mysql_num_fields

Last updated: Fri, 20 May 2011

mysql_list_tables

(PHP 4, PHP 5)

mysql_list_tables — Lista las tablas de una base de datos MySQL

Descripción

resource mysql_list_tables ( string $database [, resource $link_identifier ] )

Devuelve una lista de el nombre de las tablas de la base de datos MySQL.

Esta función ha sido obsoleta. Es preferible usar mysql_query() para ejecutar el SQL SHOW TABLES [FROM db_name] [LIKE 'pattern'] en su lugar.

Parámetros

database: El nombre de la base de datos
link_identifier: La conexión MySQL. Si el identificador de enlace no se especifica, el último enlace abierto por mysql_connect() es asumido. Si no se encuentra dicho enlace, la función intentará establecer un nuevo enlace como si mysql_connect() fuese invocado sin parámetros. Si no se encuentra o establece una conexión, un error de nivel E_WARNING es generado.

Valores devueltos

El puntero resultante resource en caso de éxito o FALSE en caso de error.

Use la función mysql_tablename() para atravesar este puntero resultante o cualquier otra función para mostrar tablas, como mysql_fetch_array().

Historial de cambios

Versión	Descripción
4.3.7	Esta función quedó obsoleta.

Ejemplos

Example #1 Ejemplo alternativo de mysql_list_tables()


<?php
$dbname = 'mysql_dbname';

if (!mysql_connect('mysql_host', 'mysql_user', 'mysql_password')) {
    echo 'No se pudo conectar a mysql';
    exit;
}

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

if (!$result) {
    echo "DB Error, no se pudo listar las tablas\n";
    echo 'MySQL Error: ' . mysql_error();
    exit;
}

while ($row = mysql_fetch_row($result)) {
    echo "Tabla: {$row[0]}\n";
}

mysql_free_result($result);
?>

Notas

Note:
Por razones de compatibilidad con versiones anteriores, los siguientes alias obsoletos podrían usarse: mysql_listtables()

Ver también

mysql_list_dbs() - Lista de las bases de datos disponibles en un servidor MySQL
mysql_tablename() - Obtiene el nombre de la tabla de un campo

mysql_num_fields

mysql_list_processes

Last updated: Fri, 20 May 2011

add a note User Contributed Notes mysql_list_tables

daniel at eguest dot net 24-Jul-2008 09:13


Just a simple note:





mysql-list-tables change the context of your active database.





for example:


<?php


//set databse context to db1


mysql_select_db("db1");





//list database tables from db2


mysql-list-tables("db2");


?>





-> Now your active database is db2 and not db1. You have to consider using your sql statements.

kroczu at interia dot pl 08-Dec-2006 11:01


<?

// 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."'")))

{

 //...

}



?>

arturclemente at gmail dot com 03-Apr-2006 08:09


<?php



    // Search any world in entire database



    set_time_limit(0);



    $database = 'embratur';

    $word = 'tempo';

    

    echo "Search for word: <b>".$word."</b><br><br>";

    

    $connect = mysql_connect('localhost','root','123456');

    mysql_selectdb( $database, $connect );

    

    $j=0;

    $tables = mysql_list_tables( $database );

    

    while( $line = mysql_fetch_row( $tables ) )

    {

        $fields = mysql_list_fields( $database, $line[0], $connect );

        $columns = mysql_num_fields( $fields );

        

        for ($i = 0; $i < $columns; $i++) {

               $nomField = mysql_field_name( $fields, $i );

            $sql = "SELECT ".$nomField." FROM " . $line[0] .

                " WHERE " .

                    $nomField . " LIKE '%".$word."' LIMIT 1";

            

            $query = mysql_query( $sql );

            if( mysql_num_rows( $query ) > 0)

            {

                $result = mysql_fetch_array( $query );

                echo "Table: <b>".$line[0]."</b><br>";

                echo "Field: <b>".$nomField."</b><br>";

                echo "SQL: <b>".$sql."</b><br>";

                echo "Result: <b>" . str_replace( $word, "<font color='blue'>". $word . "</font>", $result[0] ) . "</b>";

                echo "<br><br>";

                $j++;

            }

        }

    }

    echo "<br><br><br>";    

    echo "Results: ".$j;

    

?>

mrkvomail at centrum dot cz 29-Jan-2006 07:48


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));

}

Anonymous 15-Dec-2005 02:38


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)) { die ('Error getting table list (' . $sql . ' :: ' . mysql_error() . ')'); }



// Make the list of tables an array

$tablerow = array();

while ($row = mysql_fetch_array($result)) { $tablerow[] = $row; }



// 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...

