mysql_tablename

(PHP 4, PHP 5)

mysql_tablenameGet table name of field

Avviso

Questa enstensione deprecata da PHP 5.5.0, e sarĂ  rimossa in futuro. Al suo posto, usare l'estensione MySQLi o PDO_MySQL. Vedere anche la guida MySQL: scelta dell'API e le FAQ relative per ulteriori informazioni. Le alternative a questa funzione includono:

  • SQL Query: SHOW TABLES

Descrizione

mysql_tablename(resource $result, int $i): string|false

Retrieves the table name from a result.

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

Elenco dei parametri

result

A result pointer resource that's returned from mysql_list_tables().

i

The integer index (row/table number)

Valori restituiti

The name of the table on success o false in caso di fallimento.

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

Esempi

Example #1 mysql_tablename() example

<?php
mysql_connect
("localhost", "mysql_user", "mysql_password");
$result = mysql_list_tables("mydb");
$num_rows = mysql_num_rows($result);
for (
$i = 0; $i < $num_rows; $i++) {
echo
"Table: ", mysql_tablename($result, $i), "\n";
}

mysql_free_result($result);
?>

Note

Nota:

The mysql_num_rows() function may be used to determine the number of tables in the result pointer.

Vedere anche:

add a note add a note

User Contributed Notes 2 notes

up
5
Haseldow
20 years ago
Another way to check if a table exists:

if(mysql_num_rows(mysql_query("SHOW TABLES LIKE '".$table."'"))==1) echo "Table exists";
else echo "Table does not exist";
up
-15
pl at thinkmetrics dot com
20 years ago
A simple function to check for the existance of a table:

function TableExists($tablename, $db) {
   
    // Get a list of tables contained within the database.
    $result = mysql_list_tables($db);
    $rcount = mysql_num_rows($result);

    // Check each in list for a match.
    for ($i=0;$i<$rcount;$i++) {
        if (mysql_tablename($result, $i)==$tablename) return true;
    }
    return false;
}
To Top