I'm writing a query that returns a bunch of things from multiple tables. The main query is against Table_1. I need to return a substring from a field in table 7. But I'm getting an error that Substring_Index is an invalid identifier. How can I achieve the intended result?
I have a field COLUMN_1 of TABLE_1 that has 3+ pieces of data, separated by " : " (space colon space) and I need to strip out the text before the first delimiter, and return the rest of it (regardless of length).
A simplified example:
SELECT t1.name
,t1.address
,t1.phone
,t2. fave_brand
,SUBSTRING_INDEX(t3.fave_product, ' : ', -1) AS Fave Product
FROM table_1 t1
INNER JOIN table_2 t2
ON t2.brand_SK = t1.fave_brand_FK
INNER JOIN table_3 t3
ON t3.product_list_SK = t1.fave_products
WHERE <a series of constraints>;
Please note, I am NOT normally an SQL developer, but the back-end dev is on vacation and I've been tasked with cobbling this fix together. I'm a beginner at best.
In oracle you could use regexp_replace():
regexp_replace(t3.fave_product, '^[^:]*:', '') "Fave Product"
regexp_replace() replaces the part of the string that matches the regexp given as second argument with the value given as third argument. Here, we use the empty string as third argument, meaning that the matching part of the string is suppressed.
Regexp breakdown:
^ beginning of the string
[^:]* as many characters as possible other than ":" (possibly, 0 characters)
: character ":"
NB: identifiers that contain special characters (such as space) need to be double quoted.
Oracle does not support substring_index(). That is a MySQL function.
You can use regexp_substr(). Without sample data it is a little hard to be 100% sure, but I think the logic you want is:
regexp_substr(t3.fave_product, '[^:]+$') as fave_product
Related
I need to find rows where col have special characters or numbers (except hyphen,apostrophe and space) in Oracle SQL.
I am doing like below:
SELECT *
FROM test
WHERE Name_test LIKE '%[^A-Za-z _]%'
But It is not working and I also need to exclude any apostrophe.
Kindly help.
If you need to find all rows where column have ONLY numbers and special characters (and you can specify all of required special characters):
SELECT *
FROM test
WHERE regexp_like(Name_test, q['^[0-9'%##]+$]')
as you can see you just need to add your special characters after 0-9.
^ - start
$ - end
About format q'[SOMETHING]' please see TEXT LITERALS here: https://docs.oracle.com/en/database/oracle/oracle-database/19/sqlrf/Literals.html#GUID-1824CBAA-6E16-4921-B2A6-112FB02248DA
If you need to find all rows where column have no alpha-characters:
SELECT *
FROM test
WHERE regexp_like(Name_test, '^[^a-zA-Z]*$');
or
SELECT *
FROM test
WHERE regexp_like(Name_test, '^\W*$');
about \W - please see "Table 8-5 PERL-Influenced Operators in Oracle SQL Regular Expressions" here:
https://docs.oracle.com/database/121/ADFNS/adfns_regexp.htm#ADFNS235
I need to find rows where col have special characters or numbers (except hyphen, apostrophe and space [and presumably single quotes]) in Oracle SQL.
You can use double single quotes to put a single quote in:
WHERE Name_test LIKE '%[^-A-Za-z _'']%'
However, this is not Oracle syntax. If the above works, then I would guess you are using SQL Server. In Oracle:
WHERE REGEXP_LIKE(Name_test, '[^A-Za-z _'']')
I have the following SQL code
SELECT State.CODE Code,
State.CODE1 Code1,
State.CODE2 Code2,
YYY.1003."BRANCH" "Branch"
FROM XXX.1002 State
LEFT JOIN YYY.1003
ON State.CODE= YYY.1003.CODE
Now I need a new column that shows me the field/ column "Country" from table ZZZ.100.
The key is 1002.CODE1 - ZZZ.100.CODE1
Unfortunately, ZZZ.100.CODE1 has spaces before the values (4 spaces).
How can I use the trim function (is this the right one) to get a join on 1002.CODE1 - ZZZ.100.CODE1
It looks like you want:
select x.code, x.code1, x.code2, y.branch, z.country
from xxx.1002 x
left join yyy.1003 y on x.code= y.code
left join zzz.100 z on trim(z.code1) = x.code1
trim() removes spaces on both ends of the string. If you want to remove only leading spaces, you can do: trim(leading ' ' from z.code1).
Note that I used more meaningful table aliases, in order to make the query easier to write and read.
I would also reommend against using all-digits table names: in Oracle, non-quoted identifiers must begin begin with an alphabetic character.
First thing is why a Key column contains spaces on it? A key column should never contain spaces, it is terrible for performance.
The point to think about here is to remove those spaces from this column
The use of TRIM, considering you are using SQL Server
In SQL Server you can use LTRIM (for left trim, remove left spaces) and/or RTRIM (for right trim, remove right spaces)
Ex:
Select
LTRIM(RTRIM(Table.Column))
from Table
I hope this can help you
Newbie here. Been searching for hours now but I can seem to find the correct answer or properly phrase my search.
I have thousands of rows (orderids) that I want to put on an IN function, I have to run a LIKE at the same time on these values since the columns contains json and there's no dedicated table that only has the order_id value. I am running the query in BigQuery.
Sample Input:
ORD12345
ORD54376
Table I'm trying to Query: transactions_table
Query:
SELECT order_id, transaction_uuid,client_name
FROM transactions_table
WHERE JSON_VALUE(transactions_table,'$.ordernum') LIKE IN ('%ORD12345%','%ORD54376%')
Just doesn't work especially if I have thousands of rows.
Also, how do I add the order id that I am querying so that it appears under an order_id column in the query result?
