(PHP 4, PHP 5)
eval — Evaluate a string as PHP code
Evaluates the given
code as PHP.
The eval() language construct is very dangerous because it allows execution of arbitrary PHP code. Its use thus is discouraged. If you have carefully verified that there is no other option than to use this construct, pay special attention not to pass any user provided data into it without properly validating it beforehand.
Valid PHP code to be evaluated.
The code mustn't be wrapped in opening and closing PHP tags, i.e. 'echo "Hi!";' must be passed instead of '<? echo "Hi!"; >'. It is still possible to leave and reenter PHP mode though using the appropriate PHP tags, e.g. 'echo "In PHP mode!"; ?>In HTML mode!<? echo "Back in PHP mode!";'.
Apart from that the passed code must be valid PHP. This includes that all statements must be properly terminated using a semicolon. 'echo "Hi!"' for example will cause a parse error, whereas 'echo "Hi!";' will work.
A return statement will immediately terminate the evaluation of the code.
The code will be executed in the scope of the code calling eval(). Thus any variables defined or changed in the eval() call will remain visible after it terminates.
NULL unless return is called in the evaluated code, in which case the value passed to return is returned. If there is a parse error in the evaluated code, eval() returns
FALSE and execution of the following code continues normally. It is not possible to catch a parse error in eval() using set_error_handler().
Example #1 eval() example - simple text merge
$string = 'cup';
$name = 'coffee';
$str = 'This is a $string with my $name in it.';
echo $str. "\n";
eval("\$str = \"$str\";");
echo $str. "\n";
The above example will output:
This is a $string with my $name in it. This is a cup with my coffee in it.
Note: Because this is a language construct and not a function, it cannot be called using variable functions.
In case of a fatal error in the evaluated code, the whole script exits.