Select data from “show tables” MySQL query

问题:

Is it possible to select from show tables in MySQL?

SELECT * FROM (SHOW TABLES) AS `my_tables`

Something along these lines, though the above does not work (on 5.0.51a, at least).

回答1:

I think you want SELECT * FROM INFORMATION_SCHEMA.TABLES

See http://dev.mysql.com/doc/refman/5.0/en/tables-table.html

回答2:

Not that I know of, unless you select from INFORMATION_SCHEMA, as others have mentioned.

However, the SHOW command is pretty flexible, E.g.:

SHOW tables like '%s%'

回答3:

To count:

SELECT COUNT(*) as total FROM (SELECT TABLE_NAME as tab, TABLES.* FROM INFORMATION_SCHEMA.TABLES WHERE TABLE_SCHEMA='database_name' GROUP BY tab) tables;

To list:

SELECT TABLE_NAME as table, TABLES.* FROM INFORMATION_SCHEMA.TABLES WHERE TABLE_SCHEMA='database_name' GROUP BY table;

回答4:

Have you looked into querying INFORMATION_SCHEMA.Tables? As in

SELECT ic.Table_Name,
    ic.Column_Name,
    ic.data_Type,
    IFNULL(Character_Maximum_Length,'') AS `Max`,
    ic.Numeric_precision as `Precision`,
    ic.numeric_scale as Scale,
    ic.Character_Maximum_Length as VarCharSize,
    ic.is_nullable as Nulls, 
    ic.ordinal_position as OrdinalPos, 
    ic.column_default as ColDefault, 
    ku.ordinal_position as PK,
    kcu.constraint_name,
    kcu.ordinal_position,
    tc.constraint_type
FROM INFORMATION_SCHEMA.COLUMNS ic
    left outer join INFORMATION_SCHEMA.key_column_usage ku
        on ku.table_name = ic.table_name
        and ku.column_name = ic.column_name
    left outer join information_schema.key_column_usage kcu
        on kcu.column_name = ic.column_name
        and kcu.table_name = ic.table_name
    left outer join information_schema.table_constraints tc
        on kcu.constraint_name = tc.constraint_name
order by ic.table_name, ic.ordinal_position;

回答5:

I think what you want is MySQL's information_schema view(s): http://dev.mysql.com/doc/refman/5.0/en/tables-table.html

回答6:

SELECT * FROM INFORMATION_SCHEMA.TABLES

That should be a good start. For more, check INFORMATION_SCHEMA Tables.

回答7:

SELECT column_comment FROM information_schema.columns WHERE table_name = 'myTable' AND column_name = 'myColumnName'

This will return the comment on: myTable.myColumnName

回答8:

You may be closer than you think — SHOW TABLES already behaves a lot like SELECT:

$pdo = new PDO("mysql:host=$host;dbname=$dbname",$user,$pass);
foreach ($pdo->query("SHOW TABLES") as $row) {
    print "Table $row[Tables_in_$dbname]\n";
}

回答9:

You can't put SHOW statements inside a subquery like in your example. The only statement that can go in a subquery is SELECT.

As other answers have stated, you can query the INFORMATION_SCHEMA directly with SELECT and get a lot more flexibility that way.

MySQL's SHOW statements are internally just queries against the INFORMATION_SCHEMA tables.

User @physicalattraction has posted this comment on most other answers:

This gives you (meta)information about the tables, not the contents of the table, as the OP intended. – physicalattraction

On the contrary, the OP's question does not say that they want to select the data in all the tables. They say they want to select from the result of SHOW TABLES, which is just a list of table names.

If the OP does want to select all data from all tables, then the answer is no, you can't do it with one query. Each query must name its tables explicitly. You can't make a table name be a variable or the result of another part of the same query. Also, all rows of a given query result must have the same columns.

So the only way to select all data from all tables would be to run SHOW TABLES and then for each table named in that result, run another query.

回答10:

Yes, SELECT from table_schema could be very usefull for system administration. If you have lot of servers, databases, tables... sometimes you need to DROP or UPDATE bunch of elements. For example to create query for DROP all tables with prefix name "wp_old_...":

SELECT concat('DROP TABLE ', table_name, ';') FROM INFORMATION_SCHEMA.TABLES
WHERE table_schema = '*name_of_your_database*'
AND table_name LIKE 'wp_old_%';

回答11:

in MySql 5.1 you can try

show tables like 'user%';

output:

mysql> show tables like 'user%';

+----------------------------+

| Tables_in_test (user%) |

+----------------------------+

| user                       |

| user_password              |

+----------------------------+

2 rows in set (0.00 sec)

回答12:

I don't understand why you want to use SELECT * FROM as part of the statement.

12.5.5.30. SHOW TABLES Syntax

回答13:

You can create a stored procedure and put the table names in a cursor, then loop through your table names to show the data.

Getting started with stored procedure: http://www.mysqltutorial.org/getting-started-with-mysql-stored-procedures.aspx

Creating a cursor: http://www.mysqltutorial.org/mysql-cursor/

For example,

CREATE PROCEDURE `ShowFromTables`()
BEGIN

DECLARE v_finished INTEGER DEFAULT 0;
DECLARE c_table varchar(100) DEFAULT "";

DECLARE table_cursor CURSOR FOR 
SELECT table_name FROM information_schema.tables WHERE table_name like 'wp_1%';

DECLARE CONTINUE HANDLER 
    FOR NOT FOUND SET v_finished = 1;

OPEN table_cursor;

get_data: LOOP

FETCH table_cursor INTO c_table;

IF v_finished = 1 THEN 
LEAVE get_data;
END IF;

SET @s=CONCAT("SELECT * FROM ",c_table,";");

PREPARE stmt FROM @s;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;

END LOOP get_data;

CLOSE table_cursor;

END

Then call the stored procedure:

CALL ShowFromTables();