I have a numbers column in my sql table and I want to pull all the numbers that start with 3 specific digits; how would I query this?
Use the "LIKE" query:
123 is the prefix. I think you have to store the numbers as strings though, just try it out on your data set :-)
SELECT * from TableName Where ColumnName LIKE '123%'
See also this Q&A: In MySql, find strings with a given prefix
Related
I am trying to write a SQL query that helps me find out the unique amount of "Numbers" that show up in a specific column. Example, in a select * query, the column I want can look like this
Num_Option
9000
9001
9000,9001,9002
8080
8080,8000,8553
I then have another field of "date_available" which is a date/time.
Basically, what want is something where I can group by the "date_available" while combing all the Num_Options on that date, so something like this..
Num_Option date_available
9000,9001,9002,8080 10/22/2020
9000,9002,8080,8000,8553 10/23/2020
I am struggling to figure this out. I have gotten to the possible point of using a python script and matplotlib instead... but I am hoping there is a SQL way of handling this as well.
In Postgres, you can use regexp_split_to_table() in a lateral join to turn the csv elements to rows, then string_agg() to aggregate by date:
select string_agg(x.num, ',') num_option, t.date_available
from mytable t
cross join lateral regexp_split_to_table(t.num_option, ',') x(num)
group by date_available
Of course, this assumes that you want to avoid duplicate nums on the same data (otherwise, there is not need to split, you can directly aggregate).
You may just be able to use string_agg():
select date_available, string_agg(num_option, ',')
from t
group by date_available;
first you have to split the strings into multiple rows with something like split_part('9000,9001,9002',',',1) etc. (use UNION ALL to append the 2nd number etc.), then group them back by availability date with string_agg
if you don't want to hardcode split_part part there is an answer here on how to dynamically split strings in Redshift, look for it
i have an array of strings and a column which may contain one or more of those strings(seperated by space) i want to get all rows where this column contains one of the strings. Since the values all have 3 letters and therefore can't contain each other, i know i could just write
SELECT * FROM table WHERE
column LIKE '%val1%' OR
column LIKE '%val2%' OR
column LIKE '%val3%' OR
column LIKE '%val4%'
But i'm wondering if there isn't an easier statement, like column IN ('val1', 'val2', 'val3', 'val4') (This one seems only to work when the entry is equal to one of the values, but not if it just contains them)
Try reading this Is there a combination of "LIKE" and "IN" in SQL? and Combining "LIKE" and "IN" for SQL Server , this will solve you question.
Something like this from the first link.
SQL Server:
WHERE CONTAINS(t.something, '"bla*" OR "foo*" OR "batz*"')
Ist oracle you could use regular expressions
select *
from table
where regexp_like (column,'val(1|2|3|4)')
I need help for sql query LIKE.
Value for column in database is same below:
record 1 : "3,13,15,20"
record 2 : "13,23,14,19"
record 3 : "3,14,15,19,20"......
for now I want to get the most accurate record with a value of 3
This is my query :
SELECT * FROM accounts where type like '%3%'
This query will find all record with value exist is '3' eg: 13,23 ....
And It does not solve my problem.
Try this:
SELECT *
FROM accounts
WHERE CONCAT(',', type, ',') LIKE '%,3,%';
Demo
This trick places commas around the end of the type CSV string, so that we all we have to do is then check for ,3, anywhere in that string.
By the way, it is generally not desirable to store CSV data like this in your SQL tables. Instead, consider normalizing your data and storing those CSV values across separate rows.
I would like to ask if it is possible to do this:
For example the search string is '009' -> (consider the digits as string)
is it possible to have a query that will return any occurrences of this on the database not considering the order.
for this example it will return
'009'
'090'
'900'
given these exists on the database. thanks!!!!
Use the Like operator.
For Example :-
SELECT Marks FROM Report WHERE Marks LIKE '%009%' OR '%090%' OR '%900%'
Split the string into individual characters, select all rows containing the first character and put them in a temporary table, then select all rows from the temporary table that contain the second character and put these in a temporary table, then select all rows from that temporary table that contain the third character.
Of course, there are probably many ways to optimize this, but I see no reason why it would not be possible to make a query like that work.
It can not be achieved in a straight forward way as there is no sort() function for a particular value like there is lower(), upper() functions.
But there is some workarounds like -
Suppose you are running query for COL A, maintain another column SORTED_A where from application level you keep the sorted value of COL A
Then when you execute query - sort the searchToken and run select query with matching sorted searchToken with the SORTED_A column
I have a table in Postgres. In that table more than 1000 names are there. Most of the names are start with SHRI or SMT. I want to delete this SHRT and SMT from the names and to save original name only. How can I do that with out any database function?
I'll step you through the logic:
Select left(name,3) from table
This select statement will bring back the first 3 chars of a column (the 'left' three). If we are looking for SMT in the first three chars, we can move it to the where statement
select * from table where left(name,3) = 'SMT'
Now from here you have a few choices that can be used. I'm going to keep to the left/right style, though replace could likely be used. We want the chars to the right of the SMT, but we don't know how long each string is to pick out those chars. So we use length() to determine that.
select right(name,length(name)-3) from table where left(name,3) = 'SMT'
I hope my syntax is right there, I'm lacking a postgres environment to test it. The logic is 'all the chars on the right of the string except the last 3 (the minus 3 excludes the 3 chars on the left. change this to 4 if you want all but the last 4 on the left)
You can then change this to an update statement (set name = right(name,length(name)-3) ) to update the table, or you can just use the select statement when you need the name without the SMT, but leave the SMT in the actual data.