PHP 4.1.0 and MySQL 3.23.41-log

The function mysql_select_db set the db for other link
identifiers than specified.

Doc:
bool mysql_select_db (string database_name, resource [link_identifier])
mysql_select_db() sets the current active database on the server that's associated with the specified link identifier. If no link identifier is specified, the last opened link is assumed. If no link is open, the function will try to establish a link as if mysql_connect() was called without arguments, and use it. 

The table my_table is in db1 but the active database
seems to be test on link #8.

Here the result:
mysql_pconnect() id1=Resource id #8
mysql_select_db(db1,Resource id #8)
mysql_pconnect() id2=Resource id #9
mysql_select_db(test,Resource id #9)
mysql_query(select * from my_table,Resource id #8)
failed: Table 'test.my_table' doesn't exist

Here the source:
$id1=mysql_pconnect($host,$user,$pass);
if($id1==false){
	die("<br>mysql_pconnect() failed: ".mysql_error());
}
echo "<br>mysql_pconnect() id1=$id1";

$b=mysql_select_db($db1,$id1);
if($b==false){
	die("<br>mysql_select_db($db1,$id1)
failed:".mysql_error());
}
echo "<br>mysql_select_db($db1,$id1)";

$id2=mysql_pconnect($host,$user,$pass);
if($id2==false){
	echo "<br>mysql_pconnect() failed: ".mysql_error();
}
echo "<br>mysql_pconnect() id2=$id2";

$b=mysql_select_db($db2,$id2);
if($b==false){
	die("<br>mysql_select_db($db2,$id2)
failed:".mysql_error());
}
echo "<br>mysql_select_db($db2,$id2)";

$sql='select * from my_table';
$result=mysql_query($sql,$id1);
if($result==false){
	die("<br>mysql_query($sql,$id1)
failed:".mysql_error());
}