I am trying to extract the following from the text field using Regrex in Oracle.
For example
"This is example,
and this really a example :h,j,j,j,j,
l //Updated question , as this letter is on the next line
now this is a disease:yes"
I am expecting a result as h,j,j,j,j,l, but if I use
REGEXP_SUBSTR(text_field,'example :[^:]+,') AS Result
I am getting example:h,j,j,j,j
But I am not getting the last letter 'l' like above and I am guessing that's because it's on the next line.Also, if I want the string "disease:yes" only, that will be so helpful as well. Thank you much!
The result you are getting is because your pattern includes the word 'example' and ends with a comma, leaving out the ending 'l'. Try this form instead. Note the example is shown using a Common table Expression (CTE). The WITH statement creates the table called tbl which just sets up test data, kind of like a temp table. This is also a great way to set up data when asking a question. This form of the REGEXP_SUBSTR() function uses a captured group, which is the set of characters after the string 'example:' until the end of that line in the multi-line field. From this you should be able to get the other string you are after. Give it a go.
WITH tbl(text_field) AS (
SELECT 'This is example,
and this really a example :h,j,j,j,j,l
now this is a disease:yes' FROM dual
)
SELECT REGEXP_SUBSTR(text_field,'example :(.*)', 1, 1, NULL, 1) AS Result
FROM tbl;
RESULT
-----------
h,j,j,j,j,l
1 row selected.
Edit based on new info. Since that last letter could be on it's own line, you'll need to allow for the newline. Use the 'n' flag to REGEXP_REPLACE() which allows the newline to match in the usage of the dot (match any character) symbol in regex. We switch to REGEXP_REPLACE as we'll need to return multiple capture groups. Here the WITH sets up 2 rows, one with an embedded newline in the data and one without. The capture groups are (going left to right) 1-the data after "example :" and ending in a comma, 2-the optional newline and 3-the next single character. Then replace the entire data with captured groups 1 and 3 (leaving out the newline).
NOTE this is very specific to the case of only 1 character on the following line.
WITH tbl(ID, text_field) AS (
SELECT 1, 'This is example,
and this really a example :h,j,j,j,j,
l
now this is a disease:yes' FROM dual UNION ALL
SELECT 2, 'This is example,
and this really a example :h,j,j,j,j,l
now this is a disease:yes' FROM dual
)
SELECT ID,
REGEXP_REPLACE(text_field, '.*example :(.*,)('||CHR(10)||')?(.).*', '\1\3', 1, 1, 'n') AS Result
FROM tbl;
ID RESULT
---------- ------------
1 h,j,j,j,j,l
2 h,j,j,j,j,l
2 rows selected.
I'm using the following code to get a user's recovery_token and store it in a variable:
Connect To Database psycopg2 ${DB_NAME}
... ${DB_USER_NAME}
... ${DB_USER_PASSWORD}
... ${DB_HOST}
... ${DB_PORT}
${RECOVERY_TOKEN}= Query select recovery_token FROM public."system_user" where document_number like '57136570514'
Looking at the log, the recovery_token is being saved as follows:
${RECOVERY_TOKEN} = [('eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJpZCI6ImU3ZGM4MmNjLTliMGQtNDc3OC1hMzM0LWEyNjM4MDU1Mzk1MSIsImlhdCI6MTYyMzE5NjM4NSwiZXhwIjoxNjIzMTk2NDQ1fQ.mdsrQlgaWUol02tZO8dXlL3KEwY6kqwj5T7gfRDYVfU',)]
But I need what is saved in the variable ${RECOVERY_TOKEN} to be just the token, without the special characters [('',)]
${RECOVERY_TOKEN} = eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJpZCI6ImU3ZGM4MmNjLTliMGQtNDc3OC1hMzM0LWEyNjM4MDU1Mzk1MSIsImlhdCI6MTYyMzE5NjM4NSwiZXhwIjoxNjIzMTk2NDQ1fQ.mdsrQlgaWUol02tZO8dXlL3KEwY6kqwj5T7gfRDYVfU
Is there any way I can remove the special characters?
Thanks in advance!!
The returned value is a list of tuples, a two-dimensional matrix (e.g. a table); if you have queried for 3 columns for example, the inner tuple would have 3 members. And if there were 5 records that match it, the list would have 5 tuples in it.
