Powershell: join two lines in a array

Question

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"
       "@

I am trying to write a powershell code which searches for lines that start with " but does not end with ". If it matches the condition next line is joined to the matched line with a comma delimiter. Ex: second line matches the condition, so third line is joined with the second line with a comma in between. like the one below

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

So far I have this with me. But it only adds a , to EOL but does not join it. Please let me know what am I missing

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

Hm, this may be an x-y problem. Can you explain why you want to do that? — Martin Brandl
– Martin Brandl, Commented Nov 8, 2016 at 14:13
I am trying search lines that does not end with ". if it matches the case, then I need to join the matched line with next line (with a comma in between) to make it as single line. And this variable result has multiple lines which does not end with " I have highlighted the desired output in second code section. — clicks010
– clicks010, Commented Nov 8, 2016 at 14:32

Mathias R. Jessen · Accepted Answer · 2016-11-08 15:20:56Z

3

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)}

No, you have a multi-line string - there's a difference!

You can use a regular expression pattern to match linebreaks surrounded by optional whitespace, but no " characters:

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

The pattern consists of:

(?<!") - no preceding " (negative lookbehind assertion)
\s* - 0 or more whitespace characters
\r?\n - line break (CRLF or LF)
\s* - 0 or more whitespace characters
(?!") - no following " (negative lookahead assertion)

And finally replaces the thing with a ,

edited Nov 8, 2016 at 15:20

answered Nov 8, 2016 at 14:27

Mathias R. Jessen

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2 Comments

Djee Over a year ago

Brilliant. Very good reminder of us of negative lookahead and lookbehind assertions in regular expressions

Mathias R. Jessen Over a year ago

@clicks010 you're most welcome. Please consider accepting my answer if it solves your problem, by clicking the checkmark on the left

Djee · Accepted Answer · 2016-11-08 14:53:42Z

0

I don't know if this is a real XY problem but the question is fuzzy (confusion between a multi-line string and an array for example). Solution from Mathias is fine. However I think the original poster, according to the expected result, missed the correct syntax for the @""@ Here string which is very verbatim: all leading spaces were not intentional. Hence this:

$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

which brings this:

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

answered Nov 8, 2016 at 14:53

Djee

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Comments

Esperento57 · Accepted Answer · 2016-11-08 19:56:34Z

0

with multiple replace

$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"

answered Nov 8, 2016 at 19:56

Esperento57

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