I have a row who contains this value :
account :
adm.ahrgrst001
adm.ns2dhdujhd
adm.ff2hdjhh
adm.haidhidh103
adm.hshiksh122
adm.cn3ehuioe
i want to extract two different values:
when it ends like adm.hshiksh122 i want to extract hshiksh
and with start with adm.cn3ehuioe i want ehuioe
both without the adm. at the beginning
I have thinked this
IIF(isnumeric(RIGHT (account,3)),LEFT(account,LEN(account)-4),RIGHT (account,LEN(account)-7))
the value that are like adm.cn3ehuioe i got wrong like adm.cn3ehui
and adm.ahrgrst001 is correct ahrgrst
Thanks to everyone who will read
The correct way to get the 2 types of values is with:
account LIKE 'adm.[!0-9][!0-9]#[!0-9]*'
for the values that have 2 letters, 1 digit and letters after the dot, and:
account LIKE 'adm.*###'
for the values that end in 3 digits.
So use this:
SELECT IIF(
account LIKE 'adm.*###' ,
MID(account, 5, LEN(account) - 7),
MID(account, 8)
) AS result
FROM tablename
WHERE account LIKE 'adm.*###' OR account LIKE 'adm.[!0-9][!0-9]#[!0-9]*'
If there are no other values than these 2 types then you may remove the WHERE clause.
Results for your sample data:
result
ahrgrst
dhdujhd
hdjhh
haidhidh
hshiksh
ehuioe
Related
Currently I have this field in my database that is a (16, 10) double. I need to change this to a ( 16, 4 ) double. the only problem is that I have a lot of records and I need to check if some any of those records are using more than 4 decimal places. Something like 1500.3333333. The problem is that by changing just changing the field to (16,4) I get zeros to the right, like 14,5500.
Is there any way, using Mysql that I can check if the double is using more than 4 decimal places that are not 0?
Use CHAR_LENGTH() and LIKE functions, like this:
SELECT * FROM table
WHERE CHAR_LENGTH(column) > 4
AND column NOT LIKE "%0"
I wasn't able to find this anywhere, here's my problem:
I have a string like '1 2 3 4 5' and then I have a mysql table that has a column, let's call it numbers, that look like this:
numbers
1 2 6 8 9 14
3
1 5 3 6 9
7 8 9 23 44
10
I am trying to find the easiest way (hopefully in a single query) to find the rows, where any of the numbers in my search string (1 or 2 or 3 or 4 or 5) is contained in the numbers column. In the give example I am looking for rows with 1,2 and 3 (since they share numbers with my search string).
I am trying to do this with a single query and no loops.
Thanks!
The best solution would be to get rid of the column containing a list of values, and use a schema where each value is in its own row. Then you can use WHERE number IN (1, 2, 3, 4, 5) and join this with the table containing the rest of the data.
But if you can't change the schema, you can use a regular expression.
SELECT *
FROM yourTable
WHERE numbers REGEXP '[[:<:]](1|2|3|4|5)[[:<:]]'
[[:<:]] and [[:<:]] match the beginning and end of words.
Note that this type of search will be very slow if the table is large, because it's not feasible to index it.
Here is a start point (split string function) : http://blog.fedecarg.com/2009/02/22/mysql-split-string-function/ := SplitString(string,delimiter,position)
Create a function so it converts a string to an array := stringSplitted(string,delimiter)
Create a function so it compares two arrays :=arrayIntersect(array1, array2)
SELECT numbers
FROM table
WHERE arrayIntersect(#argument, numbers)
Two function definitions with loops and one single query without any loop
SELECT * FROM MyTable WHERE (numbers LIKE '%1%' OR numbers LIKE '%2%')
or you can also use REGEX something like this
SELECT * FROM events WHERE id REGEXP '5587$'
I've got a database with a column that contains the following data:
aaa-1
aaa-2
aaa-3
...
aaa-10
aaa-11
...
aaa-100
aaa-101
...
aaa-1000
When I query and sort the data in ascending order, I get:
aaa-1
aaa-10
aaa-11
...
aaa-100
aaa-101
...
aaa-1000
...
aaa-2
...
aaa-3
Is this actually the correct (machine) way of sorting? Is the order being screwed up because of the aaa- prefix? How do I go about sorting this the way a human would (ie something that looks like the first snippet)?
