I have a table with person's information.
I need to extract all person's who have their FirstName column's first character starting with lowercase.
-- Persons
Mike
Peter
andrew
jason
Elena
-- Output
andrew
jason
Thanx in advance. Any suggestion would be helpful.
Here you go...
select *
from PersonsTable
where ascii(left(Name, 1)) between 97 and 122
Something like this, by checking the first character of the string.
SELECT DISTINCT p.FirstName
FROM dbo.tblPersons p
WHERE LEFT(p..FirstName, 1) = LOWER(LEFT(p..FirstName, 1)) Collate SQL_Latin1_General_CP1_CS_AS
I found it on this link
How to find rows that have a value that contains a lowercase letter
IN MS SQL server use the COLLATE clause as mentioned in this post:
http://blog.sqlauthority.com/2007/04/30/case-sensitive-sql-query-search/
you can use IF test not in ( ABCDEF...) in your main programme :
list = { A,B ....
if(! list.contains(name.substring(1)){
}
Related
I have a SQL Table in that i use BETWEEN Operater.
The BETWEEN Operater selects values within range. The values can be numbers, text , dates.
stu_id name city pin
1 Raj Ranchi 123456
2 sonu Delhi 652345
3 ANU KOLKATA 879845
4 K.K's Company Delhi 345546
5 J.K's Company Delhi 123456
I have a query like this:-
SELECT * FROM student WHERE stu_id BETWEEN 2 AND 4 //including 2 & 4
SELECT * FROM `student` WHERE name between 'A' and 'K' //including A & not K
Here My Question is why not including K.
but I want K also in searches.
Don't use between -- until you really understand it. That is just general advice. BETWEEN is inclusive, so your second query is equivalent to:
WHERE name >= 'A' AND
name <= 'K'
Because of the equality, 'K' is included in the result set. However, names longer than one character and starting with 'K' are not -- "Ka" for instance.
Instead, be explicit:
WHERE name >= 'A' AND
name < 'L'
Of course, BETWEEN can be useful. However, it is useful for discrete values, such as integers. It is a bit dangerous with numbers with decimals, strings, and date/time values. That is why I encourage you to express the logic as inequalities.
In supplement to gordon's answer, one way to get what you're expecting is to turn your name into a discrete set of values:
SELECT * FROM `student` WHERE LEFT(name, 1) between 'A' and 'K'
You need to appreciate that K.K's Company is alphabetically AFTER the letter K on its own so it is not BETWEEN, in the same way that 4.1 is not BETWEEN 2 and 4
By stripping it down to just a single character from the start of the string it will work like you expect, but take cautionary note, you should always avoid running functions on values in tables, because if you had a million names, thats a million strings that mysql has to strip out to just the first letter and it might no longer be able to use an index on name, battering the performance.
Instead, you could :
SELECT * FROM `student` WHERE name >= 'A' and name < 'L'
which is more likely to permit the use of an index as you aren't manipulating the stored values before comparing them
This works because it asks for everything up to but not including L.. Which includes all of your names starting with K, even kzzzzzzzz. Numerically it is equivalent to saying number >= 2 and number < 5 which gives you all the numbers starting with 2, 3 or 4 (like the 4.1 from before) but not the 5
Remember that BETWEEN is inclusive at both ends. Always revert to a pattern of a >= b and a < c, a >= c and a < d when you want to specify ranges that capture all possible values
Compare in lexicographical order, 'K.K's Company' > 'K'
We should convert the string to integer. You can try that mysql script with CAST and SUBSTRING. I've updated your script here. It will include the last record as well.
SELECT * FROM student WHERE name CAST(SUBSTRING(username FROM 1) AS UNSIGNED)
BETWEEN 'A' AND 'K';
The script will work. Hope it will helps to you.
Here I've attached my test sample.
Like my table consist of name street and city .
My query is to find customer name whose street address match any string of 1)exactly 3 character 2)at least 3 character
You can use String function - Length.
