Obtaining the maximum value with Order By and Limit 1 - mysql

First of all, I am using MySQL. When I make the following query:
SELECT CodE,sum(tiempo) AS 'tiempo total'
FROM Participa
GROUP BY CodE
ORDER BY 'tiempo total' DESC LIMIT 1;
it shows me the first line of my table instead of the MAX value. However, If I make the following query:
SELECT CodE,sum(tiempo)
FROM Participa
GROUP BY CodE
ORDER BY 2 DESC LIMIT 1
I get the correct result.
I have just changed the alias 'tiempo total' for somthing that should be equivalent.
How it´s possible?

Only use single quotes for string and date constants -- never for column aliases. So:
SELECT CodE, sum(tiempo) AS `tiempo total`
FROM Participa
GROUP BY CodE
ORDER BY `tiempo total` DESC
LIMIT 1;
You are ordering by a constant string, not the name of a column. Hence, if you get the maximum in your query, it would be a total accident.
Note: You can get around these issues by giving columns names that never need to be escaped:
SELECT CodE, sum(tiempo) AS tiempo_total
FROM Participa
GROUP BY CodE
ORDER BY tiempo_total DESC
LIMIT 1;
Easier to type, too.

Related

Mysql "Like" not return two digit values like LAMO10

I have an mysql query like this:
select empcode from employee where empcode like 'LAMO%' order by empcode desc limit 1;
and it's result like this,
But my actual records is,
I need output like this LAMO10
The field empcode is having a data type corresponding to characters.
Here, LAMO1 is less than LAMO2 and so on, but observe that the string is compared character by character. Therefore, LAMO10 is smaller than LAMO2 because while comparing from the left, the first 4 characters LAMO are equal, the 5th character 1 in LAMO10 is smaller than the 5th character 2 in LAMO2. So, the order that you would get (if you removed the limit in your query) is:
LAMO9
LAMO8
LAMO7
LAMO6
LAMO5
LAMO4
LAMO3
LAMO2
LAMO10
LAMO1
This explains why you aren't getting your desired output LAMO10. To generate it, you need to order by only the numbers in your string. In this particular dataset, that number you are looking for appears to be everything onward from the 5th character. The corresponding query segment for ordering would be:
ORDER BY CAST(SUBSTR(empcode,5) AS UNSIGNED) DESC
So, putting it in your query:
SELECT empcode
FROM employee
WHERE empcode like 'LAMO%'
ORDER BY CAST(SUBSTR(empcode,5) AS UNSIGNED) DESC
LIMIT 1;
should get you your desired result.
You may use an order by clause which sorts on the numeric component of the employee code:
SELECT empcode
FROM employee
WHERE empcode LIKE 'LAMO%'
ORDER BY CAST(REPLACE(empcode, 'LAMO', '') AS UNSIGNED) DESC
LIMIT 1;

MySQL INSERT... SELECT column count and virtual/aliased columns

I'm trying to insert using a select statement. However, I need to order the sub-select results using a ranking equation. If I create an alias, it throws off the column count. Can I somehow order my results using an equation?
INSERT INTO draft
( fk_contrib_id , end_time )
SELECT pk_contrib_id, UNIX_TIMESTAMP(), (X+Y+Z) AS ranking
FROM contrib
ORDER BY ranking DESC
LIMIT 1
I need the 'ranking' column for sorting, but if I do, the column count is off for the insert. Do I have to use two queries for this?
You could simply change your query to directly use the expression in the ORDER BY clause, like so:
INSERT INTO draft
( fk_contrib_id , end_time )
SELECT pk_contrib_id, UNIX_TIMESTAMP()
FROM contrib
ORDER BY (X+Y+Z) DESC
LIMIT 1
Remove the expression from the SELECT list. And use the expression in the ORDER BY clause.
ORDER BY X+Y+Z
It's perfectly valid to ORDER BY expressions that are not in the SELECT list.

MySQL sort by not sorting?

I am fetching rows from my DB using this request:
SELECT * FROM {$db_sales} WHERE date = '{$date}' ORDER BY 'amount' DESC
So, obviously, i expected the returned values to be sorted in descending order by the amount column in my DB, but it doesn't? it still fetches them, but just doesn't sort them?
Any ideas here? is my SQL statement wrong?
remove single quote around amount like this and try:
SELECT * FROM {$db_sales} WHERE date = '{$date}' ORDER BY amount DESC
Use below query
SELECT * FROM {$db_sales} WHERE date = '{$date}' ORDER BY amount DESC
ORDER BY clause uses column name.
Column name should not give in quotes.
there fore the query becomes as follows
SELECT * FROM {$db_sales} WHERE date = '{$date}' ORDER BY amount DESC

SQL First() Function

I am using phpMyAdmin to write some SQL code that I thought was simple but proving to be a headache. I'm using this tutorial to help me out. My goal is to get the first and last columns id's from a result set. When I do this query I get 5 rows starting at 15 and going through 11.
SELECT id
FROM boardPost
WHERE recipientId = 1
ORDER BY id DESC
LIMIT 0,5
However, when I try this query I get an error #1064: "You have an error in your SQL syntax."
SELECT FIRST(id)
FROM boardPost
WHERE recipientId = 1
ORDER BY id DESC
LIMIT 0,5
Something like this maybe?
SELECT min(id), max(id)
from (
select id
from boardPost
where recipientId = 1
order by id desc
limit 0,5
) t
I think that is what you want?
Select id from boardPost order by id asc limit 1
and
Select id from boardPost order by id desc limit 1
If you just want the first and the last id of a result set, you could consider this:
SELECT MIN(id) firstId, MAX(id) lastId FROM someTable WHERE aField = 1;
Note that it'll only work if you do use and ORDER BY an AUTO_INCREMENT field, else you might get unexpected results.
It'll only work with the full set. If you need the first and last id of a limited one, you're probably better of using 2 queries with ASC and DESC order and LIMIT 1.
MySQL does not support the FIRST() function. You will need to use the workaround they specified in that tutorial (using ORDER BY and LIMIT)
In some situations (like mine, that first brought me here), there are some rarely-used MySQL "windowing functions" such as FIRST_VALUE and LAST_VALUE that may provide the functionality you're seeking.
(Here's more info on Window Function Concepts and Syntax, and the Wikipedia description of the term.)

ORDER BY FIELD() and duplicate data

Lets say I have the following query.
SELECT stringdata FROM table ORDER BY FIELDS( stringdata, 'tg','nk','mg','pl') asc;
For some reason I'm getting the results at the very bottom. How can I get the query to put the results starting from 'tg' at row 1 rather than the last row in the results?
Not only that but there's more than one 'tg' in the data, I'd like it to sort it in this expected output:
stringdata
__________
'tg'
'tg'
'tg'
'nk'
'nk'
'mg'
'mg'
'mg'
'pl'
So far using ORDER BY Fields() is only sorting one instance of the data rather than all.
Using desc instead of asc in the query works as expected. I get 'pl' on the first row, then 'mg', 'nk', etc.
Normally the FIELD clause in ORDER BY works something like
SELECT * FROM table ORDER BY FIELD(field, high_priority, second_high,
,....., low_priority);
So in your query, the sorting took place as you mentioned and when you gave the ASC it printed from the lowest_priority. So, for your case, if you want tg at the top, you can either reorder the priority in the FIELDS or as you have already tried use desc
If you want first rows with 'tg', then rows with .... then with 'pl' and then all the rest sorted (ASC or DESC) use this:
SELECT stringdata
FROM table
ORDER BY FIELD( stringdata, 'pl','mg', 'nk', 'tg') DESC
, stringdata ASC --- or DESC