I have a postgres table which contains rows that each hold multiple lines of text (split by new lines), for example...
The table name is formats, column is called format, an example format (1 table row) would look like the following:
list1=text1;
list2=text2;
list3=text3;
etc etc
I would like a way to identify the list2 string and then append additional text to the end of the same line.
So the outcome would be:
list1=text1;
list2=test2;additionaltext
list3=text3;
I have tried the below to try and pull in the 'capture string' into the replace string but have been unsuccessful so far.
regexp_replace(format, 'list2=.*', '\1 additionaltext','n');
To capture a pattern, you must enclose it in parenthesis.
regexp_replace(format, '(list2=.*)', '\1additionaltext', 'n')
Related
I have a table that has some rows with normal JSON and some with escaped values in the JSON field (backslashes)
id
obj
1
{"is_from_shopping_bag":true,"products":[{"price":{"amount":"18.00","currency":"USD","offset":100,"amount_with_offset":"1800"},"product_id":"1234","quantity":1}],"source":"cart"}
2
{"is_from_shopping_bag":"","products":"[{\ "product_id\ ":\ "2345\ ",\ "price\ ":{\ "currency\ ":\ "USD\ ",\ "amount\ ":\ "140.00\ ",\ "offset\ ":100},\ "quantity\ ":1}]"}
(Note: I needed to include a space after the backslashes in the above table so that they would show up in the github generated markdown table -- my actual table does not include those spaces between the backslash and the quote character)
I am doing a sql query in Hive to get the 'currency' field.
Currently I can run
SELECT
id,
JSON_EXTRACT(obj, '$.products[0].price.currency')
FROM my_table
Which will give me the correct output for the first row, but gives me a NULL in the second row
id
obj
1
"USD"
2
NULL
What is the best way to get currency field from the second row? Is there a way to clean up the field and remove the backslashes before trying to JSON_EXTRACT the relevant data?
I could use REPLACE to swap the '\ ' for '', but is that the most efficient method?
Replace \" with " using regexp_replace like this:
regexp_replace(obj,'\\\\"','"')
I would like to extract the string using where clause in SAP HANA.For an example,these are 3 strings for name column.
123._SYS_BIC.meag.app.qthor.cidwh_eingangsschicht.backend.dblayer.l2.checks/MasterData_Holdings.
153._SYS_BIC.meag.app.qthor.centralAdministration.backend.dblayer.l2.checks/AuditAndSecurities.
meag.app.qthor.centralAdministration.backend.dblayer.l2.checks/GeneralLedger
After filter the name column using where clause, output in the name column would be shown only the last portion of the string. So, output will be like this. That means whatever we have, just remove from the beginning till '/'.
"MasterData_Holdings"
"AuditAndSecurities"
"GeneralLedger"
You can try using the REPLACE_REGEXPR
I'm not familiar myself with Hana but the function is pretty straight forward and it should be:
select REPLACE_REGEXPR('.+/(.+)' IN fieldName WITH '\1' OCCURRENCE ALL) as field
...
where
... -- your filter
Be aware that this regex '.+/(.+)' will eat everything until the last / so for instance if you have ....checks/MasterData_Holdings/Something it will return only Something
Database I'm using: https://uploadfiles.io/72wph
select acnum, field.fieldnum, title, descrip
from field, interest
where field.fieldnum=interest.fieldnum and trim(ID) like 'B.1._';
What will the output be from the above query?
Does trim(ID) like 'B.1._' mean that it will only select items from B.1._ column?
trim removes spaces at the beginning and end.
"_" would allow representing any character. Hence query select any row that starts with "B.1."
For eg.
'B.1.0'
'B.1.9'
'B.1.A'
'B.1.Z'
etc
Optional Wildcard characters allowed in like are % (percent) and _ (underscore).
A % matches any string with zero or more characters.
An _ matches any single character.
I don't know about the DB you are using but trim usually remove spaces around the argument you give to it.
The ID is trimmed to be sure to compare the ID without any white-space around it.
About your second question, Only the ROWS with an ID like 'B.1.' will be selected.
SQL like
SQL WHERE
Using Oracle db,
Select name from name_table where name like 'abc%';
returns one row with value "abc, cd" but when I do a select query with a comma before % in my like query, it fails to return any value.
Select name from name_table where name like 'abc,%';
returns no row. How can I handle a comma before % in the like query?
Example:
Database has "Sam, Smith" in the name column when the like has "Sam%" it returns one row, when i do "Sam,%" it doesn't return any row
NOT AN ANSWER but posting it as one since I can't format in a comment.
Look at this and use DUMP() on your own machine... see if this helps.
SQL> select dump('Smith, Stan') from dual;
DUMP('SMITH,STAN')
-----------------------------------------------------
Typ=96 Len=11: 83,109,105,116,104,44,32,83,116,97,110
If you count, the string is 11 characters (including the comma and the space). The comma is character 44, and the space is character 32. If you look at YOUR string and you don't see 44 where the comma should be, you will know that's the problem. You could then let us know what you see there (just for that character, I understand posting "Leno, Jay" would be a violation of privacy).
Also, make sure you don't have any extra characters (perhaps non-printable ones!) right before the comma. Just compare the two strings you are using as inputs and see where the differences may be.
The regex I want to use is: ^(?=.*[,])(,?)ABC(,?)$
What I want to get out is:
^ // start
(?=.*[,]) // contains at least one comma (,)
(,?)ABC(,?) // The comma is either in the beginning or in the end of the string "ABC"
$ // end
Of course ABC is ought to be a variable based on my search term.
So if ABC = 'abc' then ",abc", "abc,", ",abc," will match but not "abc" or "abcd"
Better way to do this is also welcome.
The value in the record looks like "abc,def,ghi,ab,cde..." and I need to find out if it contains my element (i.e. 'abc'). I cannot change the data structure. We can assume that in no case the record will contain only one sub-value, so it is correct to assume that there always is a comma in the value.
If you want to know if a comma delimited string contains abc, then I think like is the easiest method in any database:
where ',' + col + ',' like '%,abc,%'