I have ,for example, this table in a Microsoft Access database:
id numeric
context text
numberfield numeric
I want to select every record that ends with 9 in the column"numberfield". This gives a problem because it is a numeric field and as a result I can not use the following SQL:
select * from table where numberfield like "%9"
A solution is that I change the numberfield to a text. But this gives a problem because there are several users and the change might give a problem in the future. Is there an option to select on the ending when it is a number field?
That sound a little fishy.. are you sure you can use that query? Don't know about Access but almost any other DBMS allows it.
If it really doesn't work, you can do this:
select * from table where STR(numberfield) like "*9"
EDIT: Maybe it didn't work because you used % which is used with * in Access :
select * from table where numberfield like "*9"
Numbers are numbers, so use Mod for this:
select * from table where numberfield mod 10 = 9
Instead of casting to string and comparing, just extract the rightmost digit with a MOD operation.
Edit your query as follows:
SELECT *
FROM table
WHERE ((([numberfield] Mod 10)=9));
Related
I need for a SQL query to transform an int with a value between 1 to 300000 to a number which has this pattern : always 8 number.
For example:
1 becomes 00000001,
123 becomes 00000123,
123456 becomes 00123456.
I have no idea how to do that... How can I do it?
In Standard SQL, you can use this trick:
select substring(cast( (num + 100000000) as varchar(255)) from 2)
Few databases actually support this syntax. Any given database can do what you want, but the method depends on the database you are using.
For MS SQL Server
You could use FORMAT function, like this:
SELECT FORMAT(123,'00000000')
https://database.guide/how-to-format-numbers-in-sql-server/#:~:text=Starting%20from%20SQL%20Server%202012,the%20output%20should%20be%20formatted.
Read at the link Leading Zeroes
For MySql/Oracle
You could use LPAD, like this:
SELECT LPAD('123',8,'0')
https://database.guide/how-to-add-leading-zeros-to-a-number-in-mysql/
I have a table with about 200 million records. One of the columns is defined as varchar(100) and it's included in a full text index. Most of the values are numeric. Only few are not numeric.
The problem is that it's not working well. For example if a row contains the value '123456789' and i look for '567', it's not returning this row. It will only return rows where the value is exactly '567'.
What am I doing wrong?
sql server 2012.
Thanks.
Full text search doesn't support leading wildcards
In my setup, these return the same
SELECT *
FROM [dbo].[somelogtable]
where CONTAINS (logmessage, N'28400')
SELECT *
FROM [dbo].[somelogtable]
where CONTAINS (logmessage, N'"2840*"')
This gives zero rows
SELECT *
FROM [dbo].[somelogtable]
where CONTAINS (logmessage, N'"*840*"')
You'll have to use LIKE or some fancy trigram approach
The problem is probably that you are using a wrong tool since Full-text queries perform linguistic searches and it seems like you want to use simple "like" condition.
If you want to get a solution to your needs then you can post DDL+DML+'desired result'
You can do this:
....your_query.... LIKE '567%' ;
This will return all the rows that have a number 567 in the beginning, end or in between somewhere.
99% You're missing % after and before the string you search in the LIKE clause.
es:
SELECT * FROM t WHERE att LIKE '66'
is the same as as using WHERE att = '66'
if you write:
SELECT * FROM t WHERE att LIKE '%66%'
will return you all the lines containing 2 'sixes' one after other
I have data like this
1234500010
1234500020
1234500021
12345600010
12345600011
123456700010
123456700020
123456710010
The pattern is
1-data(varian 3-7 digit number) + 2-data(any 3 digit number) + 3-data (any 2 digit number)
I want to create SQL to get 1-data only.
For example I want to get data 12345
I want the result only
1234500010
1234500020
1234500021
If I using "like",
select *
FROM data
where ID like '12345%' `
I will get all the data with 12345, 123456 and 1234567
If I using equal, I will only get one specific data.
Can I combine like and equal together to get result like what I want?
select * FROM data where data = '12345 + any 2-data(3 digit) + any 3-data(2 digit)'
Anyone can help?
Addition : Sorry if I didn't mention the data type and make some miss communication. The data type is in char. #Gordon answers and the others not wrong. It works for number and varchar. but not works for char type. Here I post some pic for char data type. Oracle specification for char data type is a fixed lenght. So if I input less than lenght the remain of it will be change into a space.
Thank you very much. Hope someone can help for this
Since your datatype is CHAR, Gordon's answer is not working for you. CHAR adds trailing spaces for the strings less than maximum limit. You could use TRIM to fix this as shown. But, you should preferably store numbers in the NUMBER type and not CHAR or VARCHAR2, which will create other problems sooner or later.
select *
from data
where trim(ID) like '12345_____';
I think you want:
select *
from data
where ID like '12345_____' -- exactly 5 _
Here is a rextester demonstrating the answer.
