Making a variable a variable

Anon

I'm passing a variable from a user/login script, but here is my problem.... I have the username field named "user" and the password field "pass". I have information saved in another file that I have include() with information on people's names. Such as age, date and so on... This information is in an array per user.

ie.

$bobbob[age] = "24";
$bobbob[city] = "sometown";
$bobbob[pass] = "encryptedpass";

I want "bobbob" to be a valid username and to pull the password from $bobbob[pass].

Here is my problem:

the $user is set to bobbob, and I dont know how I can basically get ${$user}[pass] to check on the password.

In TCL I could do a join[$, $user] and get what I want. but join(); for php isn't the same.

Any thoughts? and yes I know, pulling information from a database would be much safer and easier, but I'm just trying this.

Thanks
Craig

Anon

simple: in php, just write

$user_name = 'bobbob';

$$user['age'] = '24';
$$user['city'] = 'sometown';
$$user['pass'] = 'encryptedpass';

print($bobbob['age']); //prints '24'.

That should work.

Anon

that didnt work for me....

I tried what you had exactly... but then changed $$user['age'] to $$user_name['age'] and still didnt work...

I didnt get a parse error or anything, but $bobbob doesnt get set, but I can print $$user_name['age'] just fine.

print($$user_name['age']);

with

print('$bobbob['age']'); it prints that exact phrase.

but it doesnt print anything if I dont use any quote, or double quotes.

print("bobbob['age']"); // returns nothing.
print(bobbob['age']); // returns nothing.

This is My current setup:

$user_name = "bobbob";

$$user_name[age] = "24";

$$user_name[city] = "sometown";

$$user_name[pass] = "encryptedpass";

print("$bobbob[age]");

Anon

Update: This works... but isnt an array.

$user_name = 'bobbob';
$$user_name = '24';

print($bobbob);

now if I can make it an array with keys.