How can I find the first word and second word in a string separated by unknown number of spaces in SQL Developer? I need to run a query to get the expected result.
String:
Hello Monkey this is me
Different sentences have different number of spaces between the first and second word and I need a generic query to get the result.
Expected Result:
Hello
Monkey
I have managed to find the first word using substr and instr. However, I do not know how to find the second word due to the unknown number of spaces between the first and second word.
select substr((select ltrim(sentence) from table1),1,
(select (instr((select ltrim(sentence) from table1),' ',1,1)-1)
from table1))
from table1
Since you seem to want them as separate result rows, you could use a simple common table expression to duplicate the rows, once with the full row, then with the first word removed. Then all you have to do is get the first word from each;
WITH cte AS (
SELECT value FROM table1
UNION ALL
SELECT SUBSTR(TRIM(value), INSTR(TRIM(value), ' ')) FROM table1
)
SELECT SUBSTR(TRIM(value), 1, INSTR(TRIM(value), ' ') -1) word
FROM cte
Note that this very simple example assumes that there is a second word, if there isn't, NULL will be returned for both words.
An SQLfiddle to test with.
While Joachim Isaksson's answer is a robust and fast approach, you can also consider splitting the string and selecting from the resulting pieces set. This is just meant as hint for another approach, if your requirements alter (e.g. more than two string pieces).
You could split finally by the regex /[ ]+/, and so getting the words between the blanks.
Find more about splitting here: How do I split a string so I can access item x?
This will strongly depend on the SQL dialect you are using.
Try this with REGEXP_SUBSTR:
SELECT
REGEXP_SUBSTR(sentence,'\w+\s+'),
REGEXP_SUBSTR(sentence,'\s+(\w+)\s'),
REGEXP_SUBSTR(sentence,'\s+(\w+)\s+(\w+)'),
REGEXP_SUBSTR(REGEXP_SUBSTR(sentence,'\s+(\w+)\s+(\w+)'),'\w+$'),
REGEXP_SUBSTR(sentence,'\s+(\w+)\s+$')
FROM table1;
result:
1 2 3 4 5
Hello Monkey Monkey this this is_me
Learn more about REGEXP_SUBSTR reference to Using Regular Expressions With Oracle Database
Test use SqlFiddle: http://sqlfiddle.com/#!4/8e9ef/9
If you only want to get the first and the second word, use REGEXP_INSTR to get second word start position :
SELECT
REGEXP_SUBSTR(sentence,'\w+\s+') AS FIRST,
REGEXP_SUBSTR(sentence,'\w+\s',REGEXP_INSTR(sentence,'\w+\s+')+length(REGEXP_SUBSTR(sentence,'\w+\s+'))) AS SECOND
FROM table1;
Related
I am doing this in Impala or Hive. Basically let say I have a string like this
f-150:aa|f-150:cc|g-210:dd
Each element is separated by the pipe |. Each has prefix f-150 or whatever. I want to be able to remove the prefix and keep only element that matches specific prefix. For example, if the prefix is f-150, I want the final string after regex_replace is
aa|cc
dd is removed because g-210 is different prefix and not match, therefore the whole element is removed.
Any idea how to do this using string expression in one SQL?
Thanks
UPDATE 1
I tried this in Impala:
select regexp_extract('f-150:aa|f-150:cc|g-210:dd','(?:(?:|(\\|))f-150|keep|those):|(?:^|\\|)\\w-\\d{3}:\\w{2}',0);
But got this output:
f-150:aa
In Hive, I got NULL.
The regexyou in question could look like this:
(?:(?:|(\\|))f-150|keep|those):|(?:^|\\|)\\w-\\d{3}:\\w{2}
I have added some pseudo keywords to retain, but I am sure you get the idea:
Wholy match elements that should be dropped but only match the prefix for those that should be retained.
To keep the separator intact, match | at the beginning of an element in group 1 and put it back in the replacement with $1.
Demo
According to the documentation, your query should be written like a Java regex; likewise, this should perform like this code sample in Java.
You could match the values that you want to remove and then replace with an empty string:
f-150:|\|[^:]+:[^|]+$|[^|]+:[^|]+\|
f-150:|\\|[^:]+:[^|]+$|[^|]+:[^|]+\\|
Explanation
f-150: Match literally
| Or
\|[^:]+:[^|]+$ Match a pipe, not a colon one or more times followed by not a pipe one or more times and assert the end of the line
| Or
[^|]+:[^|]+\| Match not a pipe one or more times, a colon followed by matching not a pipe one or more times and then match a pipe
Test with multiple lines and combinations
You may have to loop through the string until the end to get the all the matching sub string. Look ahead syntax is not supported in most sql so above regexp might not be suitable for SQL syntax. For you purpose you can do something like creating a table to loop through just to mimic Oracle's level syntax and join with your table containing the string.