    $sql = "SHOW TABLE STATUS LIKE '{$tablerow[$i][0]}';";

    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];

}



// Function to calculate size of the file

function c2s($bs) {

         if ($bs < 964)     { return round($bs)           . " Bytes"; }

    else if ($bs < 1000000) { return round($bs/1024,2)    . " KB"   ; }

    else                    { return round($bs/1048576,2) . " MB"   ; }

}



// Echo the result!!!!!!!!!

echo "{$total_rows} rows in {$total_tables} tables";

echo "<br>Average size in each row: " . c2s($total_rows_average/$total_tables);

echo "<br>Average size in each table: " . c2s($sizeo/$total_tables);

echo "<br>Database size: " . c2s($sizeo);



// Close the connection

mysql_close();

?>

sindijs at apollo dot lv 06-Sep-2005 12:21


Get next auto_increment value:

<?

 

  mysql_connect("localhost", "login", "passwd") or

         die("Cannot connect: " . mysql_error());

    

    mysql_select_db("db_name");

 

  $res = mysql_query("SHOW TABLE STATUS LIKE 'table_name' ") or die(mysql_error());

 

  $row_res = mysql_fetch_array($res);

  echo $row_res[Auto_increment];

 ?>

daveheslop (dave heslop) 07-May-2005 11:45


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.

wbphfox at xs4all dot nl 09-Sep-2003 04:55


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



<?php



echo "<p align=\"left\">";

//this is the connection file for the database....

$connectfile = "connect.php";

require $connectfile;



$dbname = 'DATABASE NAME';



$result = mysql_list_tables($dbname);



echo "<table width=\"75%\" border=\"0\">";

echo  "<tr bgcolor=\"#993333\"> ";

echo    "<td><font face=\"Verdana, Arial, Helvetica, sans-serif\" size=\"-1\" color=\"#FFFFFF\">Table name:</font></td>";

echo    "<td><font face=\"Verdana, Arial, Helvetica, sans-serif\" size=\"-1\" color=\"#FFFFFF\">Delete?</font></td>";

echo  "</tr>";

  

    if (!$result) {

        print "DB Error, could not list tables\n";

        print 'MySQL Error: ' . mysql_error();

        exit;

    }

    

    while ($row = mysql_fetch_row($result)) {

        echo "<tr bgcolor=\"#CCCCCC\">";

echo    "<td>";

           print "$row[0]\n";

echo    "</td>";



echo    "<td>";

echo    "<a href=\"$PHP_SELF?action=delete&table=";

         print "$row[0]\n";

echo    "\">Yes?</a>";



echo    "</td>";



echo "</tr>";

        

        

    }



    mysql_free_result($result);



//Delete

if($action=="delete")

{

$deleteIt=mysql_query("DROP TABLE $table"); 

if($deleteIt)

{

echo "The table \"";

echo "$table\" has been deleted with succes!<br>";

}

else

{

echo "An error has occured...please try again<br>";

}

}

  

?>

thebitman at attbi dot com 07-May-2003 08:49


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!

daevid at daevid dot com 17-Dec-2002 12:36


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:



$V2DB = "V2_SL".$CompanyID;



$result = mysql_create_db($V2DB, $linkI);

if (!$result) $errorstring .= "Error creating ".$V2DB." database<BR>\n".mysql_errno($linkI).": ".mysql_error($linkI)."<BR>\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."<BR>\n";

    $tresult = mysql_query($tsql,$linkI);

    if (!$tresult) $errorstring .= "Error creating ".$V2DB.".".$row[0]." table<BR>\n".mysql_errno($linkI).": ".mysql_error($linkI)."<BR>\n"; 

}

cdarklock at darklock dot com 06-Dec-2002 04:03


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.

30-Oct-2002 01:42


<?

/*

   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;

    }

  }



?>



<?

/*

   Implementation example

*/



  $tmp = getdbsize("DATABASE_NAME");

  if (!$tmp) { echo "ERROR!"; }

  else { echo $tmp; }

?>

mail at thomas-hoerner dot de 04-Oct-2002 04:24


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



Tables_in_xxxxx



where xxxxx is the name of the database.



i.e. use 



$result = mysql_list_tables($dbname);

$varname="Tables_in_".$dbname;

while ($row = mysql_fetch_object($result)) {

   echo $row->$varname; 

};

NewToPHP_Guy at Victoria dot NOSPAM dot com 02-Oct-2002 09:06


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:



return(in_array(strtolower($tableName), $tables));

coffee at hayekheaven dot net 18-Jun-2002 12:48


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)");

}

add a note