You really can't combine equality and LIKE. But you can use a regular expression to do this kind of searching, with the REGEXP_LIKE function:
SELECT *
FROM DATA
WHERE REGEXP_LIKE(ID, '^12345[0-9]{3}[0-9]{2}');
But if I understand correctly, for your 1-data you really want a 3 to 7 digit number:
SELECT *
FROM DATA
WHERE REGEXP_LIKE(ID, '^[0-9]{3,7}[0-9]{3}[0-9]{2}');
Oracle regular expression docs here
SQLFiddle here
Best of luck.
I think this gives you the solution you want,
create table data(ID number(15));
insert into data values(1234500010);
insert into data values(1234500020);
insert into data values(1234500021);
insert into data values(12345600010);
insert into data values(12345600011);
insert into data values(123456700010);
insert into data values(123456700020);
insert into data values(123456710010);
select * from data where ID like '12345_____'
// After 5_ underscore are exactly 5 , any 3 digits from 2-data(3 underscores) and 2 digits from 3-data(2 underscores)
You'll be getting(OUTPUT) :
ID
1234500010
1234500020
1234500021
3 rows returned in 0.00 seconds
BigQuery Standard SQL does not seems to allow period "." in the select statement. Even a simple query (see below) seems to fail. This is a big problem for datasets with field names that contain "." Is there an easy way to avoid this issue?
select id, time_ts as time.ts
from `bigquery-public-data.hacker_news.comments`
LIMIT 10
Returns error...
Error: Syntax error: Unexpected "." at [1:27]
This also fails...
select * except(detected_circle.center_x )
from [bigquery-public-data:eclipse_megamovie.photos_v_0_2]
LIMIT 10
It depends on what you are trying to accomplish. One interpretation is that you want to return a STRUCT named time with a single field named ts inside of it. If that's the case, you can use the STRUCT operator to build the result:
SELECT
id,
STRUCT(time_ts AS ts) AS time
FROM `bigquery-public-data.hacker_news.comments`
LIMIT 10;
In the BigQuery UI, it will display the result as id and time.ts, where the latter indicates that ts is inside a STRUCT named time.
BigQuery disallows columns in the result whose names include periods, so you'll get an error if you run the following query:
SELECT
id,
time_ts AS `time.ts`
FROM `bigquery-public-data.hacker_news.comments`
LIMIT 10;
Invalid field name "time.ts". Fields must contain only letters, numbers, and underscores, start with a letter or underscore, and be at most 128 characters long.
Elliot's answer great and addresses first part of your question, so let me address second part of it (as it is quite different)
First, wanted to mention that select modifiers like SELECT * EXCEPT are supported for BigQuery Standard SQL so, instead of
SELECT * EXCEPT(detected_circle.center_x )
FROM [bigquery-public-data:eclipse_megamovie.photos_v_0_2]
LIMIT 10
you should rather tried
#standardSQL
SELECT * EXCEPT(detected_circle.center_x )
FROM `bigquery-public-data.eclipse_megamovie.photos_v_0_2`
LIMIT 10
and of course now we are back to issue with `using period in standard sql
So, above code can only be interpreted as you try to eliminate center_x field from detected_circle STRUCT (nullable record). Technically speaking, this makes sense and can be done using below code
SELECT *
REPLACE(STRUCT(detected_circle.radius, detected_circle.center_y ) AS detected_circle)
FROM `bigquery-public-data.eclipse_megamovie.photos_v_0_2`
LIMIT 10
... still not clear to me how to use your recommendation to remove the entire detected_circle.*
SELECT * EXCEPT(detected_circle)
FROM `bigquery-public-data.eclipse_megamovie.photos_v_0_2`
LIMIT 10
I'm trying to retrieve all columns that start with any non alpha characters in SQlite but can't seem to get it working. I've currently got this code, but it returns every row:
SELECT * FROM TestTable WHERE TestNames NOT LIKE '[A-z]%'
Is there a way to retrieve all rows where the first character of TestNames are not part of the alphabet?
Are you going first character only?
select * from TestTable WHERE substr(TestNames,1) NOT LIKE '%[^a-zA-Z]%'
The substr function (can also be called as left() in some SQL languages) will help isolate the first char in the string for you.
edit:
Maybe substr(TestNames,1,1) in sqllite, I don't have a ready instance to test the syntax there on.
Added:
select * from TestTable WHERE Upper(substr(TestNames,1,1)) NOT in ('A','B','C','D','E',....)
Doesn't seem optimal, but functionally will work. Unsure what char commands there are to do a range of letters in SQLlite.
I used 'upper' to make it so you don't need to do lower case letters in the not in statement...kinda hope SQLlite knows what that is.
try
SELECT * FROM TestTable WHERE TestNames NOT LIKE '[^a-zA-Z]%'
SELECT * FROM NC_CRIT_ATTACH WHERE substring(FILENAME,1,1) NOT LIKE '[A-z]%';
SHOULD be a little faster as it is
A) First getting all of the data from the first column only, then scanning it.
B) Still a full-table scan unless you index this column.