creating variable names using eval?

Anon

Hi,

I was wondering if it is possible to create and use a variable name. For example, here's a piece of code that I'm really not using, but it'll show you what I'm trying to do:

$bg1type = "type1";
$bg2type = "type2";
$bg3type = "type3";
$bg4type = "type4";
$bg5type = "type5";

for($i = 1, $i < 6, $i++) {
$var = eval("\$bg$itype");
echo $var;
}

I'm just wondering how eval can figure out in the middle of that that the variable I'm talking about is $i, not $itype or $it or something like that. Just curious as to if and how this would work...

Anon

PHP can't say what variable you want. This would be interpreted as $itype and, therefore, as an empty string (since this variable does not exist). Use ${i}type instead.

Anon

One more thing - wouldn't array be more appropriate for something like this?

Also, if you insist, this would work as well: echo ${"bg${i}type"};

Anon

Would it be possible to set a variable like so:

$bg1type = "type1";

$i = 1;

$name = "bg".$i."type"; // $name now = "bg1type"

eval("\$$name");

to have eval return the contents of the variable $bg1type? I've tried it, and it didn't work. But then again, maybe I'm doing something wrong... again, just curious

Anon

Yes, an array would work better for the example I put together here... but for what I'm actually working with, an array would be hard to put together. You'd just have to take a look at my code, which is pretty long. But you're right, if I was doing something like that, using an array would be the way to go.

Anon

For this construction check my other post.

If you want it THIS way, here is correction:
eval("\$t = \$$name;");
Now, $t holds the value you of $bg1type.

Anon

Awesome. Thanks a million

Anon

You needn't, in fact, mess with eval() at all if you don't want to. OMG! TMTWODI! In PHP!!! =)

$foo = "bar";

$bar = "baz";

print $$foo; // prints "baz"

print ${$foo}; // also prints "baz"

You might need the second sort of syntax if you are going to use $$foo within, say double quotes (the curly braces disambiguate).

Similarly, you could load up an array like this:

$bgtype = array ("type1", "type2");
$type0= "foo";
$type1="bar";

foreach ($bgtype as $this_one) {
print $$this_one . "\n";
}
// prints "foo\nbar\n"

HTH,