PowerShell, parsing, problem using a specific string with -match/-like and -split

Question

I have a variable that is filled with several lines of text and I am trying to parse the data from it. Now around the middle of the text is a specific string "Reference(s) :" and I need to get everything from above this specific string. However every way I have tried has failed.

I tried making it a delimiter

$Var.split("Reference(s) :")

I tried the below 2 options just to try to capture that one line (because if I can do this, then I know I can pull everything before it).

$Var.split("`n") | Where-Object {$_ -match "Reference(s) :"}

and

$Var.split("`n") | Where-Object {$_ -like "*Reference(s) :*"}

and I've tried some if statements (Where $_ is a single line of text)

If ($_ -like "*Reference(s) :*") {some logic}

I cannot just match "Reference" because that word appears elsewhere in the text....and I am needing this to process several instances of this text.

I think the problem has to do with the parenthesis, the space, and the colon (special characters). I did try preceding each special character with a ` but that did not seem to work.

Anyone have any ideas? There has to be a way to match special characters, I just haven't found it yet.

Answer 1

If $var is truly a single string, you can use -split at your reference point and then retrieve the first split string ([0]). This will retrieve everything from the start of the string until the split point.

($Var -split 'Reference\(s\) :')[0]

Since -split uses regex matching, you must backslash escape regex special characters to match them literally. Here ( and ) are special.

In the future, you can just process your match string using [regex]::Escape('String'), and it will do all the escaping for you.

---

If you want the line just before the reference point, you can convert your string into an array. Then return the line above the matching line.

foreach ($line in ($Var -split '\r?\n')) {
    if ($line -match 'Reference\(s\) :') { 
        $lastLine
    } else { 
        $lastLine = $line
    } 
}