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Запрос всех баз mysql php

Содержание
  1. mysql_list_tables
  2. Описание
  3. Список параметров
  4. Возвращаемые значения
  5. Список изменений
  6. Примеры
  7. Примечания
  8. Смотрите также
  9. User Contributed Notes 13 notes

mysql_list_tables

Данная функция объявлена устаревшей в PHP 4.3.0, и, вместе с модулем MySQL, удалена PHP в 7.0.0. Вместо неё используйте активно развивающиеся модули MySQLi или PDO_MySQL. Так же смотрите раздел MySQL: выбор API. Альтернативы для этой функции:

Описание

Возвращает список имён таблиц базы данных MySQL.

Данная функция устарела. Вместо неё рекомендуется использовать mysql_query() с запросом SHOW TABLES [FROM db_name] [LIKE ‘pattern’] .

Список параметров

Соединение MySQL. Если идентификатор соединения не был указан, используется последнее соединение, открытое mysql_connect() . Если такое соединение не было найдено, функция попытается создать таковое, как если бы mysql_connect() была вызвана без параметров. Если соединение не было найдено и не смогло быть создано, генерируется ошибка уровня E_WARNING .

Возвращаемые значения

Дескриптор результата ( resource ) в случае успешного выполнения или false в случае возникновения ошибки.

Используйте функцию mysql_tablename() для работы с результатом запроса, либо любую другую функцию, способную это делать, например, mysql_fetch_array() .

Список изменений

Версия Описание
4.3.7 Функция помечена устаревшей.

Примеры

Пример #1 Пример использования mysql_list_tables()

if (! mysql_connect ( ‘mysql_host’ , ‘mysql_user’ , ‘mysql_password’ )) echo ‘Ошибка подключения к mysql’ ;
exit;
>

$sql = «SHOW TABLES FROM $dbname » ;
$result = mysql_query ( $sql );

if (! $result ) echo «Ошибка базы, не удалось получить список таблиц\n» ;
echo ‘Ошибка MySQL: ‘ . mysql_error ();
exit;
>

while ( $row = mysql_fetch_row ( $result )) echo «Таблица: < $row [ 0 ]>\n» ;
>

Примечания

Замечание:

Для обратной совместимости может быть использован следующий устаревший псевдоним: mysql_listtables()

Смотрите также

  • mysql_list_dbs() — Возвращает список баз данных, доступных на сервере
  • mysql_tablename() — Возвращает имя таблицы, содержащей указанное поле

User Contributed Notes 13 notes

Worth noting for beginners: using a row count to test for the existence of a table only works if the table actually contains data, otherwise the test will return false even if the table exists.

// here is a much more elegant method to check if a table exists ( no error generate)

if( mysql_num_rows( mysql_query(«SHOW TABLES LIKE ‘».$table.»‘»)))
//.
>

The example by PHP-Guy to determine if a table exists is interesting and useful (thanx), except for one tiny detail. The function ‘mysql_list_tables()’ returns table names in lower case even when tables are created with mixed case. To get around this problem, add the ‘strtolower()’ function in the last line as follows:

A better alternative to mysql_list_tables() would be the following mysql_tables() function.

/**
* Better alternative to mysql_list_tables (deprecated)
*/
function mysql_tables ( $database = » )
$tables = array();
$list_tables_sql = «SHOW TABLES FROM < $database >;» ;
$result = mysql_query ( $list_tables_sql );
if( $result )
while( $table = mysql_fetch_row ( $result ))
$tables [] = $table [ 0 ];
>
return $tables ;
>

# Usage example
$tables = mysql_tables ( $database_local );
?>

I was in need of a way to create a database, complete with tables from a .sql file. Well, since PHP/mySQL doesn’t allow that it seems, the next best idea was to create an empty template database and ‘clone & rename it’. Guess what? There is no mysql_clone_db() function or any SQL ‘CREATE DATABASE USING TEMPLATEDB’ command. grrr.

So, this is the hack solution I came up with:

$result = mysql_create_db($V2DB, $linkI);
if (!$result) $errorstring .= «Error creating «.$V2DB.» database\n».mysql_errno($linkI).»: «.mysql_error($linkI).»\n»;

mysql_select_db ($V2DB, $linkI) or die («Could not select «.$V2DB.» Database»);

//You must have already created the «V2_Template» database.
//This will make a clone of it, including data.

$tableResult = mysql_list_tables («V2_Template»);
while ($row = mysql_fetch_row($tableResult))
$tsql = «CREATE TABLE «.$V2DB.».».$row[0].» AS SELECT * FROM V2_Template.».$row[0];
echo $tsql.»\n»;
$tresult = mysql_query($tsql,$linkI);
if (!$tresult) $errorstring .= «Error creating «.$V2DB.».».$row[0].» table\n».mysql_errno($linkI).»: «.mysql_error($linkI).»\n»;
>

okay everybody, the fastest, most accurate, safest method:

function mysql_table_exists($table, $link)
$exists = mysql_query(«SELECT 1 FROM `$table` LIMIT 0», $link);
if ($exists) return true;
return false;
>

Note the «LIMIT 0», I mean come on, people, can’t get much faster than that! 🙂
As far as a query goes, this does absolutely nothing. But it has the ability to fail if the table doesnt exist, and that’s all you need!

Actually, the initially posted SELECT COUNT(*) approach is flawless. SELECT COUNT(*) will provide one and only one row in response unless you can’t select from the table at all. Even a brand new (empty) table responds with one row to tell you there are 0 records.

