Replace string to the left of value and to the right of quote character-CodePudding

I have a text file content.txt:

Some other text 1
"one" : "Text To Replace1:/Text To Stay.133" 
Some other text 2
"five" : "Text To Change2:/Another Text To Stay.50" 
Some other text 5

I came up with the following script:

$SRCFile = "K:\content.txt"
$DSTFile = "K:\result.txt"
$Text2Replace = "YabaDaba.du:/"

get-content $SRCFile |
ForEach-Object { $_ -replace ".*:\/", $Text2Replace } | Out-File $DSTFile

It works almost okay, but it selects the entire line to the left of the ":/" string. I want it only to select the text to the previous quotation mark (excluding it):

What regex value should I use to point the above script to select only the text up to previous quotation mark? I've been trying Regex101.com, especially LookBehind, but I couldn't come up with any idea.

CodePudding user response：

(?<=. : ").*:\/ might do what you're after. In this case you can also read the file as single multi-line string (hence the use of -Raw in the code) and the use of the (?m) flag (Multiline mode).

See https://regex101.com/r/3CXaOI/1 for details.

$Text2Replace = "YabaDaba.du:/"

(Get-Content $SRCFile -Raw) -replace '(?m)(?<=. : ").*:\/', $Text2Replace |
    Out-File $DSTFile