With loop_tab as (
Select 1 loop union all
Select 2 union all
select 3 union all
select 4 union all
select 5),
string_tab as(Select 'f-150:aa|ade|f-150:ce|akg|f-150:bb|'::varchar(40) as str)
Select regexp_substr(str,'(f\\-150\\:\\w+\\|)',1,loop)
from string_tab
join loop_tab on 1=1
Output:
regexp_substr
f-150:aa|
f-150:ce|
f-150:bb|
I have code that looks like this:
left outer join
gme_batch_header bh
on
substr(ln.lot_number,instr(ln.lot_number,'(') + 1,
instr(ln.lot_number,')') - instr(ln.lot_number,'(') - 1)
=
bh.batch_no
It works fine, but I have come across a few lot numbers that have two sections of strings that are between parenthesis. How would I compare what is between the second set of parenthesis? Here is an example of the data in the lot number field:
E142059-307-SCRAP-(74055)
This one works with the code,
58LF-3-B-2-2-2 (SCRAP)-(61448)
This one tries comparing SCRAP with the batch no, which isn't correct. It needs to be the 61448.
The result is always the last item in parenthesis.
After more research, I actually got it to work with this code:
substr(ln.lot_number,instr(ln.lot_number,'(',-1) + 1, instr(ln.lot_number,')',-1) - instr(ln.lot_number,'(',-1) - 1)
Assuming SQL2005+, and it is always the last occurrence you want, then I would suggest finding the last instance of a ( in your query and substring to there. To get the last instance you could use something like:
REVERSE(SUBSTRING(REVERSE(lot_number),0,CHARINDEX('(',REVERSE(lot_number))))
If your version of Oracle supports regular expressions try this:
substr(regexp_substr(ln.lot_number,'[0-9]+\)$'),1,length(regexp_substr(ln.lot_number,'[0-9]+\)$'))-1)
Explanation:
regexp_substr(scrap_row,'[0-9]+\)$' ==> find me just numbers in the string that ends in ). This returns the numbers but it includes the closing parenthesis.
To remove the closing parenthsis, just send it through substring and extract first number through the length of the number stopping at 1 character from the end of the string.
Query for analysis:
with scrap
as (select '58LF-3-B-2-2-2 (SCRAP)-(61448)' as scrap_row from dual)
select scrap_row,
regexp_substr(scrap_row,'[0-9]+\)$') as regex_substring,
length(regexp_substr(scrap_row,'[0-9]+\)$')) as length_regex_substring,
substr(regexp_substr(scrap_row,'[0-9]+\)$'),1,length(regexp_substr(scrap_row,'[0-9]+\)$'))-1) as regex_sans_parenthesis
from scrap
If you have 11g, this will do it pretty simply by using the subgroup argument of regexp_substr() and constructing the regex appropriately:
SQL> with tbl(data) as
(
select 'E142059-307-SCRAP-(74055)' from dual
union
select '58LF-3-B-2-2-2 (SCRAP)-(61448)' from dual
)
select data from tbl
where regexp_substr(data, '\((\d+)\)$', 1, 1, NULL, 1)
= '61448';
DATA
------------------------------
58LF-3-B-2-2-2 (SCRAP)-(61448)
The regular expression can be read as:
\( - Search for a literal left paren
( - Start a remembered subgroup
\d+ - followed by 1 more more digits
) - End remembered subgroup
\) - followed by a literal right paren
$ - at the end of the line.
The regexp_substr function arguments are:
Source - the source string
Pattern - The regex pattern to look for
position - Position in the string to start looking for the pattern
occurrence - If the pattern occurs multiple times, which occurrence you want
match_params - See the docs, not used here
subexpression - which subexpression to use (the remembered group)
So in English, look for a series of 1 or more digits surrounded by parens, where it occurs at the end of the line and save the digit part only to use to compare. IMHO a lot easier to follow/maintain than nested instr(), substr().
For re-useability, make a function called get_last_number_in_parens() that contains this code and uses an argument of the string to search. This way that logic is encapsulated and can be re-used by folks that may not be so comfortable with regular expressions, but can benefit from the power! One place to maintain code too. Then call like this:
select data from tbl
where get_last_number_in_parens(data) = '61448';
How easy is that?!
Hello you can check with this code. It works whaever the condition may be
SELECT SUBSTR('58LF-3-B-2-2-2-(61448)',instr('58LF-3-B-2-2-2-(61448)','(',-1)+1,LENGTH('58LF-3-B-2-2-2-(61448)')-instr('58LF-3-B-2-2-2-(61448)','(',-1)-1)
FROM dual;
SELECT SUBSTR('58LF-3-B-2-2-2 (SCRAP)-(61448)',instr('58LF-3-B-2-2-2 (SCRAP)-(61448)','(',-1)+1,LENGTH('58LF-3-B-2-2-2 (SCRAP)-(61448)')-instr('58LF-3-B-2-2-2 (SCRAP)-(61448)','(',-1)-1)
FROM dual;
Output
==================================
61448
==================================
I know I'm close to figuring this out but need a little help. What I'm trying to do is all grab a column from a particular table, but chop off the first 4 characters. For example if in a column the value is "KPIT08L", the result I was is 08L. Here is what I have so far but not getting the desired results.
