• 首页
  • vue
  • TypeScript
  • JavaScript
  • scss
  • css3
  • html5
  • php
  • MySQL
  • redis
  • jQuery
  • mysql_list_tables()

    (PHP 4, PHP 5)

    列出 MySQL 数据库中的表

    说明

    mysql_list_tables(string $database[,resource $link_identifier]): resource

    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()。

    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), $tables));
    A better alternative to mysql_list_tables() would be the following mysql_tables() function.
    <?php
    /**
    * Better alternative to mysql_list_tables (deprecated)
    */
    function mysql_tables($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);
    ?>
    
    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.
    <?
    // 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."'")))
    {
     //...
    }
    ?>
    
    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!
    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)");
    }
    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"; 
    }
    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) 
    {
      $notexist=FALSE;
    }
    While the increase in speed here means virtually nothing to the average user who has a database-driven backend on his personal web site to handle a guestbook and forum that might get a couple hundred hits a week, it means EVERYTHING to the professional who has to handle tens of millions of hits a day... where a single extra millisecond on the query turns into more than a full day of processing time. Developing good habits when they don't matter keeps you from having bad habits when they *do* matter.
    You can also do this with function mysql_query(). It's better because mysql_list_tables is old function and you can stop showing errors.
    function mysql_table_exists($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));
    }
    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();
    ?>
    
    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>";
    }
    }
     
    ?>
    
    <?
    /*
      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; }
    ?>
    
    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; 
    };