question about php eval() function

i 开发者_开发技巧defined a couple of variables: $day_0, $day_1...$day_8 and need to build a string with those days variable embedded in the string, here is what i did;

for($i = 0; $i <=8; $i++) {
       $d = 'day_'.$i;
       $day = eval($($d));echo $day;
       $cmd_line .= 'INPUT'.$i.'='.$quote.$day.$quote.$space;
}

but php always complained

syntax error, unexpected '(', expecting T_VARIABLE or '$' 

what's wrong with it? Thanks.

No need to use eval:

for($i = 0; $i <=8; $i++) {
       $day = ${'day_' . $i};
       echo $day;
       $cmd_line .= 'INPUT'.$i.'='.$quote.$day.$quote.$space;
}

See the PHP Documentation for more information on variable variables.

A safer and easier solution would be to use:

$d = 'day_'.$i;
$day = $$d;

Why not just use $$ ?

for($i = 0; $i <=8; $i++) {
       $d = 'day_'.$i;
       $day = $$d; //this will point to $day_0 ...
       echo $day;
       $cmd_line .= 'INPUT'.$i.'='.$quote.$day.$quote.$space;
}

I don't suggest eval() but if you want to use it pass valid php code as eval() parameter (string).

Example:

$d = 'day_1';
$phpcode = "\${$d}='something';";
$day = eval($phpcode);
echo $day_1;

This will print out string something assigned to new variable $day_1.

Don't forget to add ; at the and of every line you want to evaluate using eval() function.

Array would be better solution.