AHK:将文本从字符串中移到第一行断线处

Question

我在一家医生办公室做账单，我遇到了一个问题。由于EMR (电子病历)程序的工作方式有问题，免疫代码并不总是被列入图表。为了补偿这一点，我也有AHK的名字寻找免疫(危险，我知道)。直到最近，这项工作还很顺利，直到其中一名医生取消了免疫接种的命令。这会造成错误的结果，因为我为没有代码的图表(IMM94，和IMM97)提供了故障安全。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行中断。我创建了一个相当漂亮的字符串裁剪函数来完成大部分这类工作，但是它与找出断线有问题，因为有那么多。我使用的测试用例是：

评估和计划：

1. 需要预防接种和接种流感疫苗(选项卡)：流感、高剂量(65+) (新)
2. 糖尿病前期在减肥和运动方面的工作
3. 骨质减少运动与vit d和b 12

这上面和下面都有很多东西，但我想从我的“取消：”字串中提取所有实例，到后面的第一行中断。

这就是我正在尝试的，还有我的功能。它不起作用，实际上，它复制了以"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

条子“取消：”以及剪贴板中每一行后面的任何文本

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

示例输出：

alpha
beta 
delta
gamma omega 
theta

Answer 2

弄明白了。我将评论它，以帮助任何有这个问题的人。我喜欢你通常能有多紧凑的功能，但这件作品确实需要比我想要的多几行。如果有人能把它降到更低的程度，并且让它仍然有效，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
    }
}