How can I catch a parse error in php eval()?

Question

I'd like to use php eval() to identify potential parse errors. I'm aware of the dangers of eval, but this is a very limited use which will be fully validated beforehand.

I believe that in php 7 we should be able to catch a parse error, but it doesn't work. Here's an example:

  $one = "hello";
  $two = " world";
  $three = '';
  $cmdstr = '$three = $one . $tw;';

  try {
     $result = eval($cmdstr);
 } catch (ParseError $e) {
     echo 'error: ' . $e;
 }

echo $three;

I'm trying to cause a parse error here to see if I can catch it, but when I run it, the error (undefined variable tw) appears as it usually would. It was not being caught.

Any ideas how to catch a parse error from eval?

I would recommend using something like github.com/nikic/PHP-Parser which would be a more controlled way of doing this (I think). — Nigel Ren
– Nigel Ren, Commented Aug 10, 2020 at 18:16
One thing to point out is that a variable not being defined isn't a parse error but a runtime error. This is down to PHP being able to create variables in all sorts of weird ways that it's impossible to know at compile time is a variable may exist or not. Parse errors would be something like missing the = - $three $one . $tw; — Nigel Ren
– Nigel Ren, Commented Aug 10, 2020 at 18:28

Samuel Tallet · Accepted Answer · 2020-08-10 19:11:59Z

4

Your code doesn't work as expected because, in PHP, an undefined variable doesn't trigger a parse error but a notice instead. Thanks to set_error_handler native function, you can convert a notice to error then catch it with this PHP 7 code:

<?php

set_error_handler(function($_errno, $errstr) {
    // Convert notice, warning, etc. to error.
    throw new Error($errstr);
});

$one = "hello";
$two = " world";
$three = '';
$cmdstr = '$three = $one . $tw;';

try {
    $result = eval($cmdstr);
} catch (Throwable $e) {
    echo $e; // Error: Undefined variable: tw...
}

echo $three;

edited Aug 10, 2020 at 19:11

answered Aug 10, 2020 at 19:09

Samuel Tallet

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1 Comment

Niet the Dark Absol Over a year ago

I believe the correct way to "convert" errors into exceptions is with throw new ErrorException(...), as the ErrorException exception type has fields for the various components of a legacy error.

keidakida · Accepted Answer · 2020-08-10 19:23:23Z

Your PHP code would throw a 'Notice' kind of error and those cannot be handled by try..catch blocks. You would have to use your own error handlers using PHP's set_error_handler method. Read the document and you will understand what to do. If you want a sample of how to do it then:

<?php

function myErrorHandler($errno, $errstr)
{
    switch ($errno) {
        case E_USER_ERROR:
            die("User Error");
            break;
        default:
            die("Your own error");
            break;
    }

    /* Don't execute PHP internal error handler */
    return true;
}

$err = set_error_handler("myErrorhandler");

$one = "hello";
$two = " world";
$three = '';
$cmdstr = '$three = $one . $tw;';

$result = eval($cmdstr);

echo $three;

?>

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