I have a table column that I need to get from databricks whatever appears between the 15th and 16th appearance of the character # as follows the following example:
Column
1234##E#A#1234#01/01/4500#X#*ABCDE#7#1##N#N#N#0#Z.POIUS.LKJS_20200103#0#
Results
Z.POIUS.LKJS_20200103
how can I do this?
select reverse(substring_index(reverse(substring_index('1234##E#A#1234#01/01/4500#X#*ABCDE#7#1##N#N#N#0#Z.POIUS.LKJS_20200103#0#', '#', 16)),'#', 1))
You can just split the string and get the 15th element, eg something like this:
%sql
SELECT *,
regexp_extract( yourCol, '(?:[^#]*(#)){15}(.[^#]+)', 2 ) xregex,
split( yourCol, '#' )[15] AS xsplit
FROM tmp
I was experimenting with regex which may be appropriate for some cases too. My results:
Related
I have a column data like this in 2 formats
1)"/abc/testapp/v1?FirstName=username&Lastname=test123"
2)"/abc/testapp/v1?FirstName=username"
I want to retrieve the output as "/abc/testapp/v1?FirstName=username" and strip out the data starting with "&Lastname" and ending with "".The idea is to remove the Lastname with its value.
But if the data doesn't contain "&Lastname" then it should also work fine as per the second scenario
The value for Lastname shown in the example is "test123" but in general this will be dynamic
I have started with regexp_replace but i am able to replace "&Lastname" but not its value.
select regexp_replace("/abc/testapp/v1?FirstName=username&Lastname=test123&type=en_US","&Lastname","");
Can someone please help here how i can achieve both these with a single hive query?
Use split function:
with your_data as (--Use your table instead of this example
select stack (2,
"/abc/testapp/v1?FirstName=username&Lastname=test123",
"/abc/testapp/v1?FirstName=username"
) as str
)
select split(str,'&')[0] from your_data;
Result:
_c0
/abc/testapp/v1?FirstName=username
/abc/testapp/v1?FirstName=username
Or use '&Lastname' pattern for split:
select split(str,'&Lastname')[0] from your_data;
It will allow something else with & except starting with &Lastname
for both queries with or without last name its working in this way using split for hive no need for any table to select you can directly execute the function like select functionname
select
split("/abc/testapp/v1FirstName=username&Lastname=test123",'&')[0]
select
split("/abc/testapp/v1FirstName=username",'&')[0]
Result :
_c0
/abc/testapp/v1FirstName=username
you can make a single query :
select
split("/abc/testapp/v1FirstName=username&Lastname=test123",'&')[0],
split("/abc/testapp/v1FirstName=username",'&')[0]
_c0 _c1
/abc/testapp/v1FirstName=username /abc/testapp/v1FirstName=username
Input string: ["1189-13627273","89-13706681","118-13708388"]
Expected Output: ["14013627273","14013706681","14013708388"]
What I am trying to achieve is to replace any numbers till the '-' for each item with hard coded text like '140'
SELECT replace(value_to_replace, '-', '140')
FROM (
VALUES ('1189-13627273-77'), ('89-13706681'), ('118-13708388')
) t(value_to_replace);
check this
I found the right way to achieve that using the below regular expression.
SELECT REGEXP_REPLACE (string_to_change, '\\"[0-9]+\\-', '140')
You don't need a regexp for this, it's as easy as concatenation of 140 and the substring from - (or the second part when you split by -)
select '140'||substring('89-13706681' from position('-' in '89-13706681')+1 for 1000)
select '140'||split_part('89-13706681','-',2)
also, it's important to consider if you might have instances that don't contain - and what would be the output in this case
Use regexp_replace(text,text,text) function to do so giving the pattern to match and replacement string.
First argument is the value to be replaced, second is the POSIX regular expression and third is a replacement text.
Example
SELECT regexp_replace('1189-13627273', '.*-', '140');
Output: 14013627273
Sample data set query
SELECT regexp_replace(value_to_replace, '.*-', '140')
FROM (
VALUES ('1189-13627273'), ('89-13706681'), ('118-13708388')
) t(value_to_replace);
Caution! Pattern .*- will replace every character until it finds last occurence of - with text 140.
I have a column with assignment numbers like - 11827,27266,91717,09818-2,726252-3,8716151-0,827272,18181
Now i am selecting the records like
select assignment_number from table;
But now i want that the column detail is retreived in such a way that numbers are only retrieved without -2 -3 etc like
726252-3---> 726252 8716151-0-->8716151
I know i can use regex for this but i do not know how to use it
This will select everthing before the character -:
^([^-]+)
From 726252-3 will match 726252
You would use regexp() substr:
select regexp_substr(assignmentnumber, '[0-9]+')
This will return the first string of numbers encountered in the string.
I run the following query:
select * from my_temp_table
And get this output:
PNRP1-109/RT
PNRP1-200-16
PNRP1-209/PG
013555366-IT
How can I alter my query to strip the last two characters from each value?
Use the SUBSTR() function.
SELECT SUBSTR(my_column, 1, LENGTH(my_column) - 2) FROM my_table;
Another way using a regular expression:
select regexp_replace('PNRP1-109/RT', '^(.*).{2}$', '\1') from dual;
This replaces your string with group 1 from the regular expression, where group 1 (inside of the parens) includes the set of characters after the beginning of the line, not including the 2 characters just before the end of the line.
While not as simple for your example, arguably more powerful.
Lets say I have following string: 'product=1627;color=45;size=7' in some field of the table.
I want to query for the color and get 45.
With this query:
SELECT REGEXP_SUBSTR('product=1627;color=45;size=7', 'color\=([^;]+);?') "colorID"
FROM DUAL;
I get :
colorID
---------
color=45;
1 row selected
.
Is it possible to get part of the matched string - 45 for this example?
One way to do it is with REGEXP_REPLACE. You need to define the whole string as a regex pattern and then use just the element you want as the replace string. In this example the ColorID is the third pattern in the entire string
SELECT REGEXP_REPLACE('product=1627;color=45;size=7'
, '(.*)(color\=)([^;]+);?(.*)'
, '\3') "colorID"
FROM DUAL;
It is possible there may be less clunky regex solutions, but this one definitely works. Here's a SQL Fiddle.
Try something like this:
SELECT REGEXP_SUBSTR(REGEXP_SUBSTR('product=1627;color=45;size=7', 'color\=([^;]+);?'), '[[:digit:]]+') "colorID"
FROM DUAL;
From Oracle 11g onwards we can specify capture groups in REGEXP_SUBSTR.
SELECT REGEXP_SUBSTR('product=1627;color=45;size=7', 'color=(\d+);', 1, 1, 'i', 1) "colorID"
FROM DUAL;