1)Exactly 3 character
SELECT customer_name,street_address FROM TABLE WHERE LENGTH(street_address ) = 3
2)At least 3 character
SELECT customer_name,street_address FROM TABLE WHERE LENGTH(street_address ) >= 3
Hope this helps.
It should help you: WHERE CustomerName LIKE 'a_%_%' Finds any values that start with "a" and are at least 3 characters in length
How can I get all the values from mysql table field, having more than 10 characters without any special characters (space, line breaks, colons, etc.)
Let's say I have table name myTable and the field I want to get values from is myColumn.
myColumn
--------
1234
------
123 456
------
123:456
-------
1234
5678
--------
123-456
----------------
1234567890123
So here I would like to get all the field values except first one i.e. 1234
Any help is much appreciated.
Thanks
UPDATE:
Sorry if I was unable to give proper description of my problem. I have tried it again:
If there is count of more than 10 characters without punctuation, then retrieve that as well.
Retrieve all the values which have special characters like line break, spaces, etc.
Yes, I have primary key in this table if this helps.
The logic seems to be "more than 10 characters OR has special punctuation":
where length(mycol) > 10 or
mycol regexp '[^a-zA-Z0-9]'
SELECT MyColumn
From MyTable
WHERE MyColumn RLIKE '([a-z0-9].*){10}'
[a-z0-9] matches a normal character.
([a-z0-9].*) matches a normal character followed by anything.
{10} matches the preceding regexp 10 times.
The result is that this matches 10 normal characters with anything between them.
Hi I want to sort a table .The field contains numbers,alphabets and numbers with alphabets ie,
1
2
1a
11a
a
6a
b
I want to sort this to,
1
1a
2
6a
11a
a
b
My code is, SELECT * FROM t ORDER BY CAST(st AS SIGNED), st
But the result is,
a
b
1
1a
2
6a
11a
I found this code in this url "http://www.mpopp.net/2006/06/sorting-of-numeric-values-mixed-with-alphanumeric-values/"
Anyone please help me
This would do your required sort order, even in the presence of 0 in the table;
SELECT * FROM t
ORDER BY
st REGEXP '^[[:alpha:]].*',
st+0,
st
An SQLfiddle to test with.
As a first sort criteria, it sorts anything that starts with a letter after anything that doesn't. That's what the regexp does.
As a second sort criteria it sorts by the numerical value the string starts with (st+0 adds 0 to the numerical part the string starts with and returns an int)
As a last resort, it sorts by the string itself to get the alphabetical ones in order.
You can use this:
SELECT *
FROM t
ORDER BY
st+0=0, st+0, st
Using st+0 the varchar column will be casted to int. Ordering by st+0=0 will put alphanumeric rows at the bottom (st+0=0 will be 1 if the string starts with an alphanumeric character, oterwise it will be 0)
Please see fiddle here.
The reason that you are getting this output is that all the character like 'a', 'b' etc are converted to '0' and if you use order by ASC it will appear at the top.
SELECT CAST(number AS SIGNED) from tbl
is returning
1
2
1
11
0
6
0
Look at this fiddle:- SQL FIDDLE
I did small change in your query -
SELECT *, CAST(st AS SIGNED) as casted_column
FROM t
ORDER BY casted_column ASC, st ASC
this should work.
in theory your syntax should work but not sure why mysql doesn't accept these methods after from tag.
so created temp field and then sorted that one .
This should work as per my experience, and you can check it.
SQL FIDDLE
I'm using this query to getresults from my database:
MATCH(`Text2`) AGAINST ('$s')
I want to get only results when there is a full match of the string, like when on google when you search between quotes "".
How can I do this with Match/MySQL?
EG: Query is "ab cd"
ID | Text
1 ab cd
2 aab cda
3 aab a cd
Row 1 and 2 should be returned
SELECT FROM your_table WHERE Text2 LIKE '%yourstring%';
Try this::
If you need the wild search irrespective to the cases ::
SELECT FROM LOWER(mytable) WHERE LOWER(Text2) LIKE LOWER('%yourstring%');
You can use REGEXP for this purpose too.
SELECT *
FROM `tableName`
WHERE `columnName` REGEXP 'ab cd';