rxDot.Replace(rxWSp.Replace(strC1, " "), "")
strC2_Normal = rxDot.Replace(rxWSp.Replace(strC2, " "), "")
' Comparison (Type 2) of Normalized column 1,2 values
' ============================================
strC1_2 = rxC1.Replace(strC1_Normal, "$7$8,$1$3")
strC2_2 = rxC2.Replace(strC2_Normal, "$1$2,$3$4")
If strC1_2 = strC2_2 Then
MsgBox ("Type 2 values are EQUAL")
Else
MsgBox ("Type 2 values are NOT Equal")
End If
End Sub
Paren / Quote Response
As a side note, I will need to eliminate the quotes from the nicknames and the parentheses from the preferred names.
如果我理解正确..
是的，您可以捕获引号内的内容和括号内的括号
它只需要一些修改。以下正则表达式有能力
制定带或不带引号和/或括号的替代品，
或其他形式。
以下示例提供了制定替代品的方法。
非常重要的注意事项
如果您正在谈论从中删除引号“”和括号（） 匹配正则表达式，也可以这样做。它需要一个新的正则表达式。
唯一的问题是所有区分首选/中间/缺口
被抛出窗外，因为这些都是位置以及
分隔（即:(首选）中间“缺口”）。
考虑到这一点需要像这样的正则表达式子表达式
(?:[ ]([^ ,]+))? # optional preferred
(?:[ ]([^ ,]+))? # optional middle
(?:[ ]([^ ,]+))? # optional nick
并且，他们是可选的，失去所有位置参考并呈现midle-initial
表达无效。
结束注释
正则表达式模板（用于制定替换字符串）
^
[ ]?
# (required)
# First
# $1 name
# -----------------------------------------
([^\ ,()"']+) # (1) name
# (optional)
# Parenthetical 'preferred'
# $2 all
# $3$4 name
# -----------------------------------------
(?: ( # (2) all
([ ]) \( ([^)]*) \) # (3,4) space and name
)
)?
# (optional)
# Middle
# $5 initial
# $5$6 name
# -----------------------------------------
(?: ([ ] [^\ ,()"'] ) # (5) first character
([^\ ,()"']*) # (6) remaining characters
)?
# (optional)
# Quoted nick
# $7$8$9$8 all
# $7$9 name
# -----------------------------------------
(?: ([ ]) # (7) space
(["']) # (8) quote
(.*?) # (9) name
\8
)?
# (required)
# Last
# $10 name
# -----------------------------------------
[ ] ([^\ ,()"']+) # (10) name
# (optional)
# Suffix
# $11 suffix
# -----------------------------------------
(?: [, ]* ([ ].+?) [ ]? # (11) suffix
| .*?
)?
$
VBA正则表达式（第2版，在我的VBA项目中测试过）
rxC1.Pattern = "^[ ]?([^ ,()""']+)(?:(([ ])\(([^)]*)\)))?(?:([ ][^ ,()""'])([^ ,()""']*))?(?:([ ])([""'])(.*?)\8)?[ ]([^ ,()""']+)(?:[, ]*([ ].+?)[ ]?|.*?)?$"
strC1_1a = rxC1.Replace( strC1_Normal, "$10$11,$1$2$5$6$7$8$9$8" )
strC1_1aa = rxC1.Replace( strC1_Normal, "$10$11,$1$3$4$5$6$7$9" )
strC1_1b = rxC1.Replace( strC1_Normal, "$10$11,$1$2$5$7$8$9$8" )
strC1_1bb = rxC1.Replace( strC1_Normal, "$10$11,$1$3$4$5$7$9" )
strC1_2 = rxC1.Replace( strC1_Normal, "$10$11,$1$5" )
样本输入/输出可能性
Input (raw) = 'John (Johnny) Bertrand "Abe" Smith, Jr. '
Out type 1a full middle = 'Smith Jr,John (Johnny) Bertrand "Abe"'
Out type 1aa full middle = 'Smith Jr,John Johnny Bertrand Abe'
Out type 1b middle initial = 'Smith Jr,John (Johnny) B "Abe"'
Out type 1bb middle initial = 'Smith Jr,John Johnny B Abe'
Out type 2 middle initial = 'Smith Jr,John B'
Input (raw) = 'John (Johnny) Smith, Jr.'
