mysql_list_tables

bimal at sanjaal dot com (2013-02-28 09:22:21)

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

<?php
/**
* 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);
?>

daniel at eguest dot net (2008-07-24 02:13:43)

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 (2006-12-08 03:01:52)

<?
// 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."'")))
{
 //...
}

?>

mrkvomail at centrum dot cz (2006-01-29 11:48:21)

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 (2005-12-15 06:38:42)

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 (2005-09-06 05:21:01)

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) (2005-05-07 16:45:32)

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 (2003-09-09 09:55:04)

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 (2003-05-07 01:49:37)

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 (2002-12-17 04:36:55)

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 (2002-12-06 08:03:03)

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.

(2002-10-29 17:42:01)

<?
/*
   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 (2002-10-04 09:24:12)

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 (2002-10-02 14:06:35)

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 (2002-06-17 17:48:37)

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)");
}