Question

我得到了一个具有值的数组（每个“text”之后的新行）

$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，用于搜索以“但不以...结尾”开头的行。如果匹配条件，则下一行使用逗号分隔符连接到匹配的行。例如：第二行与条件匹配，因此第三行与第二行连接，中间使用逗号。喜欢下面的那个

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

到目前为止，我和我有这个。但它只为EOL添加了一个，但没有加入它。请让我知道我错过了什么

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

Answer 1

我得到了一个具有值的数组（每个＆＃34之后的新行;文本＆＃34;）
$result=@"
        "first","line","of","text"
        "second","line","of","text
        third","line","of","text"
        "fourth","line","of","text"
"@
_{* （语法错误已修复）}

不，你有一个多行字符串 - 这有区别！

您可以使用正则表达式模式来匹配可选空格所包围的换行符，但不包含"个字符：

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

该模式包括：

(?<!") - 没有先前的"（负面的背后断言）
\s* - 0个或更多空格字符
\r?\n - 换行符（CRLF或LF）
\s* - 0个或更多空格字符
(?!") - 没有关注"（负面前瞻断言）

最后用,

替换该东西

Answer 2

我不知道这是否是一个真正的 XY问题，但问题是模糊的（例如，多行字符串和数组之间的混淆）。 Mathias的解决方案很好。但是我认为根据预期的结果，原始海报错过了@＆＃34;＆＃34; @ Here 字符串的正确语法， 非常逐字 em> ：所有前导空格都是不故意。因此：

$result=@" "first","line","of","text" "second","line","of","text third","line","of","text" "fourth","line","of","text" "@ $result=$result -replace '(?<!")\r?\n(?!")', ',' write-output $result

带来了这个：

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

Answer 3

多次替换

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

(((($result -split "`n") -join """ ") -replace """ """, """`n""") -replace """ ", " ") -replace "\""`n", "`n"

Powershell：在数组中连接两行

3 个答案: