I have a problem in SQL Oracle, I'm trying to create a view that contains values with letters and numbers and I want to sort them in a specific order.
Here is my query:
create or replace view table1_val (val, msg_text) as
select
val, msg_text
from
table_val
where
val in ('L1','L2','L3','L4','L5','L6','L7','L8','L9','L10','L11','L12','L13','L14','G1','G2','G3','G4')
order by lpad(val, 3);
The values are displayed like this:
G1,G2,G3,G4,L1,L2,L3,L4,L5,L6,L7,L8,L9,L10,L11,L12,L13
The thing is that I want to display the L values first and then the G values like in the where condition. The 'val' column is VARCHAR2(3 CHAR). The msg_text column is irrelevant. Can someone help me with that? I use Oracle 12C.
You must interpret the second part of the val column as a number
order by
case when val like 'L%' then 0 else 1 end,
to_number(substr(val,2))
This work fine for your current data, but may fail in future if a new record is added with non-numeric structure.
More conservative (and more hard to write), but safe would be to used a decode for all the current keys, ordering unknown keys on the last position (id = 18 in the example):
order by
decode(
'L1',1,
'L2',2,
'L3',3,
'L4',4,
'L5',5,
'L6',6,
'L7',7,
'L8',8,
'L9',9,
'L10',10,
'L11',11,
'L12',12,
'L13',13,
'G1',14,
'G2',15,
'G3',16,
'G4',17,18)
You can't do anything based on the order of the WHERE condition
But you can use a CASE on the ORDER BY
ORDER BY CASE
WHEN SUBSTR(val, 1, 1) = 'L' THEN 1
WHEN SUBSTR(val, 1, 1) = 'G' THEN 2
ELSE 3
END,
TO_NUMBER (SUBSTR(val, 2, 10));
Another option to consider might be using regular expressions, such as
SQL> with table1_val (val) as
2 (select 'L1' from dual union all
3 select 'L26' from dual union all
4 select 'L3' from dual union all
5 select 'L21' from dual union all
6 select 'L11' from dual union all
7 select 'L4' from dual union all
8 select 'G88' from dual union all
9 select 'G10' from dual union all
10 select 'G2' from dual
11 )
12 select val
13 from table1_val
14 order by regexp_substr(val, '^[[:alpha:]]+') desc,
15 to_number(regexp_substr(val, '\d+$'));
VAL
---
L1
L3
L4
L11
L21
L26
G2
G10
G88
9 rows selected.
SQL>
Related
I have some strings in my table. They are like 1101-1, 1101-2, 1101-10, 1101-11 pulse, shock, abc, 1104-2, 1104-11, 2201-1, 2202-4. I tried to sort them like below:
1101-1
1101-2
1101-10
1101-11
1104-2
1104-11
2201-1
2202-4
abc
pulse
shock
But I can't get the sort correctly. Below is my codes:
select column from table
order by regexp_substr(column, '^\D*') nulls first,
to_number(substr(regexp_substr(column, '\d+'),1,4)) asc
Sort numbers as numbers:
first the ones in front of the hyphen (line #16)
then the ones after it (line #17),
then the rest (line #18)
Mind the to_number function! Without it, you'll be sorting strings! and get the wrong result.
SQL> with test (col) as
2 ( select '1101-1' from dual union all
3 select '1101-2' from dual union all
4 select '1101-10' from dual union all
5 select '1101-11' from dual union all
6 select 'pulse' from dual union all
7 select 'shock' from dual union all
8 select 'abc' from dual union all
9 select '1104-2' from dual union all
10 select '1104-11' from dual union all
11 select '2201-1' from dual union all
12 select '2202-4' from dual
13 )
14 select col
15 from test
16 order by to_number(regexp_substr(col, '^\d+')),
17 to_number(regexp_substr(col, '\d+$')),
18 col;
COL
-------
1101-1
1101-2
1101-10
1101-11
1104-2
1104-11
2201-1
2202-4
abc
pulse
shock
11 rows selected.
