get size of mysql database

 SELECT table_schema "DB Name",

        ROUND(SUM(data_length + index_length) / 1024 / 1024, 1) "DB Size in MB" 

FROM information_schema.tables 

GROUP BY table_schema; 

get the sizes of the tables of a MySQL database

 SELECT

  TABLE_NAME AS `Table`,

  ROUND((DATA_LENGTH + INDEX_LENGTH) / 1024 / 1024) AS `Size (MB)`

FROM

  information_schema.TABLES

WHERE

  TABLE_SCHEMA = "dbname"

ORDER BY

  (DATA_LENGTH + INDEX_LENGTH)

DESC;

Get record counts for all tables in MySQL database

SELECT table_name, table_rows

FROM INFORMATION_SCHEMA.TABLES 

WHERE TABLE_SCHEMA = YOUR SCHEMA;

SQL - Rounding off to 2 decimal places

select 
    round(630/60.0,2), 
    cast(round(630/60.0,2) as numeric(36,2))

Sử dụng để thay đổi string

Substring()

STUFF

STUFF ( character_expression , start , length , replaceWith_expression )  

MySQL WEEK() function

MySQL WEEK() returns the week number for a given date.
The argument allows the user to specify whether the week starts on Sunday or Monday and whether the return value should be in the range from 0 to 53 or from 1 to 53. If no argument is included with the function, it returns the default week format.
Syntax :

WEEK(date[,mode]);

Arguments :

Name	Description
date	A date value.
mode	An integer indicating the starting of the week.

The following table describes how the mode arguments work :

Mode	First day of week	Range	Week 1 is the first week …
0	Sunday	0-53	with a Sunday in this year
1	Monday	0-53	with more than 3 days this year
2	Sunday	1-53	with a Sunday in this year
3	Monday	1-53	with more than 3 days this year
4	Sunday	0-53	with more than 3 days this year
5	Monday	0-53	with a Monday in this year
6	Sunday	1-53	with more than 3 days this year
7	Monday	1-53	with a Monday in this year

SQL Select only rows with Max Value on a Column

I have this table for documents (simplified version here):

+------+-------+--------------------------------------+
| id   | rev   | content                              |
+------+-------+--------------------------------------+
| 1    | 1     | ...                                  |
| 2    | 1     | ...                                  |
| 1    | 2     | ...                                  |
| 1    | 3     | ...                                  |
+------+-------+--------------------------------------+

How do I select one row per id and only the greatest rev?
With the above data, the result should contain two rows: [1, 3, ...] and [2, 1, ..]. I'm using MySQL.
Currently I use checks in the while loop to detect and over-write old revs from the resultset. But is this the only method to achieve the result? Isn't there a SQL solution?
Update
As the answers suggest, there is a SQL solution, and here a sqlfiddle demo.
Update 2
I noticed after adding the above sqlfiddle, the rate at which the question is upvoted has surpassed the upvote rate of the answers. That has not been the intention! The fiddle is based on the answers, especially the accepted answer.
 

----------------------------

At first glance...

All you need is a GROUP BY clause with the MAX aggregate function:

SELECT id, MAX(rev)
FROM YourTable
GROUP BY id

It's never that simple, is it?

I just noticed you need the content column as well.
This is a very common question in SQL: find the whole data for the row with some max value in a column per some group identifier. I heard that a lot during my career. Actually, it was one the questions I answered in my current job's technical interview.
It is, actually, so common that StackOverflow community has created a single tag just to deal with questions like that: greatest-n-per-group.
Basically, you have two approaches to solve that problem:

Joining with simple group-identifier, max-value-in-group Sub-query

In this approach, you first find the group-identifier, max-value-in-group (already solved above) in a sub-query. Then you join your table to the sub-query with equality on both group-identifier and max-value-in-group:

SELECT a.id, a.rev, a.contents
FROM YourTable a
INNER JOIN (
    SELECT id, MAX(rev) rev
    FROM YourTable
    GROUP BY id
) b ON a.id = b.id AND a.rev = b.rev

Left Joining with self, tweaking join conditions and filters

In this approach, you left join the table with itself. Equality, of course, goes in the group-identifier. Then, 2 smart moves:

1. The second join condition is having left side value less than right value
2. When you do step 1, the row(s) that actually have the max value will have NULL in the right side (it's a LEFT JOIN, remember?). Then, we filter the joined result, showing only the rows where the right side is NULL.

So you end up with:

SELECT a.*
FROM YourTable a
LEFT OUTER JOIN YourTable b
    ON a.id = b.id AND a.rev < b.rev
WHERE b.id IS NULL;

Conclusion

Both approaches bring the exact same result.
If you have two rows with max-value-in-group for group-identifier, both rows will be in the result in both approaches.
Both approaches are SQL ANSI compatible, thus, will work with your favorite RDBMS, regardless of its "flavor".
Both approaches are also performance friendly, however your mileage may vary (RDBMS, DB Structure, Indexes, etc.). So when you pick one approach over the other, benchmark. And make sure you pick the one which make most of sense to you.