Find and replace a string containing both double quotes and brackets - powershell

Question

Let's say I have a test file named testfile.txt containing the below line:
one (two) "three"
I want to use PowerShell to say that if the entire string exists, place a line directly underneath it with the value:
four (five) "six"
(Notice that it includes both spaces, brackets and double quotes. This is important, as the problem I am having is I think with escaping the brackets and double quotes).
So the result would be:
one (two) "three"
four (five) "six"
I thought the easiest way of doing it would be to say that if the first string is found, replace it with the first string itself again, and the new string forming a new line included in the same command. I had difficulty putting the strings in line so I tried using a herestring variable whereby an entire text block with formatting is read. It still does not parse the full string with quotes into the pipeline. I'm new to powershell so don't hold back if you see something stupid.
$herestring1 = #"
one (two) "three"
"#
$herestring2 = #"
one (two) "three"
four (five) "six"
"#
if((Get-Content testfile.txt) | select-string $herestring1) {
"Match found - replacing string"
(Get-Content testfile.txt) | ForEach-Object { $_ -replace $herestring1,$herestring2 } | Set-Content ./testfile.txt
"Replaced string successfully"
}
else {
"No match found"}
The above just gives "No match found" every time. This is because it does not find the first string in the file.
I have tried variations using backtick [ ` ] and doubling quotes to try to escape, but I thought the point in a here string was that it should parse the text block including all formatting so I should not have to.
If I change the file to contain only:
one two three
and then change the herestring accordingly to:
$herestring1 = #"
one two three
"#
$herestring2 = #"
one two three
four five six
"#
Then it works ok and I get the string replaced as I want.

Answer 1

As Martin points out, you can use -SimpleMatch with Select-String to avoid parsing it as a regular expression.
But -replace will still be using a regex.
You can escape the pattern for RegEx using [RegEx]::Escape():
$herestring1 = #"
one (two) "three"
"#
$herestring2 = #"
one (two) "three"
four (five) "six"
"#
$pattern1 = [RegEx]::Escape($herestring1)
if((Get-Content testfile.txt) | select-string $pattern1) {
"Match found - replacing string"
(Get-Content testfile.txt) | ForEach-Object { $_ -replace $pattern1,$herestring2 } | Set-Content ./testfile.txt
"Replaced string successfully"
}
else {
"No match found"}
Regular expressions interpret parentheses () (what you are calling brackets) as special. By default, spaces are not special, but they can be with certain regex options. Double quotes are no problem.
In regex, the escape character is backslash \, and this is independent of any escaping you do for the PowerShell parser using backtick `.
[RegEx]::Escape() will ensure anything special to regex is escaped so that a regex pattern will interpret it as literal, so your pattern will end up looking like this: one\ \(two\)\ "three"

Answer 2

Just use the Select-String cmdlet with the -SimpleMatch switch:
# ....
if((Get-Content testfile.txt) | select-string -SimpleMatch $herestring1) {
# ....
-SimpleMatch
Indicates that the cmdlet uses a simple match rather than a regular
expression match. In a simple match, Select-String searches the input
for the text in the Pattern parameter. It does not interpret the value
of the Pattern parameter as a regular expression statement.
Source.