Thus to get the value you are after, get it from the matrix by its indexes (which are 0-based, e.g. the first element is with index "0"):
${RECOVERY_TOKEN}= Set Variable ${RECOVERY_TOKEN[0][0]}
I'm having a problem that I thought to be rather common, but trying to look it up in the "Oracle Database 10g2 Utilities_b14215.pdf" didn't help. After that I've surfed through numerous threads but no luck so far.
I'm having a tab-delimited file (x'09') e. g. name, userid, persnr. The values for the userids begin with either P, R or T e. g. P2198, P2199, R7288, T1229.
I want to load only the records with userids beginning with P.
Isolating a single record with a controlfile like this works splendidly:
OPTIONS (SKIP=1)
LOAD DATA
INFILE UserlistLoader.dat
APPEND
INTO TABLE Z_USERLIST
WHEN USERID = 'P2198'
FIELDS TERMINATED BY x'09'
TRAILING NULLCOLS
(name, userid, persnr)
But every attempt at using SUBSTR in the when-clause fails.
This:
OPTIONS (SKIP=1)
LOAD DATA
INFILE UserlistLoader.dat
APPEND
INTO TABLE Z_USERLIST
WHEN SUBSTR(USERID, 1, 1) = 'P'
FIELDS TERMINATED BY x'09'
TRAILING NULLCOLS
(name, userid, persnr)
ends in an SQL*Loader-350: Syntax-Error.
This
OPTIONS (SKIP=1)
LOAD DATA
INFILE UserlistLoader.dat
APPEND
INTO TABLE Z_USERLIST
WHEN "SUBSTR(:USERID, 1, 1)" = 'P'
FIELDS TERMINATED BY x'09'
TRAILING NULLCOLS
(name, userid, persnr)
ends in an SQL*Loader-403: Referenced column USERID not present in table Z_USERLIST.
But IT IS PRESENT - as the first example proves. I've found that the column should be preceded by : but that obviously isn't the issue.
What am I doing wrong?
From SQL Loader docs the left-hand side of a WHEN condition can only be a full field name e.g. USERID or a position spec e.g. (3:5).
The docs aren't very clear though on what is allowed - e.g. can LIKE be used as the operator?
USERID LIKE 'P%'
I strongly suspect it can't though.
I would load the entire file into a staging table that matches the file layout, then run a procedure that inserts the rows you want from there into the production table. That is a more common way to handle loads with criteria like this without having to edit source data.
If you can preprocess the source file, move the userid to the first field or copy the first letter of the userid to it's own field and construct the WHEN like this so sqlldr looks at the first position (this will cause sqlldr to return non-zero though, as not all rows meet WHEN clause criteria):
WHEN (1) = 'P'
I have read the docs for MariaDB's REGEX_REPLACE but cannot get my query to work. I am storing links in a column, link and want to change the end of the link:
From www.example.com/<code> to www.example.com/#/results/<code> where <code> is some hexidecimal hash, e.g. 55770abb384c06ee00e0c579. What I am trying is:
SELECT REGEX_REPLACE("link", "www\\.example\\.com\\/(.*)", "www\\.example\\.com\\/#\\/results\\/\\1");
The result is:
Showing rows 0 - 0.
I wasn't able to figure out what the first argument was--the documentation says "subject". Turns out it's just the column name. So this works:
UPDATE my_table
SET my_link = REGEXP_REPLACE(
my_link,
"http:\\/\\/www\\.example\\.com\\/(.*)",
"http:\\/\\/www\\.example\\.com\\/#\\/results\\/\\1")
WHERE my_link IS NOT NULL
I have a large text file containing the following records:
024567808 name date etc..
0376567 dfu ugig etc..
0388888 dtg hii etc..
032357 tuth gug etc..
025789 gkh kjkjk etc..
Using the grep command, I am able to retrieve the line containing a keyword. How do I retrieve the next records if it starts with 03?
So for example I search for name, and it returns the first row. I want this record along with all the 03 rows until the next 02 record.
output should be:
024567808 name date etc..
0376567 dfu ugig etc..
0388888 dtg hii etc..
032357 tuth gug etc..
I think grep is not powerful enough for this sort of task. Perhaps awk could do it. What is the best way to do this kind of thing?
This might be what you want:
awk '/^02/{f=0} /name/{f=1} f' file
It all depends if you want "name" to only occur in a specific position on a line and whether or not you want "name" to only match if its not part of a longer word, e.g. do you want a search for "Jo" to match on "John"?