mysql_list_tables

(PHP 4, PHP 5)

mysql_list_tables — List tables in a MySQL database

Avertizare

Această funcție a devenit învechită în PHP 4.3.0, și a fost eliminată în PHP 7.0.0, împreună cu întreaga extensie MySQL originală. În locul ei trebuie utilizată extensia activ dezvoltată MySQLi sau PDO_MySQL. Accesați de asemenea ghidul MySQL: selectarea unei API și FAQ asociat pentru informații suplimentare. Variante alternative pentru această funcție includ:

SQL Query: SHOW TABLES FROM dbname

Descrierea

mysql_list_tables ( string $database , resource $link_identifier = NULL ) : resource|false

Retrieves a list of table names from a MySQL database.

This function is deprecated. It is preferable to use mysql_query() to issue an SQL SHOW TABLES [FROM db_name] [LIKE 'pattern'] statement instead.

Parametri

database: The name of the database
link_identifier: Conexiunea MySQL. Dacă identificatorul legăturii nu este specificat, se presupune că este ultima legătură deschisă cu ajutorul mysql_connect(). Dacă nu este găsită nici o astfel de legătură, se va încerca crearea uneia prin apelul mysql_connect () fără argumente. În caz că nici o conexiune nu este găsită sau stabilită, se va genera o eroare de nivelul E_WARNING.

Valorile întoarse

A result pointer resource on success sau false în cazul eșecului.

Use the mysql_tablename() function to traverse this result pointer, or any function for result tables, such as mysql_fetch_array().

Istoricul schimbărilor

Versiune	Descriere
4.3.7	This function became deprecated.

Exemple

Example #1 mysql_list_tables() alternative example


<?php
$dbname = 'mysql_dbname';

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

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

if (!$result) {
    echo "DB Error, could not list tables\n";
    echo 'MySQL Error: ' . mysql_error();
    exit;
}

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

mysql_free_result($result);
?>

Note

Notă:
Pentru compatibilitatea cu versiunile anterioare, următorul pseudonim învechit poate fi utilizat: mysql_listtables()

A se vedea și

mysql_list_dbs() - List databases available on a MySQL server
mysql_tablename() - Get table name of field

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User Contributed Notes 13 notes

down

daveheslop (dave heslop) ¶

19 years ago


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.

down

kroczu at interia dot pl ¶

18 years ago


<?
// 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."'")))
{
 //...
}

?>

down

-1

daevid at daevid dot com ¶

22 years ago


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"; 
}

down

-1

NewToPHP_Guy at Victoria dot NOSPAM dot com ¶

22 years ago


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));

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-2

bimal at sanjaal dot com ¶

11 years ago


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);
?>

down

-2

thebitman at attbi dot com ¶

21 years ago


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!

down

-3

cdarklock at darklock dot com ¶

22 years ago


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.

down

-3

mrkvomail at centrum dot cz ¶

18 years ago


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));
}

down

-3

Anonymous ¶

22 years ago


<?
/*
   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; }
?>

down

-3

Anonymous ¶

19 years ago


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();
?>

down

-5

wbphfox at xs4all dot nl ¶

21 years ago


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>";
}
}
  
?>

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-4

coffee at hayekheaven dot net ¶

22 years ago


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)");
}

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-6

mail at thomas-hoerner dot de ¶

22 years ago


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; 
};

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