Edit Report a Bug

odbc_data_source

(PHP 4 >= 4.3.0, PHP 5, PHP 7)

odbc_data_source — Returns information about a current connection

설명

array odbc_data_source ( resource $connection_id , int $fetch_type )

This function will return the list of available DSN (after calling it several times).

인수

connection_id: ODBC 접속 식별자. odbc_connect()를 참고하십시오.
fetch_type: The fetch_type can be one of two constant types: SQL_FETCH_FIRST, SQL_FETCH_NEXT. Use SQL_FETCH_FIRST the first time this function is called, thereafter use the SQL_FETCH_NEXT.

반환값

Returns FALSE on error, and an array upon success.

add a note

User Contributed Notes 2 notes

down

angelo [at] spaceblue [dot] com ¶

20 years ago


// Example usage:

// Connect to a ODBC database that exists on your system
$link = odbc_connect("some_dsn_name", "user", "password") or die(odbc_errormsg() );

$result = @odbc_data_source( $link, SQL_FETCH_FIRST );
while($result)
{
    echo "DSN: " . $result['server'] . " - " . $result['description'] . "<br>\n";
    $result = @odbc_data_source( $link, SQL_FETCH_NEXT );
}

odbc_close($link);

down

-1

critmas at hotmail dot com ¶

19 years ago


Thank you Angelo for point out the code.  Though the function doesn't really do what it is supposed to.  It returns the entire list of ODBC DSNs.
If you really are looking to get the server type based on just the ODBC, username and password in an environment where the application needs to be aware of different types of databases, use the following code:

$link2 = odbc_connect($dsn , $DBUser, $DBPwd ) or die(odbc_errormsg() );
$result = @odbc_data_source( $link2, SQL_FETCH_FIRST );
while($result)
{
    if (strtolower($dsn) == strtolower($result['server'])) {
        echo $result['description'] . "<br>\n";
        break;
    }
       else
        $result = @odbc_data_source( $link2, SQL_FETCH_NEXT );
}

odbc_close($link2);

// Hope it saves your precious time

add a note