Duplicate Variable Name

The_PHP_Newb

This one has me stumped.

In my database, I have a column named varname and the value is lets say '$pay_amount'.

This is a variable customers will setup on our site. 'varname' might be different for each user.

Now, we have a page the customer goes to that passes a value such as 5.95 and this variable name is the one defined in our database (varname) '$pay_amount'.

I need to say:
$varname = 5.95;

But, obviously $varname is going to = "$pay_amount" (bacause this value is setup in the database and $varname will be reassigned when we query it. It goes from 5.95 (passed into the page) then is reassigned to "pay_amount" when I query the database to get the name to look for.

Hope this makes sense, can anyone help?

yelvington

In English, what are you trying to do?

The_PHP_Newb

I have my user enter a variable NAME into my database so we know what the name is. Example: $total_amount

Later, when a certain page is hit... the page will already include this variable name and value in it (comes from the customer). Example: $total_amount=5.50

Now we need to write 5.50 to our database but we need to ask our own database what the variable name is to look for first.

Very basic (incoming page) example
(http://www.yoursite.com/index.php?total_amount=5.50):

On this page:
We already know $total_amount equals 5.50 but we don't know to look for the name $total_amount so I query our database.

$getname = mysql_query("select varname from config");
$name = mysql_fetch_array($getname);
$actualvariablename=$name[varname];

Now I'm stuck, how can I make $result = 5.50 now that I know the variable name to look for. I keep overwriting the 5.50 with the actual data that is already present in the database.

swr

$result = $_GET[$actualvariablename];

Is that what you want? If you don't want to pull from $_GET you can use variable variables:

$result = ${$actualvariablename};

Using variable variables is described here:
http://www.php.net/manual/en/language.variables.variable.php

By the way, it looks like your code depends on [man]register_globals[/man], which is not recommended. Be sure to read the [man]register_globals[/man] manual page if you haven't already.

The_PHP_Newb

Completely baffled now. Could you provide a code snippet for the following?

index.php?total_sale=5.00

$getname = mysql_query("select varname from config");
$name = mysql_fetch_array($getname);
$actualvariablename=$name[varname];

Note: $actualvariablename equals "$total_sale" in the database.

I've looked all the resource boards and php.net, etc. I just can't figure this one out.

Moonglobe

make them not enter the $ with the variable name. then just use:

 $$varname = 'blah';

if $varname == 'pay_amount' then $$varname == '$pay_amount' so what you're really saying is $pay_amount = 'blah'; , except that the 'pay_amount' can change depending on $varname

swr

What Moonglobe said. If you can't get rid of the $ from the database, strip it off after you pull it out of the DB. Or, use [man]eval[/man]:

$result = eval("return $actualvariablename;");

With that, assuming $actualvariablename = "$total_sale" then the string passed to eval will become "return $total_sale;" which is evaluated, and returns $total_sale into $result.

Be aware that there can be security issues using eval. If $actualvariablename = "unlink('/some/very/important/file')" or something like that you'd be in trouble.

Using eval can sometimes result in very convoluted code. It's almost always better to avoid it, though sometimes there is no other reasonable way.

The_PHP_Newb

Excellent, I'm going to give this a try this afternoon, I'll report back with the result.

Thanks much for the replys.

The_PHP_Newb

Using $total_sale as the database entry, I can't get this to work:

$a = mysql_query("select varname from config");
$b = mysql_fetch_array($a);

$varname = $b[varname];

Result Is Displayed As: $total_sale

HOWEVER, manually entering it like this WORKS:

$varname = "$total_sale";

Result Is Displayed As: 10.00 (CORRECT!)

What is the difference here? How can I get the above code to produce 10.00?

The page this code is on is this:
index.php?total_sale=10.00

I can remove the $ from the database entry but couldn't get it to work with the above given examples.

Moonglobe

after $varname = $b['varname']; put $varname = $$varname;

The_PHP_Newb

$a = mysql_query("select varname from config");
$b = mysql_fetch_array($a);
$varname = $b['varname'];
$varname = $$varname;
print $varname;

Like this? The output is nothing now. No $total_sale or 10.00.

The_PHP_Newb

I actually figured this out... THANKS TO ALL THAT HELPED! This board is the BEST!