SQL>
For your examples, this should do:
order by regexp_substr(column, '^[^-]+'), -- everything before the hyphen
len(column),
column
To get the number after '-' specifically:
with ttt (col) as (
select cast(column_value as varchar2(10)) as second_str
from table(sys.dbms_debug_vc2coll
( '1101-1'
, '1101-2'
, '1101-10'
, '1101-11'
, '1104-2'
, '1104-11'
, '2201-1'
, '2202-4'
, 'abc'
, 'pulse'
, 'shock'
))
)
select col
, regexp_substr(col, '(^\d+-)(\d+)', 1, 1, '', 2)
from ttt;
COL SECOND_STR
---------- ----------
1101-1 1
2201-1 1
1101-10 10
1101-11 11
1104-11 11
1101-2 2
1104-2 2
2202-4 4
abc
pulse
shock
11 rows selected
This treats the text string as two values, (^\d+-) followed by (\d+), and takes the second substring (the final '2' parameter). As only positional parameters are allowed for built-in SQL functions, you also have to specify occurrence (1) and match param (null, as we don't care about case etc).
For the above query, I am using regexp_count but in SQL command line I am getting regexp_count :invaild identifier;
select * from table_name WHERE (REGEXP_COUNT(column_name, 'A')) >2;
Is this query works?
I'm not sure what you really want; is it to return names whose 2nd letter is "a", or names that have two or more letters "a" within.
Anyway, pick the one you find appropriate.
SQL> create table test as
2 (select 'saritha' col from dual union all
3 select 'mamatha' from dual union all
4 select 'vaisnavi' from dual union all
5 select 'sai' from dual union all
6 select 'vijaya' from dual union all
7 select 'kumar' from dual
8 );
Table created.
2 or more letters "a":
SQL> select col
2 from test
3 where regexp_count(col, 'a') >= 2;
COL
--------
saritha
mamatha
vaisnavi
vijaya
2nd letter is "a":
SQL> select col
2 from test
3 where substr(col, 2, 1) = 'a';
COL
--------
saritha
mamatha
vaisnavi
sai
SQL>
On Oracle 10g, which doesn't support REGEXP_COUNT function, one option is to replace all letters a with an empty string (basically, you'd remove all letters a) and fetch rows whose difference of full column length and "replaced" column length is >= 2. Something like this:
SQL> select col
2 from test
3 where length(col) - length(replace(col, 'a', '')) >= 2;
COL
--------
saritha
mamatha
vaisnavi
vijaya
SQL>
I have a column with a letter followed by either numbers or letters:
ID_Col
------
S001
S1001
S090
SV911
SV800
Sfoofo
Szap
Sbart
How can I order it naturally with the numbers first (ASC) then the letters alphabetically? If it starts with S and the remaining characters are numbers, sort by the numbers. Else, sort by the letter. So SV911should be sorted at the end with the letters since it also contains a V. E.g.
ID_Col
------
S001
S090
S1001
Sbart
Sfoofo
SV800
SV911
Szap
I see this solution uses regex combined with the TO_NUMBER function, but since I also have entries with no numbers this doesn't seem to work for me. I tried the expression:
ORDER BY
TO_NUMBER(REGEXP_SUBSTR(ID_Col, '^S\d+$')),
ID_Col
/* gives ORA-01722: invalid number */
Would this help?
SQL> with test (col) as
2 (select 'S001' from dual union all
3 select 'S1001' from dual union all
4 select 'S090' from dual union all
5 select 'SV911' from dual union all
6 select 'SV800' from dual union all
7 select 'Sfoofo' from dual union all
8 select 'Szap' from dual union all
9 select 'Sbart' from dual
10 )
11 select col
12 from test
13 order by substr(col, 1, 1),
14 case when regexp_like(col, '^[[:alpha:]]\d') then to_number(regexp_substr(col, '\d+$')) end,
15 substr(col, 2);
COL
------
S001
S090
S1001
Sbart
Sfoofo
SV800
SV911
Szap
8 rows selected.
SQL>
Can some one please help me in writing a sql query that should do a oracle min function based on the following conditions.
