Output referral and special characters

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1

Here is some test code to help put my question in context:

	OutputTo "_var0"
	Say "{\"key0\":\"Scripting is \\\\\"hard\\\\\" work\","
	SayNewline
	Say "\"key1\":\"Scripting is \\\\\"HARD\\\\\" work\"}"
	RegexpReplace "^.*" "%_var0%"
	json "ON" "current"
	json_select "key1"
	SayRest

With this I can get "Scripting is "HARD" work" in _var0 output. What's interesting is the use of RegexpReplace to load the output's value:

Script-Line    : ???
Command        : regexpreplace
Parameter 1    : >^.*<
Parameter 2    : >{"key0":"Scripting is \\"hard\\" work",
"key1":"Scripting is \\"HARD\\" work"}<

Output         : >{"key0":"Scripting is \\"hard\\" work",
"key1":"Scripting is \\"HARD\\" work"}<

Line and position:
{"key0":"Scripting is \"hard\" work",
"key1":"Scripting is \"HARD\" work"}

Notice that in ["HARD"] the double quote " special character -as part of the buffer string- is escaped, BUT inside the output memory there are 2 backslashes \\, and when loaded into the buffer there is only one \.

As I understand, the escape backslash for " is 'consumed' in order to preserve the special character, when loaded into the buffer.

My question is: is there a way to keep *ALL* characters when calling an output into the current buffer i.e. keeping the string exactly as it sits in the output's memory?

Thank you in advance.

2

[After a few hours of testing]: Yes, there is!
Instead of a RegexReplace, a Replace command can do the same.... with some creativity:

RegexpReplace "^.*" "#"
Replace "#" "%_var0%"

This works. I used "#", but any suitable uncommon character will suffice.