AHK：从字符串中删除文本这个词到第一个换行符

Question

我在一家医生办公室工作，负责结算，但遇到了问题。由于 EMR（电子病历）程序的工作方式存在问题，免疫代码并不总是出现在图表中。为了弥补这一点，我让 AHK 也搜索了免疫的名称（危​​险，我知道）。直到最近，这一切都很好，直到其中一名医生取消了免疫订单。这引发了误报，因为我为没有代码的图表（IMM94 和 IMM97）包含了故障保护。我需要从一个非常大的字符串中删除单词“CANCELED:”之后的所有内容，直到该单词之后出现的第一个换行符。我创建了一个相当漂亮的字符串剪辑函数来完成大部分此类工作，但是由于太多了，所以在查找换行符时遇到了问题。我正在使用的测试用例是：

评估和计划：

1. 需要预防性疫苗接种和流感疫苗接种（选项卡）已取消：流感，高剂量 (65+)（换行）
2. 糖尿病前期致力于减肥和锻炼
3. 骨质减少运动和 vit d 和 b 12

上面和下面都有很多东西，但我想从我的单词“CANCELED:”字符串中提取所有实例到它之后的第一个换行符。

这是我正在尝试的，以及我的功能。它不起作用，实际上，它正在复制以“1.”开头的行：

^L::
lString := ClipBoard
If lString Contains Canceled
fString := ClipString(lString, "CANCELED: ", "`n")
Clipboard := fString
Return


ClipString(lString, aMarker, bMarker, Only := 0)
{ 
If (Only = 1)
Return SubStr(SubStr(lString, 1, InStr(lString, bMarker)+StrLen(bMarker)-1), InStr(lString, aMarker))
Else
Return SubStr(lString, 1, InStr(lString, aMarker)-StrLen(aMarker)) . SubStr(lString, InStr(lString, bMarker)+StrLen(bMarker))
}

Answer 1

从剪贴板中的每一行删除“已取消：”及其后面的任何文本

计划：

Clipboard := RegExReplace(Clipboard, "im)\s*CANCELLED:.*\R?", "`r`n")

示例输入：

alpha
beta cancelled: ABC asldkfalsd 
delta
gamma omega CANCELLed: zeta
theta

样本输出：

alpha
beta 
delta
gamma omega 
theta

Answer 2

想通了。我会评论它以帮助遇到此问题的任何人。我喜欢你通常可以获得功能的紧凑程度，但这件作品确实需要比我想要的多几行。如果有人能把它降低到更少，并且仍然可以工作，请告诉我！

^L::
ClipBoard := ClipString(ClipBoard, "Canceled: ", "`n",2) ;This is the function call here.
;The first parameter is the string to do stuff to, the second is what to cut out of the 
;string, and the third parameter is how far after that piece to cut, so if you wrote 
;Canceled: thenawholebunchofstuffthena newline, it'd remove all of it except for the 
;newline. The third parameter is which mode to put ClipString to.
Return


ClipString(lString, aMarker, bMarker, Mode := 0) ;already explained above. 
;Mode defaults to zero because I call the zero mode a lot.
{ 
If (Mode = 0) ;Mode Zero returns the ONLY the section from the original string 
;between aMarker and bMarker.
    Return SubStr(lString, 1, InStr(lString, aMarker)-StrLen(aMarker)) . SubStr(lString, InStr(lString, bMarker)+StrLen(bMarker))
Else If (Mode = 1) ;Mode One returns the original string with the section between aMarker 
;and bMarker removed.
    Return SubStr(SubStr(lString, 1, InStr(lString, bMarker)+StrLen(bMarker)-1), InStr(lString, aMarker))
Else If (Mode = 2) ;Mode Two returns the original string with all instances of aMarker 
;to bMarker removed. I.E. Every occurrence of aMarker will be removed to bMarker.
    {
    aString := lString
    StringReplace, aString, aString, %aMarker%, %aMarker%, UseErrorLevel ;Count how many 
;instances of aMarker are in the original string.
    CanCnt := Errorlevel ;actual count of # of aMarkers
    Loop, %CanCnt% ;loop as many times as there are aMarkers
        {
        aString := SubStr(aString, 1, InStr(aString, aMarker)-1) . SubStr(SubStr(aString, InStr(aString, aMarker)), InStr(SubStr(aString, InStr(aString, aMarker)), bMarker)-1) 
;this is a tad complicated. The first part before the concatenate takes the string, and
;keeps from the start point to the first aMarker. The concatenate adds the substring of the 
;substring between aMarker and the end of the string, and the location of the substring in 
;aString from the occurrence of aMarker to one position before the start of bMarker. 
;An easier way to read this would be to replace "SubStr(aString, InStr(aString, aMarker)) 
;with a variable like bString. It shortens it up quite a bit.
        }
    Return aString
    }
}