另一个eval（）受害者

Some code uses eval() to execute code stored in an array like this:

// $oOwner is an object
$strId = "abc";
$strClass = "someClass";
$aParams = array('a' => 'atext', 'b' => 'btext');

$this->menu = array(
"Entry 1" => ' openForm(\$oOwner,\$strId,\$strClass,\$aParams); ',
// ...
);

The values of the array's keys will be given directly into an eval() function.

Now I get an error: Parse error: syntax error, unexpected T_VARIABLE, expecting T_STRING ... eval()'d code on line 1

What is the problem?

EDIT:

Loop through the array and eval() the values:

eval( $this->menu[$param[0]] );

EDIT 2:

Now: "Entry 1" => " openForm(\$this->owner,\$strId,\$strClass,\$aParams); "

Using double quotes "..." leads to PHP Notice: Undefined variable: strId in ... : eval()'d code on line 1. Also for the other variables.

`

The way I tried always led to errors and I came up with the idea of putting the array in there as a string via var_export().

There are 2 attempts posted here: Store array in a string to eval() later

You are using single quote ' so you don't need to use \$ in your variables

Prof of Concept use both single and double quotes

function openForm(){
    var_dump(func_get_args());
}

$strId = "abc";
$strClass = "someClass";
$aParams = array('a' => 'atext', 'b' => 'btext');
$oOwner = "ABC" ;

$menu = array(
"Entry 1" => ' openForm($oOwner,$strId,$strClass,$aParams); ',
"Entry 2" => " openForm(\$aParams,\$strId); "
);

eval($menu["Entry 1"]);

Output

array
  0 => string 'ABC' (length=3)
  1 => string 'abc' (length=3)
  2 => string 'someClass' (length=9)
  3 => 
    array
      'a' => string 'atext' (length=5)
      'b' => string 'btext' (length=5)


array
  0 => 
    array
      'a' => string 'atext' (length=5)
      'b' => string 'btext' (length=5)
  1 => string 'abc' (length=3)

Have you tried double quotes when declaring code to be eval?

// ...
$this->menu = array(
"Entry 1" => " openForm(\$oOwner,\$strId,\$strClass,\$aParams); ",
// ...
);