Out type 1a full middle = 'Smith Jr,John (Johnny)'
Out type 1aa full middle = 'Smith Jr,John Johnny'
Out type 1b middle initial = 'Smith Jr,John (Johnny)'
Out type 1bb middle initial = 'Smith Jr,John Johnny'
Out type 2 middle initial = 'Smith Jr,John'
Input (raw) = 'John (Johnny) "Abe" Smith, Jr.'
Out type 1a full middle = 'Smith Jr,John (Johnny) "Abe"'
Out type 1aa full middle = 'Smith Jr,John Johnny Abe'
Out type 1b middle initial = 'Smith Jr,John (Johnny) "Abe"'
Out type 1bb middle initial = 'Smith Jr,John Johnny Abe'
Out type 2 middle initial = 'Smith Jr,John'
Input (raw) = 'John "Abe" Smith, Jr.'
Out type 1a full middle = 'Smith Jr,John "Abe"'
Out type 1aa full middle = 'Smith Jr,John Abe'
Out type 1b middle initial = 'Smith Jr,John "Abe"'
Out type 1bb middle initial = 'Smith Jr,John Abe'
Out type 2 middle initial = 'Smith Jr,John'
回复：4/17关注
last names that have 2 or more words. Would the allowance for certain literal names, rather than generic word patterns, be the solution?
实际上，不，不会。在这种情况下，对于您的表单，允许姓氏中的多个单词
将空间字段分隔符注入到姓氏字段中。
然而，对于你的特定形式，它可以完成，因为唯一的障碍是什么时候
"nick"字段缺失。如果它缺失并且假设中只有一个单词
中间名，2个排列。
希望您可以从下面的3个正则表达式和测试用例输出中获得解决方案。
正则表达式已经从捕获中清除了空间分隔符。所以，你可以写作
使用Replace方法替换，或者只存储捕获缓冲区以与
进行比较
其他专栏的捕获方案的结果。
Nick_rx.Pattern (template)
* This pattern is multi-word last name, NICK is required
^
[ ]?
# First (req'd)
([^\ ,()"']+) # (1) first name
# Preferred first
(?: [ ]
( # (2) (preferred), -or-
\( ([^)]*?) \) # (3) preferred
)
)?
# Middle
(?: [ ]
( # (4) full middle, -or-
([^\ ,()"']) # (5) initial
[^\ ,()"']*
)
)?
# Quoted nick (req'd)
[ ]
( # (6) "nick",
(["']) # (7) n/a -or-
(.*?) # (8) nick
\7
)
# Single/Multi Last (req'd)
[ ]
( # (9) multi/single word last name
[^\ ,()"']+
(?:[ ][^\ ,()"']+)*
)
# Suffix
(?: [ ]? , [ ]? (.*?) )? # (10) suffix
[ ]?
$
-----------------------------------
FLs_rx.Pattern (template)
* This pattern has no MIDDLE/NICK, is single-word last name,
* and has no permutations.
^
[ ]?
# First (req'd)
([^\ ,()"']+) # (1) first name
# Preferred first
(?: [ ]
( # (2) (preferred), -or-
\( ([^)]*?) \) # (3) preferred
)
)?
# Single Last (req'd)
[ ]
([^\ ,()"']+) # (4) single word last name
# Suffix
(?: [ ]? , [ ]? (.*?) )? # (5) suffix
[ ]?
$
-----------------------------------
FLm_rx.Pattern (template)
* This pattern has no NICK, is multi-word last name,
* and has 2 permutations.
* 1. Middle as part of Last name.
* 2. Middle is separate from Last name.
^
[ ]?