Desired Output:
Option one
WITH transf as (Select order_id, transaction_uuid,client_name , JSON_VALUE(transactions_table,'$.ordernum') as o_num from transactions_table)
Select * from transf where o_num like '%ORD12345%' or o_num like '%ORD54376%'
Option two
split o_num by "-" as separator , create table of orders like (select 'ORD12345' as num
Union
Select 'ORD54376' aa num) and inner join it with transf.o_num
One method uses OR:
WHERE JSON_VALUE(transactions_table, '$.ordernum') LIKE IN '%ORD12345%' OR
JSON_VALUE(transactions_table, '$.ordernum') LIKE '%ORD54376%'
An alternative method uses regular expressions:
WHERE REGEXP_CONTAINS(JSON_VALUE(transactions_table, '$.ordernum'), 'ORD12345|ORD54376')
According to the documentation, here, the LIKE operator works as described:
Checks if the STRING in the first operand X matches a pattern
specified by the second operand Y. Expressions can contain these
characters:
A percent sign "%" matches any number of characters or
bytes.
An underscore "_" matches a single character or byte.
You can escape "\", "_", or "%" using two backslashes. For example, "\%". If
you are using raw strings, only a single backslash is required. For
example, r"\%".
Thus , the syntax would be like the following:
SELECT
order_id,
transaction_uuid,
client_name
FROM
transactions_table
WHERE
JSON_VALUE(transactions_table,
'$.ordernum') LIKE '%ORD12345%'
OR JSON_VALUE(transactions_table,
'$.ordernum') LIKE '%ORD54376%
Notice that we specify two conditions connected with the OR logical operator.
As a bonus information, when querying large datasets it is a good pratice to select only the columns you desire in your out output ( either in a Temp Table or final view) instead of using *, because BigQuery is columnar, one of the reasons it is faster.
As an alternative for using LIKE, you can use REGEXP_CONTAINS, according to the documentation:
Returns TRUE if value is a partial match for the regular expression, regex.
Using the following syntax:
REGEXP_CONTAINS(value, regex)
However, it will also work if instead of a regex expression you use a STRING between single/double quotes. In addition, you can use the pipe operator (|) to allow the searched components to be logically ordered, when you have more than expression to search, as follows:
where regexp_contains(email,"gary|test")
I hope if helps.
I have a column in sql which holds value inside double quotes like "P1234567" , "P1234" etc..
I need to identify only columns which start with letter P and is followed by seven digits (numbers) only. I tried where column like'"P[0-9][0-9][0-9][0-9][0-9][0-9][0-9]"' but it doesn't seem to work.
Can someone please correct me or point me to a thread which can help me out?
Thanks
Standard SQL has no regex support, but most SQL engines have regex extensions added to them on top of the standard SQL. So, for example, if you're using MySQL then you'd do this:
... WHERE column REGEXP '^"P[0-9]{7}"'
And if you're using Postgres then that would be:
... WHERE column ~ '^"P[0-9]{7}"'
(updated to match the double-quote part of the question, I'd misunderstood that to begin with)
How about using length and isnumeric:
Select
*
from
mytable
where
mycolumn like '"P%'
and len(mycolumn) = 10 --2 chars for quotes + 1 for 'P' + 7 for the digits
and isnumeric(substring(mycolumn, 3, 7))=1
This answer is for SQL Server, other DBMS's may have a different syntax for length
I have a table with one column " otname "
table1.otname contains multiple rows of alpha-numeric string resembling the following data sample:
11.10.32.12.U.A.F.3.2.21.249.1
2001.1.1003.8281.A.LE.P.P
2010.1.1003.8261.A.LE.B.B
I want to read the fourth number in every string ( part of the string in bold ) and write a query in Oracle 10g
to read its description stored in another table. My dilemma is writing the first part of the query.i.e. choosing the fourth number of every string in a table
My second query will be something like this:
select description_text from table2 where sncode = 8281 -- fourth part of the data sample in every string
Many thanks.
novice
Works with 9i+:
WITH portion AS (
SELECT SUBSTR(t.otname, INSTR(t.otname, ".", 1, 3)+1, INSTR(t.otname, ".", 1, 4)) 'sncode'
FROM TABLE t)
SELECT t.description_text
FROM TABLE2 t
JOIN portion p ON p.sncode = t.sncode
The use of SUBSTR should be obvious; INSTR is being used to find location the period (.), starting at the first character in the string (parameter value 1), on the 3rd and 4th appearance in the string. You might have to subtract one from the position returned for the 4th instance of the period - test this first to be sure you're getting the right values:
SELECT SUBSTR(t.otname, INSTR(t.otname, ".", 1, 3)+1, INSTR(t.otname, ".", 1, 4)) 'sncode'
FROM TABLE t
I used subquery factoring so the substring happens before you join to the second table. It can be done as a subquery, but subquery factoring is faster.
Newer versions of oracle (including 10g) have various regular expression functions. So you can do something like this:
where sncode = to_number(regexp_replace(otname, '^(\d+\.\d+\.\d+\.(\d+))?.+$', '\2'))
This matches 3 sets of digits-followed-by-a-dot, and a fourth grouped set of digits, followed by the rest of the string, and returns a string consisting of all that entirely replaced by the first group (the fourth set of digits).
Here's a complete query (if I understood your description of the two tables correctly):
select t2.description_text
from table1 t1, table2 t2
where t2.sncode = to_number(regexp_replace(t1.otname, '^(\d+\.\d+\.\d+\.(\d+))?.+$', '\2'))
Another slightly shorter alternative regex:
where t2.sncode = to_number(regexp_replace(t1.otname, '^((\d+\.){3}(\d+))?.+$', '\3'))