For eg for column values
0,0,0,0 then output should be 0
0,null,0,null then output should be o
0,2,4,5,6 then output should be 2 (Note that we are excluding Zero here)
0,2,null,4,5 then output should be 2 (same here we are excluding zero)
null,null,null, null then output should be null.
I wrote query already that satisfies all the above cases but failing for last case when all the column values are null. Instead of returning null it is returning 0. Can some one modify the below query to fit for the last case as well?
select NVL(MIN(NULLIF(columnname,0)),0) from tablename;
Please also keep in mind that the query should be runnable in oracle as well as hsqldb as we are using hsql db for running junits.
If all 4 cases satisfied by your query then just a case will solve your problem.
SELECT CASE WHEN MIN(COLUMNNAME) IS NULL THEN NULL ELSE NVL(MIN(NULLIF(COLUMNNAME,0)),0) END FROM TABLENAME;
Note:- assuming all the cases satisfied by your query except 5th.
I will show below an input table with two columns, ID and VAL, to illustrate the various possibilities. You want a single result per ID (or even for the entire table), so this must be a job for GROUP BY and some aggregate function. You want to distinguish between three types of values: Greater than zero, zero, and null (in this order); you want to pick the "highest priority group" that exists for each ID (in this order of priority), and for that priority group only, you want to pick the min value. This is exactly what the aggregate FIRST/LAST function does. To order by the three "classes" of values, we use a CASE expression in the ORDER BY clause of the aggregate LAST function.
The WITH clause below is not part of the solution - I only include it to create test data (in your real life situation, use your actual table and column names and remove the entire WITH clause).
with
inputs ( id, val ) as (
select 1, 0 from dual union all
select 1, 0 from dual union all
select 1, 0 from dual union all
select 2, 0 from dual union all
select 2, null from dual union all
select 2, 0 from dual union all
select 3, 0 from dual union all
select 3, 2 from dual union all
select 3, 5 from dual union all
select 4, 0 from dual union all
select 4, 3 from dual union all
select 4, null from dual union all
select 5, null from dual union all
select 5, null from dual
)
select id,
min(val) keep (dense_rank last order by case when val > 0 then 2
when val = 0 then 1
else 0
end
) as min_val
from inputs
group by id
order by id
;
ID MIN_VAL
---------- ----------
1 0
2 0
3 2
4 3
5
For these strings
RSLR_AIRL19_ID3454_T20030913091226
RSLR_AIRL19_ID3122454_T20030913091226
RSLR_AIRL19_ID34_T20030913091226
How to get the number after ID ?
Or how to get the content between two characters but not include them ?
I use this '/\_ID([^_]+)/' got matches like Array ( [0] => _ID3454 [1] => 3454 )
Is this the right way?
To extract a number after an ID, you could write a similar query.
SQL> with t1 as(
2 select 'RSLR_AIRL19_ID3454_T20030913091226' as col from dual union all
3 select 'RSLR_AIRL19_ID3122454_T20030913091226' from dual union all
4 select 'RSLR_AIRL19_ID34_T20030913091226' from dual
5 )
6 select regexp_substr(col, '^([[:alnum:]]+_){2}ID([[:digit:]]+)_([[:alnum:]]+){1}$', 1, 1, 'i', 2) as ID
7 from t1
8 ;
ID
-------------
3454
3122454
34
Or, if you want to extract digits from a first occurrence of the pattern without verifying if an entire string matches a specific format:
SQL> with t1 as(
2 select 'RSLR_AI_RL19_ID3454_T20030913091226' as col from dual union all
3 select 'RSLR_AIRL19_ID3122454_T20030913091226' from dual union all
4 select 'RSLR_AIRL19_ID34_T20030913091226' from dual
5 )
6 select regexp_substr(col, 'ID([[:digit:]]+)', 1, 1, 'i', 1) as ID
7 from t1
8 ;
ID
--------------
3454
3122454
34
With pcre & perl engines :
ID\K\w+
NOTE
\K "restart" the match.
See http://www.phpfreaks.com/blog/pcre-regex-spotlight-k (php use pcre)