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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.

Regards
Thomos

ASKER CERTIFIED SOLUTION

slightwv (䄆 Netminder)

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Swadhin Ray

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_thomos

ASKER

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
begin
return replace(replace(replace(replace(replace(replace(replace(p_column_id, '\', '\\'), '"','\"'),
CHR(9),'\t'),CHR(8),'\b'),CHR(13),'\r'),CHR(12),'\f'),CHR(10),'\n');
end rep;

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Regards
Thomos

slightwv (䄆 Netminder)

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_thomos

ASKER

THANKS A LOT