# First (req'd)
([^\ ,()"']+) # (1) first name
# Preferred first
(?: [ ]
( # (2) (preferred), -or-
\( ([^)]*?) \) # (3) preferred
)
)?
# Multi Last (req'd)
[ ]
( # (4) Multi, as Middle + Last,
# -or-
(?: # Middle
( # (5) full middle, -or-
([^\ ,()"']) # (6) initial
[^\ ,()"']*
)
[ ]
)
# Last (req'd)
( # (7) multi/single word last name
[^\ ,()"']+
(?:[ ][^\ ,()"']+)*
)
)
# Suffix
(?: [ ]? , [ ]? (.*?) )? # (8) suffix
[ ]?
$
-----------------------------------
Each of these regexes are mutually exclusive and should be checked
in an if-then-else like this (Pseudo code):
str_Normal = rxDot.Replace(rxWSp.Replace(str, " "), "")
If Nick_rx.Test(str_Normal) Then
N_1a = rxWSp.Replace( Nick_rx.Replace(str_Normal, "$9 $10 , $1 $2 $4 $6 "), " ")
N_1aa = rxWSp.Replace( Nick_rx.Replace(str_Normal, "$9 $10 , $1 $3 $4 $8 "), " ")
N_1b = rxWSp.Replace( Nick_rx.Replace(str_Normal, "$9 $10 , $1 $2 $5 $6 "), " ")
N_1bb = rxWSp.Replace( Nick_rx.Replace(str_Normal, "$9 $10 , $1 $3 $5 $8 "), " ")
N_2 = rxWSp.Replace( Nick_rx.Replace(str_Normal, "$9 $10 , $1 $5 "), " ")
' see test case results in output below
Else
If FLs_rx.Test(str_Normal) Then
FLs_1a = rxWSp.Replace( FLs_rx.Replace(str_Normal, "$4 $5 , $1 $2 "), " ")
FLs_1aa = rxWSp.Replace( FLs_rx.Replace(str_Normal, "$4 $5 , $1 $3 "), " ")
FLs_2 = rxWSp.Replace( FLs_rx.Replace(str_Normal, "$4 $5 , $1 "), " ")
Else
If FLm_rx.Test(str_Normal) Then
' Permutation 1:
FLm1_1a = rxWSp.Replace( FLm_rx.Replace(str_Normal, "$4 $8 , $1 $2 "), " ")
FLm1_1aa = rxWSp.Replace( FLm_rx.Replace(str_Normal, "$4 $8 , $1 $3 "), " ")
FLm1_2 = rxWSp.Replace( FLm_rx.Replace(str_Normal, "$4 $8 , $1 "), " ")
' Permutation 2:
FLm2_1a = rxWSp.Replace( FLm_rx.Replace(str_Normal, "$7 $8 , $1 $2 $5 "), " ")
FLm2_1aa = rxWSp.Replace( FLm_rx.Replace(str_Normal, "$7 $8 , $1 $3 $5 "), " ")
FLm2_1b = rxWSp.Replace( FLm_rx.Replace(str_Normal, "$7 $8 , $1 $2 $6 "), " ")
FLm2_1bb = rxWSp.Replace( FLm_rx.Replace(str_Normal, "$7 $8 , $1 $3 $6 "), " ")
FLm2_2 = rxWSp.Replace( FLm_rx.Replace(str_Normal, "$7 $8 , $1 $6 "), " ")
' At this point, the odds are that only one of these permutations will match
' a different column.
