mysql_list_tables

(PHP 3, PHP 4, PHP 5)

mysql_list_tables -- Возвращает список таблиц базы данных MySQL

Описание

resource mysql_list_tables ( string database [, resource link_identifier] )

mysql_list_tables(), принимая параметром имя базы данных, возвращает указатель на результат (аналогично mysql_query()), содержащий список её таблиц. Используйте функцию mysql_tablename() для обработки полученного указателя, либо любую другую функцию для обработки результатов, такую как mysql_fetch_array().

Параметр database -- имя базы данных, список таблиц которой будет возвращён. В случае ошибки, функция возвращает FALSE.

Для совместимости, как алиас, доступна устаревшая функция mysql_listtables(). Однако, использовать её крайне не рекомендуется.

Замечание: Функция устарела и не должна быть использована. Используйте SQL-запрос SHOW TABLES FROM DATABASE вместо неё.

Пример 1. Пример использования mysql_list_tables()

<?php
   $dbname
= 'mysql_dbname';

   if (!
mysql_connect('mysql_host', 'mysql_user', 'mysql_password')) {
       print
'Could not connect to mysql';
       exit;
   }

  
$result = mysql_list_tables($dbname);
  
   if (!
$result) {
       print
"DB Error, could not list tables\n";
       print
'MySQL Error: ' . mysql_error();
       exit;
   }
  
   while (
$row = mysql_fetch_row($result)) {
       print
"Table: $row[0]\n";
   }

  
mysql_free_result($result);
?>

См. также mysql_list_dbs() и mysql_tablename().



mysql_list_tables
arturclemente at gmail dot com
03-Apr-2006 01:09
<?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 11:48
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 06:38
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 05:21
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 04:45
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 09:55
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 01:49
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 04:36
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 08:03
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.
29-Oct-2002 05:42
<?
/*
   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 09:24
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 02:06
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
17-Jun-2002 05:48
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)");
}

<mysql_list_processesmysql_num_fields>
 Last updated: Tue, 15 Nov 2005