Improving error messages when using indirect references

21 years ago by Martin Jansen — view source

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Hey,

    &lt;?php
    error_reporting(E_ALL);

    echo $$foo;
    ?>

results in

    Notice: Undefined variable: foo in /foo/bar.php on line 4

    Notice: Undefined variable: in /foo/bar.php on line 4

Wouldn't it be more appropriate if the second notice was something like
"Notice: Undefined variable: <unnamed> in /foo/bar.php on line 4"?

--
- Martin Martin Jansen
http://martinjansen.com/

21 years ago by Hartmut Holzgraefe — view source

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Martin Jansen wrote:

Hey,
    &lt;?php
    error_reporting(E_ALL);

    echo $$foo;
    ?>
results in
    Notice: Undefined variable: foo in /foo/bar.php on line 4

    Notice: Undefined variable: in /foo/bar.php on line 4
Wouldn't it be more appropriate if the second notice was something like
"Notice: Undefined variable: <unnamed> in /foo/bar.php on line 4"?

i'd prefere to put the variable name in single quotes here
as the name actually is '' (empty string)

<?php
$foo = "";
$$foo = "hallo";
echo ${""};
?>

--
Hartmut Holzgraefe <hartmut@php.net

21 years ago by Derick Rethans — view source

unread

Hey,
    &lt;?php
    error_reporting(E_ALL);

    echo $$foo;
    ?>
results in
    Notice: Undefined variable: foo in /foo/bar.php on line 4

    Notice: Undefined variable: in /foo/bar.php on line 4
Wouldn't it be more appropriate if the second notice was something like
"Notice: Undefined variable: <unnamed> in /foo/bar.php on line 4"?

Would be a lie... $foo WILL evaluate to the empty string which used like
this, although it does look weird, it is right :)

Derick