Library code snippets
Get metadata on MySQL databases
Getting information about databases if essential if you want to write generic and scalable applications. This code shows you how to get information such as all databases on the server, all tables in each database and all field and field info for each table. Even if you do not need to build on this code, you might want to copy the code which prints out all databases, tables and field information plus examples. Its a great way to get an overview of the tables you are working on for a project.
<?
//getting metadata on MySQL databases
//output the structure of a table
$connection_1 = mysql_connect("localhost");
$fields = mysql_list_fields("cmphp","rights");
for($i=0;$i<mysql_num_fields($fields);$i++) {
echo mysql_field_name($fields,$i)." (".mysql_field_len($fields,$i).") - ".mysql_field_type($fields,$i)."<br>";
}
mysql_close;
//show the structure of ALL tables in ALL databases on the server
$server_connection_1 = mysql_connect("localhost");
$databases = mysql_query("SHOW DATABASES");
while($database = mysql_fetch_row($databases)) {
echo '<h2>DATABASE: '.$database[0].'</h2>';
$database_connection_1 = mysql_select_db($database[0]);
$tables = mysql_query("SHOW TABLES");
while($table = mysql_fetch_row($tables)){
echo '<table border="1" cellpadding="5" width="500">';
echo '<tr><td colspan="3" bgcolor="silver">TABLE: '.$table[0].'</td></tr>';
$fields = mysql_list_fields($database[0],$table[0]);
for($i=0;$i<mysql_num_fields($fields);$i++) {
echo '<tr>';
echo '<td>'.mysql_field_name($fields,$i)."</td>";
echo '<td>'.mysql_field_len($fields,$i)."</td>";
echo '<td>'.mysql_field_type($fields,$i)."</td>";
echo '</tr>';
}
echo '</table><br>';
}
}
mysql_close;
//show the nice structure of a particular database
$server_connection_1 = mysql_connect("localhost");
$the_database = "cmphp";
echo '<h2>DATABASE: '.$the_database.'</h2>';
$database_connection_1 = mysql_select_db($the_database);
$tables = mysql_query("SHOW TABLES");
while($table = mysql_fetch_row($tables)){
echo '<table border="1" cellpadding="5" width="600">';
echo '<tr><td colspan="4" bgcolor="silver"><b>TABLE: '.$table[0].'</b></td></tr>';
echo '<tr><td bgcolor="silver">NAME</td><td bgcolor="silver">SIZE</td><td bgcolor="silver">TYPE</td><td bgcolor="silver">EXAMPLE</td></tr>';
$fields = mysql_list_fields($the_database,$table[0]);
for($i=0;$i<mysql_num_fields($fields);$i++) {
echo '<tr>';
echo '<td>'.mysql_field_name($fields,$i)."</td>";
echo '<td>'.mysql_field_len($fields,$i)."</td>";
echo '<td>'.mysql_field_type($fields,$i)."</td>";
$rows = mysql_query("SELECT ".mysql_field_name($fields,$i)." FROM ".$table[0]." LIMIT 1");
$row = mysql_fetch_array($rows);
echo '<td bgcolor="eeeeee">'.$row[0].' </td>';
echo '</tr>';
}
echo '</table><br>';
}
mysql_close;
?>
Edward Tanguay updates his personal web site tanguay.info weekly with code, links, quotes and thoughts on web development. Sign up for the free newsletter.
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I try to show my databases but I get this:
Warning: mysqlfetchrow(): supplied argument is not a valid MySQL result resource
I use:
$databases = mysqlquery("SHOW DATABASES", $link);
while($database = mysqlfetch_row($databases))
{
echo '<h2>DATABASE: '.$database[0].'</h2>';
}
??
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