SELECT LEFT(FIELD_NAME, 4)
FROM TABLE_NAME
First up, left will give you the leftmost characters. If you want the characters starting at a specific location, you need to look into mid:
select mid (field_name,5) ...
Secondly, if you value performance,portability and scalability at all, this sort of "sub-column" manipulation should generally be avoided. It's usually far easier (and faster) to patch columns together than to split them apart.
In other words, keep the first four characters in their own column and the rest in a separate column, and do your selects on the relevant one. If you're using anything less than a full column, then it's technically not one attribute of the row.
Try with
SELECT MID(FIELD_NAME, 5) FROM TABLE_NAME
Mid is very powerfull, it let you select the starting point and all the remainder, or,
if specified, the length desidered as in
SELECT MID(FIELD_NAME, 5, 2) FROM TABLE_NAME ' gives 08 in your example text
SELECT RIGHT(FIELD_NAME,LEN(FIELD_NAME)-4)
FROM TABLE_NAME;
If it is for a generic string then the above one will work...
Don't have Access at my current location, but please try this.
SELECT RIGHT(FIELD_NAME, LEN(FIELD_NAME)-4)
FROM TABLE_NAME
The LEFT(FIELD_NAME, 4) will return the first 4 caracters of FIELD_NAME.
What you need to do is :
SELECT MID(FIELD_NAME, 5)
FROM TABLE_NAME
If you have a FIELD_NAME of 10 caracters, the function will return the 6 last caracters (chopping the first 4)!
I got this query and want to extract the value between the brackets.
select de_desc, regexp_substr(de_desc, '\[(.+)\]', 1)
from DATABASE
where col_name like '[%]';
It however gives me the value with the brackets such as "[TEST]". I just want "TEST". How do I modify the query to get it?
The third parameter of the REGEXP_SUBSTR function indicates the position in the target string (de_desc in your example) where you want to start searching. Assuming a match is found in the given portion of the string, it doesn't affect what is returned.
In Oracle 11g, there is a sixth parameter to the function, that I think is what you are trying to use, which indicates the capture group that you want returned. An example of proper use would be:
SELECT regexp_substr('abc[def]ghi', '\[(.+)\]', 1,1,NULL,1) from dual;
Where the last parameter 1 indicate the number of the capture group you want returned. Here is a link to the documentation that describes the parameter.
10g does not appear to have this option, but in your case you can achieve the same result with:
select substr( match, 2, length(match)-2 ) from (
SELECT regexp_substr('abc[def]ghi', '\[(.+)\]') match FROM dual
);
since you know that a match will have exactly one excess character at the beginning and end. (Alternatively, you could use RTRIM and LTRIM to remove brackets from both ends of the result.)
You need to do a replace and use a regex pattern that matches the whole string.
select regexp_replace(de_desc, '.*\[(.+)\].*', '\1') from DATABASE;
What would be the best way to count occurrences of all non alphanumeric characters that appear in a string in an Oracle database column.
When attempting to find a solution I realised I had a query that was unrelated to the problem, but I noticed I could modify it in the hope to solve this problem. I came up with this:
SELECT COUNT (*), SUBSTR(TITLE, REGEXP_INSTR(UPPER(TITLE), '[^A-Z,^0-9]'), 1)
FROM TABLE_NAME
WHERE REGEXP_LIKE(UPPER(TITLE), '[^A-Z,^0-9]')
GROUP BY SUBSTR(TITLE, REGEXP_INSTR(UPPER(TITLE), '[^A-Z,^0-9]'), 1)
ORDER BY COUNT(*) DESC;
This works to find the FIRST non alphanumeric character, but I would like to count the occurrences throughout the entire string, not just the first occurrence. E. g. currently my query analysing "a (string)" would find one open parenthesis, but I need it to find one open parenthesis and one closed parenthesis.
There is an obscure Oracle TRANSLATE function that will let you do that instead of regexp:
select a.*,
length(translate(lower(title),'.0123456789abcdefghijklmnopqrstuvwxyz','.'))
from table_name a
Try this:
SELECT a.*, LENGTH(REGEXP_REPLACE(TITLE, '[^a-zA-Z0-9]'), '')
FROM TABLE_NAME a
The best option, as you discovered is to use a PL/SQL procedure. I don't think there's any way to create a regex expression that will return multiple counts like you're expecting (at least, not in Oracle).
One way to get around this is to use a recursive query to examine each character individually, which could be used to return a row for each character found. The following example will work for a single row:
with d as (
select '(1(2)3)' as str_value
from dual)
select char_value, count(*)
from (select substr(str_value,level,1) as char_value
from d
connect by level <= length(str_value))
where regexp_instr(upper(char_value), '[^A-Z,^0-9]'), 1) <> 0
group by char_value;