P.S. If the problem does lie in the prefix, is there a way to remove it and sort with just the numeric component?
P.P.S. It's been suggested to me that I should just change my data and add leading zeroes like aaa-0001 and aaa-0002, etc. However, I'm loathe to go that method as each time the list goes up an order of 10, I'd have to reformat this column.
Thank you all in advance! :)
You can extract the number part, convert it to numeric data type and then do an ORDER BY:
SELECT mytable.*,
CAST(SUBSTRING_INDEX(mycolumn, '-', - 1) AS UNSIGNED) mycolumnintdata
FROM
mytable
ORDER BY mycolumnintdata;
If there are expressions which does not match number, the CAST function would return 0 and those records would be displayed first. You may handle this separately if needed.
I had a similar issue and the trick that did it for me was this one
*"ORDER BY LENGTH(column_name), column_name
As long as the non-numeric part of the value is the same length, this will sort 1 before 10, 10 before 100, etc."*
as given by Andreas Bergström on this question.
Hope that helps someone.
this is the alphabetical order,
you want numerical order,
for do this you must in the ORDER BY clause
trim the costant "aaa-" part
convert it in number
convert(SUBSTRING(val, 3), integer)
I will give you a sample sorting. Not based on your data sample, but this could help you out.
Say you have data like this :
id
----
1
2
6
10
13
when you do ORDER BY id ASC would return :
id
----
1
10
13
2
6
I suggest, use LPAD.
This query : SELECT LPAD('12',5,'0') return 00012
So when you have table data like I provide above, you can sort them like this :
SELECT * FROM TABLE
ORDER BY LPAD(ID,7,'0') ASC
Based on your data.
SELECT SUBSTR('aaa-100',5,LENGTH('aaa-100') - 3) return 100
So, SELECT LPAD( SUBSTR('aaa-100',5,LENGTH('aaa-100') - 3), 7, '0') return 00000100
So you can combine string function such as SUBSTR and LPAD. Do have any clue now?
I have a csv file that contains phone numbers, some of them have 9 digits and some of them have 10. Is there a command that would allow the transformation of the column such that numbers that have only 9 digits will have a 0 appended in front of the numbers.
For example,
if the column has values "443332332" and "0441223332", I would like to have the value of the one with 9 digits changed to "0443332332"?
Sorry, I should have elaborated.
I was wondering if there was a command to do it in SQLlite easily? I prefer not to use excel to transform the column as if I can get it to working with sqllite it would be so much easier and faster.
A more generic solution would be:
select substr('0000000000'||'1234567', -10, 10) from table_name;
The above query would always return 10 digits and add leading zeroes to the missed out number of digits.
For example, the above query would return : 0001234567
For Update, use
UPDATE TABLE_NAME SET PHONE_NO = substr('0000000000'|| PHONE_NO, -10, 10);
If you're sure that just prepending a zero on strings with length 9 will work for your application, something simple will work:
SELECT CASE WHEN LENGTH(phone_number) = 9 THEN '0'||phone_number
ELSE phone_number
END AS phone_number
FROM your_table
;
You could also update the table, depending on your needs:
UPDATE your_table
SET phone_number = '0'||phone_number
WHERE LENGTH(phone_number) = 9
;
Open the .csv using Excel,
Add a filter to the column,
Sort from A-Z to get all the columns with 9 digits,
Then follow the steps here
http://office.microsoft.com/en-au/excel-help/keep-leading-zeros-in-number-codes-HA010342581.aspx
Let's say I have a column with emails:
test1#test.com
test2#test.com
test3test#tes.com
test123#test.com
test321#test.com
test23test.com
How can I select only those that contains only one digit?
Result should be:
test1#test.com
test2#test.com
test3test#tes.com
I tried:
REGEXP '[[:digit:]]{1}' and REGEXP '[0-9]{1}' but it shows all results that contain AT LEAST one digit
try to use ^[^0-9]*[0-9]{1}[^0-9]*$
If you don't specify number of repetitions, it is one copy by default.
Try something like
REGEXP '(^[0-9]+)([0-9])#.+'