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(PHP 4, PHP 5)
mysql_list_tables — 列出 MySQL 数据库中的表
$database
[, resource $link_identifier
] )mysql_list_tables() 接受一个数据库名并返回和 mysql_query() 函数很相似的一个结果指针。用 mysql_tablename() 函数来遍历此结果指针,或者任何使用结果表的函数,例如 mysql_fetch_array() 。
database
参数是需要被取得其中的的表名的数据库名。如果失败
mysql_list_tables() 返回 FALSE
。
为向下兼容仍然可以使用本函数的别名 mysql_listtables() ,但反对这样做。
Note: 该函数已经被删除了,请不要再使用该函数。您可以用命令 SHOW TABLES FROM DATABASE 来实现该函数的功能。
Example #1 mysql_list_tables() 例子
<?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 ( $result );
?>
参见 mysql_list_dbs() 和 mysql_tablename() 。
[#1] bimal at sanjaal dot com [2013-02-28 09:22:21]
A better alternative to mysql_list_tables() would be the following mysql_tables() function.
<?php
function mysql_tables($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);
?>
[#2] kroczu at interia dot pl [2006-12-08 03:01:52]
<?php
// 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 '".$table."'")))
{
//...
}
?>
[#3] mrkvomail at centrum dot cz [2006-01-29 11:48:21]
You can also do this with function mysql_query(). It's better because mysql_list_tables is old function and you can stop showing errors.
function mysql_table_exists($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));
}
[#4] Anonymous [2005-12-15 06:38:42]
Getting the database status:
<?php
// 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();
?>
[#5] 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.
[#6] 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 "$table\" has been deleted with succes!<br>";
}
else
{
echo "An error has occured...please try again<br>";
}
}
?>
[#7] thebitman at attbi dot com [2003-05-07 01:49:37]
okay everybody, the fastest, most accurate, safest method:
function mysql_table_exists($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!
[#8] 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";
}
[#9] 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)
{
$notexist=FALSE;
}
While the increase in speed here means virtually nothing to the average user who has a database-driven backend on his personal web site to handle a guestbook and forum that might get a couple hundred hits a week, it means EVERYTHING to the professional who has to handle tens of millions of hits a day... where a single extra millisecond on the query turns into more than a full day of processing time. Developing good habits when they don't matter keeps you from having bad habits when they *do* matter.
[#10] [2002-10-29 17:42:01]
<?php
function getdbsize($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;
}
}
?>
<?php
$tmp = getdbsize("DATABASE_NAME");
if (!$tmp) { echo "ERROR!"; }
else { echo $tmp; }
?>
[#11] 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;
};
[#12] 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));
[#13] 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)");
}