Else
' The data could not be matched against a valid form
End If
-----------------------------
Test Cases
Found form 'Nick'
Input (raw) = 'John1 (JJ) Bert "nick" St Van Helsing ,Jr '
Normal = 'John1 (JJ) Bert "nick" St Van Helsing ,Jr '
Out type 1a full middle = 'St Van Helsing Jr , John1 (JJ) Bert "nick" '
Out type 1aa full middle = 'St Van Helsing Jr , John1 JJ Bert nick '
Out type 1b middle initial = 'St Van Helsing Jr , John1 (JJ) B "nick" '
Out type 1bb middle initial = 'St Van Helsing Jr , John1 JJ B nick '
Out type 2 middle initial = 'St Van Helsing Jr , John1 B '
=======================================================
Found form 'Nick'
Input (raw) = 'John2 Bert "nick" Helsing ,Jr '
Normal = 'John2 Bert "nick" Helsing ,Jr '
Out type 1a full middle = 'Helsing Jr , John2 Bert "nick" '
Out type 1aa full middle = 'Helsing Jr , John2 Bert nick '
Out type 1b middle initial = 'Helsing Jr , John2 B "nick" '
Out type 1bb middle initial = 'Helsing Jr , John2 B nick '
Out type 2 middle initial = 'Helsing Jr , John2 B '
=======================================================
Found form 'Nick'
Input (raw) = 'John3 Bert "nick" St Van Helsing ,Jr '
Normal = 'John3 Bert "nick" St Van Helsing ,Jr '
Out type 1a full middle = 'St Van Helsing Jr , John3 Bert "nick" '
Out type 1aa full middle = 'St Van Helsing Jr , John3 Bert nick '
Out type 1b middle initial = 'St Van Helsing Jr , John3 B "nick" '
Out type 1bb middle initial = 'St Van Helsing Jr , John3 B nick '
Out type 2 middle initial = 'St Van Helsing Jr , John3 B '
=======================================================
Found form 'First-Last (single)'
Input (raw) = 'John4 Helsing '
Normal = 'John4 Helsing '
Out type 1a no middle = 'Helsing , John4 '
Out type 1aa no middle = 'Helsing , John4 '
Out type 2 = 'Helsing , John4 '
=======================================================
Found form 'First-Last (single)'
Input (raw) = 'John5 (JJ) Helsing '
Normal = 'John5 (JJ) Helsing '
Out type 1a no middle = 'Helsing , John5 (JJ) '
Out type 1aa no middle = 'Helsing , John5 JJ '
Out type 2 = 'Helsing , John5 '
=======================================================
Found form 'First-Last (multi)'
Input (raw) = 'John6 (JJ) Bert St Van Helsing ,Jr '
Normal = 'John6 (JJ) Bert St Van Helsing ,Jr '
Permutation 1:
Out type 1a no middle = 'Bert St Van Helsing Jr , John6 (JJ) '
Out type 1aa no middle = 'Bert St Van Helsing Jr , John6 JJ '
Out type 2 = 'Bert St Van Helsing Jr , John6 '
Permutation 2:
Out type 1a full middle = 'St Van Helsing Jr , John6 (JJ) Bert '
Out type 1aa full middle = 'St Van Helsing Jr , John6 JJ Bert '
Out type 1b middle initial = 'St Van Helsing Jr , John6 (JJ) B '
Out type 1bb middle initial = 'St Van Helsing Jr , John6 JJ B '
Out type 2 middle initial = 'St Van Helsing Jr , John6 B '
=======================================================
Found form 'First-Last (multi)'
Input (raw) = 'John7 Bert St Van Helsing ,Jr '
Normal = 'John7 Bert St Van Helsing ,Jr '
Permutation 1:
Out type 1a no middle = 'Bert St Van Helsing Jr , John7 '
Out type 1aa no middle = 'Bert St Van Helsing Jr , John7 '
Out type 2 = 'Bert St Van Helsing Jr , John7 '
Permutation 2:
Out type 1a full middle = 'St Van Helsing Jr , John7 Bert '
Out type 1aa full middle = 'St Van Helsing Jr , John7 Bert '
Out type 1b middle initial = 'St Van Helsing Jr , John7 B '
Out type 1bb middle initial = 'St Van Helsing Jr , John7 B '
Out type 2 middle initial = 'St Van Helsing Jr , John7 B '
=======================================================
Form *** (unknown)
Input (raw) = ' do(e)s not. match ,'
Normal = ' do(e)s not match ,'
=======================================================