Replace string with pattern keeping the string intact - shell

Question

Would to replace the string either with sed or awk, where it identifies a patter as mentioned below :
Example: It looks for word starting with "XX" and ending with "XX" and replace the word with concatinating with "${hf:" at the start of "XX" and "}" at the end of "XX"
INPUT
CREATE TABLE XX_DB_XX.test_XX_YYYYMMDD_XX
AS
SELECT id
FROM XX_R_DB_XX.usr_XX_YYYYMMDD_XX
WHERE year = XX_YYYY_XX
AND month = XX_MM_XX
AND day = XX_DD_XX;
OUTPUT
CREATE TABLE ${hf:XX_DB_XX}.test_${hf:XX_YYYYMMDD_XX}
AS
SELECT id
FROM ${hf:XX_R_DB_XX}.usr_${hf:XX_YYYYMMDD_XX}
WHERE year = ${hf:XX_YYYY_XX}
AND month = ${hf:XX_MM_XX}
AND day = ${hf:XX_DD_XX};
Tried to replace the pattern matchin, but the issue is , in the output I want replace the $A , to the subsquet "XX_(*)_XX" string available over the input file.
cat test.hql | gawk '{ print gensub(/XX_+[A-Z,_]+_XX/, "${hiveconf:$A}", 1)
}' | gawk '{ print gensub(/XX_+[A-Z]+_XX/, "${hiveconf:$A}", 1) }'
OUTPUT: -> That I recived needs to be updated w.r.t the string available, So how can this be done:
CREATE TABLE ${hiveconf:$A}.test_${hiveconf:$A}
AS
SELECT id
FROM ${hiveconf:$A}.usr_${hiveconf:$A}
WHERE year = ${hiveconf:$A}
AND month = ${hiveconf:$A}
AND day = ${hiveconf:$A};

Answer 1

Following awk may help you on same.
awk '{gsub(/XX_[a-zA-Z]+_XX/,"${hf:&}")} 1' Input_file

Answer 2

That's what sed exists for,
sed 's/XX[[:alnum:]_]*XX/${hf:&}/g' file
[[:alnum:]_] stands for Alpha numeric or underscore. The append * means zero or more times of it in regular expression.

Answer 3

Or you could do
sed 's/\(XX[^'XX']*XX\)/${hf:\1}/g'
in cases where there may be non alphanumeric characters as well in between the XXs.
First an XX is matched after which waits till an XX is found.