How to make a generic function for json data in oracle ?

Hi expert,
on below query i am handling json. actually exactly i dont know which  special character handled by json and which one is not handled by json. so when i was entering data to this particular field , i have observed that these many characters are not supported by json , so i replaced this character like below. but exactly i donot know which is supported by json and which is not supported json. so i want a generic function , by which all special characters can handle by json.
   replace(replace(replace(emp_name,'\' , '\\\\' ),'''' , '\''' ),'"' , '\\"'  ) ;

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for above things is working fine but problem is i dont know  how  many specail charecters are there like this which is not handle by  json. so i want to make a generic function with all special characters which can handle by JSON.

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slightwv (䄆 Netminder) Commented:
I don't think that will work.

Those characters only need escaping in specific instances not every occurrence.

As far as the characters that need escaping, go to the source:

    \u four-hex-digits

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Swadhin RaySenior Technical Engineer Commented:
Can you check "utl_url.escape " and "utl_url.unescape"  to escape.
I am not 100% sure but this escape the URL 's but still can be used on any varchar2(4000) characters.
deve_thomosAuthor Commented:
below is my code
 I have handle for above json charecters but i am not getting idea how to handle hexadecimal digits(/U).
function rep(p_column_id  varchar2) return varchar2 as
return replace(replace(replace(replace(replace(replace(replace(p_column_id, '\', '\\'), '"','\"'),
end rep;

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slightwv (䄆 Netminder) Commented:
I don't know a lot about JSON.

I can help with the regular expression once I understand the patterns.

What do they look like in JSON?
deve_thomosAuthor Commented:
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