While other approaches here are certainly functional, the major problem comes up when you want to do something like check a database to ensure that all the tables you need exist, as I needed to do earlier today. I wrote a function called tables_needed() that would take an array of table names — $check — and return either an array of tables that did not exist, or FALSE if they were all there. With mysql_list_tables(), I came up with this in the central block of code (after validating parameters, opening a connection, selecting a database, and doing what most people would call far too much error checking):

if($result=mysql_list_tables($dbase,$conn))
< // $count is the number of tables in the database
$count=mysql_num_rows($result);
for($x=0;$x <$count;$x++)
$tables[$x]=mysql_tablename($result,$x);
>
mysql_free_result($result);
// LOTS more comparisons here
$exist=array_intersect($tables,$check);
$notexist=array_diff($exist,$check);
if(count($notexist)==0)
$notexist=FALSE;
>
>

The problem with this approach is that performance degrades with the number of tables in the database. Using the «SELECT COUNT(*)» approach, performance only degrades with the number of tables you *care* about:

// $count is the number of tables you *need*
$count=count($check);
for($x=0;$x<$count;$x++)
if(mysql_query(«SELECT COUNT(*) FROM «.$check[$x],$conn)==FALSE)
$notexist[count($notexist)]=$check[$x];
>
>
if(count($notexist)==0)
$notexist=FALSE;
>

While the increase in speed here means virtually nothing to the average user who has a database-driven backend on his personal web site to handle a guestbook and forum that might get a couple hundred hits a week, it means EVERYTHING to the professional who has to handle tens of millions of hits a day. where a single extra millisecond on the query turns into more than a full day of processing time. Developing good habits when they don’t matter keeps you from having bad habits when they *do* matter.

Even though php guy’s solution is probably the fastest here’s another one just for the heck of it.
I use this function to check whether a table exists. If not it’s created.

mysql_connect(«server»,»usr»,»pwd»)
or die(«Couldn’t connect!»);
mysql_select_db(«mydb»);

$tbl_exists = mysql_query(«DESCRIBE sometable»);
if (!$tbl_exists) mysql_query(«CREATE TABLE sometable (id int(4) not null primary key,
somevalue varchar(50) not null)»);
>

You can also do this with function mysql_query(). It’s better because mysql_list_tables is old function and you can stop showing errors.

function mysql_table_exists($dbLink, $database, $tableName)
$tables = array();
$tablesResult = mysql_query(«SHOW TABLES FROM $database;», $dbLink);
while ($row = mysql_fetch_row($tablesResult)) $tables[] = $row[0];
if (!$result) >
return(in_array($tableName, $tables));
>

/*
Function that returns whole size of a given MySQL database
Returns false if no db by that name is found
*/

function getdbsize($tdb) $db_host=’localhost’;
$db_usr=’USER’;
$db_pwd=’XXXXXXXX’;
$db = mysql_connect($db_host, $db_usr, $db_pwd) or die («Error connecting to MySQL Server!\n»);
mysql_select_db($tdb, $db);

$sql_result = «SHOW TABLE STATUS FROM » .$tdb;
$result = mysql_query($sql_result);
mysql_close($db);

if($result) $size = 0;
while ($data = mysql_fetch_array($result)) $size = $size + $data[«Data_length»] + $data[«Index_length»];
>
return $size;
>
else return FALSE;
>
>

Getting the database status:
// Get database status by DtTvB
// Connect first
mysql_connect (‘*********’, ‘*********’, ‘********’);
mysql_select_db (‘*********’);

// Get the list of tables
$sql = ‘SHOW TABLES FROM *********’;
if (!$result = mysql_query($sql))

// Make the list of tables an array
$tablerow = array();
while ($row = mysql_fetch_array($result))

// Define variables.
$total_tables = count($tablerow);
$statrow = array();
$total_rows = 0;
$total_rows_average = 0;
$sizeo = 0;

// Get the status of each table
for ($i = 0; $i < count($tablerow); $i++) // Query the status.
$sql = «SHOW TABLE STATUS LIKE »;»;
if (!$result = mysql_query($sql)) < die ('Error getting table status (' . $sql . ' :: ' . mysql_error() . ')'); >
// Get the status array of this table
$table_info = mysql_fetch_array($result);
// Add them to the total results
$total_rows += $table_info[3];
$total_rows_average += $table_info[4];
$sizeo += $table_info[5];
>

// Echo the result.
echo » rows in tables»;
echo «
Average size in each row: » . c2s($total_rows_average/$total_tables);
echo «
Average size in each table: » . c2s($sizeo/$total_tables);
echo «
Database size: » . c2s($sizeo);

// Close the connection
mysql_close();
?>

Here is a way to show al the tables and have the function to drop them.

echo «

» ;
//this is the connection file for the database.
$connectfile = «connect.php» ;
require $connectfile ;

$result = mysql_list_tables ( $dbname );

if (! $result ) print «DB Error, could not list tables\n» ;
print ‘MySQL Error: ‘ . mysql_error ();
exit;
>

//Delete
if( $action == «delete» )
$deleteIt = mysql_query ( «DROP TABLE $table » );
if( $deleteIt )
echo «The table \»» ;
echo » $table \» has been deleted with succes!
» ;
>
else
echo «An error has occured. please try again
» ;
>
>

You can also use mysql_fetch_object if you consider a specialty: The name of the object-var is

where xxxxx is the name of the database.

$result = mysql_list_tables($dbname);
$varname=»Tables_in_».$dbname;
while ($row = mysql_fetch_object($result)) echo $row->$varname;
>;

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