How do I parse a string in Oracle?
How can I parse the value of "request" in the following string in Oracle?
<!-- accountId="123" activity="add" 开发者_开发技巧request="add user" -->
The size and the position of the request is random.
You can use regular expressions to find this:
regexp_replace(str, '.*request="([^"]*)".*', '\1')
Use INSTR(givenstring, stringchartosearch,start_position)
to find the position of 'request="' and to find the position of the closing '"'.
Then use substr(string, starting_position, length)
.
You'd use a combination of instr
and substr
THIS EXAMPLE IS FOR EXAMPLE PURPOSES ONLY. DO NOT USE IT IN PRODUCTION CODE AS IT IS NOT VERY CLEAN.
substr(my_str,
-- find request=" then get index of next char.
instr(my_str, 'request="') + 9,
-- This is the second " after request. It does not allow for escapes
instr(substr(my_str,instr(my_str, 'request="')), 2))
Below is my tested variations from cwallenpoole and Craig. For the regexp - note that if "request=" does not exist, the result will be the entire string. user349433 was partly there too, a space before "request=" in the search works just as well:
SET serveroutput ON
DECLARE
l_string VARCHAR2(100) := '<!-- accountId="123" activity="add" request="add user" -->';
l_result_from_substr VARCHAR2(50);
l_result_from_regexp VARCHAR2(50);
BEGIN
SELECT SUBSTR(l_string, instr(l_string, 'request="') + 9, instr(SUBSTR(l_string,instr(l_string, 'request="')), '"', 2)-1),
regexp_replace(l_string, '.* request="([^"]*)".*', '\1')
INTO l_result_from_substr,
l_result_from_regexp
FROM dual;
dbms_output.put_line('Result from substr: '||l_result_from_substr);
dbms_output.put_line('Result from regexp: '||l_result_from_regexp);
END;
/
Please note the equal sign "=" does not necessarily have to come immediately after the request variable in the assignment. As such, it is not entirely correct to search for "request=". You should create a basic finite state machine using INSTR to first find "request", then find "=", ...
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