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.
PHP - Manual: mysql_list_tables
2024-12-21
mysql_list_tables
(PHP 4, PHP 5)
mysql_list_tables — 列出 MySQL 数据库中的表
警告
本函数自 PHP 4.3.0 起已废弃，并且它和整个MySQL 扩展自 PHP 7.0.0 开始被移除。 可以选择出于活跃开发中的MySQLi或PDO_MySQL扩展来作为替代。 参见MySQL：选择 API 指南 来获取更多信息。用以替代本函数的有：
- SQL Query:
SHOW TABLES FROM dbname
说明
mysql_list_tables(string
$database, resource $link_identifier = ?): resource|false列出 MySQL 数据库中的表。
此函数已废弃。推荐使用
mysql_query() 来执行 SQL SHOW TABLES
[FROM db_name] [LIKE 'pattern'] 来实现同样的操作。
参数
-
database -
数据库名称
-
link_identifier -
MySQL 连接。如不指定连接标识，则使用由 mysql_connect() 最近打开的连接。如果没有找到该连接，会尝试不带参数调用 mysql_connect() 来创建。如没有找到连接或无法建立连接，则会生成
E_WARNING级别的错误。
返回值
成功时返回结果指针 resource 或者在失败时返回 false。
可以使用 mysql_tablename() 函数来遍历该结果指针，或者使用任何针对表操作的函数，类似 mysql_fetch_array()。
更新日志
| 版本 | 说明 |
|---|---|
| 4.3.7 | 此函数已废弃。 |
范例
示例 #1 mysql_list_tables() 例子
<?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($result);
?>
注释
注意:
为了向下兼容，可以使用下列已废弃的别名： mysql_listtables()
add a note
User Contributed Notes 13 notes
daveheslop (dave heslop) ¶
17 years ago
NewToPHP_Guy at Victoria dot NOSPAM dot com ¶
19 years ago
The example by PHP-Guy to determine if a table exists is interesting and useful (thanx), except for one tiny detail. The function 'mysql_list_tables()' returns table names in lower case even when tables are created with mixed case. To get around this problem, add the 'strtolower()' function in the last line as follows:
return(in_array(strtolower($tableName), $tables));
bimal at sanjaal dot com ¶
9 years ago
A better alternative to mysql_list_tables() would be the following mysql_tables() function.
<?php
/**
* Better alternative to mysql_list_tables (deprecated)
*/
function mysql_tables($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);
?>
coffee at hayekheaven dot net ¶
20 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)");
}
kroczu at interia dot pl ¶
15 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 '".$table."'")))
{
//...
}
?>
mrkvomail at centrum dot cz ¶
16 years ago
You can also do this with function mysql_query(). It's better because mysql_list_tables is old function and you can stop showing errors.
function mysql_table_exists($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));
}
daevid at daevid dot com ¶
19 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($linkI)."<BR>\n";
}
thebitman at attbi dot com ¶
19 years ago
okay everybody, the fastest, most accurate, safest method:
function mysql_table_exists($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!
cdarklock at darklock dot com ¶
19 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)
{
$notexist=FALSE;
}
While the increase in speed here means virtually nothing to the average user who has a database-driven backend on his personal web site to handle a guestbook and forum that might get a couple hundred hits a week, it means EVERYTHING to the professional who has to handle tens of millions of hits a day... where a single extra millisecond on the query turns into more than a full day of processing time. Developing good habits when they don't matter keeps you from having bad habits when they *do* matter.
Anonymous ¶
19 years ago
<?
/*
Function that returns whole size of a given MySQL database
Returns false if no db by that name is found
*/
function getdbsize($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; }
?>
Anonymous ¶
16 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();
?>
wbphfox at xs4all dot nl ¶
18 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>";
}
}
?>
mail at thomas-hoerner dot de ¶
19 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;
};
官方地址：https://www.php.net/manual/en/function.mysql-list-tables.php
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