powershell - Powershell:将两行连接成一个数组

Question

我得到了一个具有值的数组(每个“文本”后都有新行)

$result=@"
        "first","line","of","text"
        "second","line","of","text
        third","line","of","text"
        "fourth","line","of","text"
       "@

我正在尝试编写一个Powershell代码，该代码搜索以“开头但不以”结尾的行。如果符合条件，则下一行用逗号定界符连接到匹配的行。例如:第二行符合条件，因此第三行与第二行之间用逗号隔开。像下面的那个

"first","line","of","text"
"second","line","of","text,third","line","of","text"
"fourth","line","of","text"

到目前为止，我已经有了这个。但它只会在EOL中添加，但不会加入。请让我知道我在想什么

$result=@()
foreach ($item in $result){
if ($item -notlike '*"') 
{ $result+=$item+"," } 
else
{ $result+=$item}
}

Answer 1

I got a array which has the value(new line after every "text")

$result=@"
        "first","line","of","text"
        "second","line","of","text
        third","line","of","text"
        "fourth","line","of","text"
"@

_{* (syntax error fixed)}

不，您有多行字符串-有所不同!

您可以使用正则表达式模式来匹配由可选空白包围的换行符，但不能包含 "字符:

$result = $result -replace '(?<!")\s*\r?\n\s*(?!")', ','

该模式包括:

(?<!")-没有在前的"(负向后置断言)

\s*-0个或更多空白字符

\r?\n-换行符(CRLF或LF)

\s*-0个或更多空白字符

(?!")-无后续"(负超前断言)

最后用 ,代替