ID: 49061
User updated by: majuki at yahoo dot com
Reported By: majuki at yahoo dot com
Status: Bogus
Bug Type: *General Issues
Operating System: CenTOS
PHP Version: 5.2.10
New Comment:
wow - harsh. Guess you're having a bad day.
The code provided, while not my original, does represent the exact
syntax that is used and executes as expected. Yes, there's a syntax
error, I put the ; in the wrong place when typing it in. I'm sorry for
not making it so you can cut and paste, it was a typo.
As to my original it is far more complex than this simple extraction
and difficult to "sum up" in 10-20 lines. I will attempt to extrapolate
the core logic for you and provide cut and paste code.
File1:
<?php
$var = 1;
include(file2.php);
?>
File2:
<div <?php if (empty($var)) echo 'value="' . $var . '"'; ?>></div>
<div <?php echo (empty($var)) ? '' : 'value="' . $var. '"'; ?>></div>
The above generate the notice where as the following do not:
<div <?php if (empty($var)) { echo $var; } ?>></div>
<div <?php echo (empty($var)) ? 'value="' . $var. '"' : ''; ?>></div>
I hope this is sufficient.
Please understand that this is my time too - I'm volunteering my time
to report what I've observed to be strange and unexpected behaviour, not
because it's a problem I need fixed. All I had to do was turn off
E_NOTICE and forget about it, but I thought it might warrant looking
into. So please, next time when someone is reporting something,
remember that they didn't have to take the time and if they don't have
the time (or in some cases the skill) to provide you with perfect code,
be a little understanding.
I hope tomorrow is a better day for you. Sincerely.
Previous Comments:
------------------------------------------------------------------------
[2009-07-26 21:29:45] [email protected]
Once again your script doesn't run. There is a syntax error.
Plus, of course you get a warning with:
if (!isset($var)) echo $var;
The warning is on the echo of a variable that hasn't been set, not on
the check itself. Until you provide an actual sensible script that you
have run before submitting it, please don't waste any more of our time.
------------------------------------------------------------------------
[2009-07-26 19:39:14] majuki at yahoo dot com
You're right, my mistake - my original script had it more like the
following:
<?php
function doX(){ return 'ok' };
if (!isset($var)) echo $var;
if (empty($var)) echo $var;
echo (!isset($var)) ? doX() : $var;
echo (empty($var)) ? doX() : $var;
?>
Both generated the Notice)
------------------------------------------------------------------------
[2009-07-26 19:09:49] [email protected]
That is a completely bogus test script. Did you even bother to try
running it? You can't have an echo there and you are claiming an
undefined index warning in a script that doesn't even have an index.
------------------------------------------------------------------------
[2009-07-26 18:58:03] majuki at yahoo dot com
Already included it... but I'll expand on it...
<?php
function doX(){};
(!isset($var)) ? doX() : echo $var;
(empty($var)) ? doX() : echo $var;
?>
Above code with E_NOTICE enabled.)
------------------------------------------------------------------------
[2009-07-26 12:02:07] [email protected]
Thank you for this bug report. To properly diagnose the problem, we
need a short but complete example script to be able to reproduce
this bug ourselves.
A proper reproducing script starts with <?php and ends with ?>,
is max. 10-20 lines long and does not require any external
resources such as databases, etc. If the script requires a
database to demonstrate the issue, please make sure it creates
all necessary tables, stored procedures etc.
Please avoid embedding huge scripts into the report.
------------------------------------------------------------------------
The remainder of the comments for this report are too long. To view
the rest of the comments, please view the bug report online at
http://bugs.php.net/49061
--
Edit this bug report at http://bugs.php.net/?id=49061&edit=1
