I have a problem with SimpleTest / xdebug test classes - specifically, I have registered a shutdown function to close my test coverage and save the results.
However, this still (correctly) shows up as never reached:
if (error_condition) {
die();
} // This line is never actually reached, and shows up in the report as unreached, confusing the automatic analyzers.
Short of avoiding using 'die' for exits, and using exceptions instead, is there a way to mark a line up as 'do not test' ?
I don't think SimpleTest has this, but, it should not even consider it as "not covered". It's just unreachable, and that is not a problem.
Related
In a php script I have some test and after the script the html page.
When a test fail i call die("Test 1 failed");
If no test fail the php script reach the end ?> and then load the html code after the php script.
Is this a good procedure? Or I need to write die() or exit() before the end of php script?
No you don't have to write that and this is not best practice. If the script reaches the end without fatal errros it will exit.
If this means "testing" for you, you're wrong. Testing should be done using unit tests. For php there is phpunit. Give it a try, that's the proper way of testing your code.
Edit: As CompuChip says in a comment, the only useful use case for exit is when you're writing a php based shell script that should return an error code. See the parameter section of the documentation for the exit() function.
You should never be using die() or exit in your production PHP scripts except in very specific cases. Instead, re-work your code paths to simply show an error message to the user rather than exiting the script early.
No you don't need that, but when writing console PHP scripts, you might want to check with for example Bash if the script completed everything in the right way. That's when you use exit() or die()
Is the die() or exit() function needed in the end of a php script?
No, PHP will end the script itself. If the script is an included file (called from another file) then it will end script in the included file and then continue with any code in the original file after where you included (if there is any code).
So you put die() or exit() where ever you want or need it.
For testing, put it after each block of code you test. I use them in some parts of testing if I just want PHP to show me something then stop, such as print out an array to make sure it's being constructed correctly etc.
eg:
print_r($array);
exit();
For other code tests, I sometimes just echo "Section A worked", etc, such as within if/else. If I want to know if a particular part of code is working or if some criteria is being met or not (basically, it lets you trace where PHP itself is going within your code).
All that said, don't use die() or exit() in production code. You should use a more friendly and controlled messaging setup. For security reasons and visual, as you could potentially give them some info like "ERROR Failed to load SomethingSecret". Also it doesn't look pretty when you page only half loads and then puts out an on screen error message which likely means nothing to the end user.
Have a read through this:
PHP Error handling: die() Vs trigger_error() Vs throw Exception
No !
This is not recommanded to use it
Use trigger_error or error_log to log the tests in your error.log. Then check it.
No you don't have to use these functions at the end of the script, because it exists anyway at the end of the script.
No need to put a die or an exit at the end of the scipt.
But you may use exit to terminate your script with a specific exit code (by default it's 0).
E.g
$ php -r "/* does nothing */;"
$ echo $?
0
$ php -r "exit(123);"
$ echo $?
123
http://php.net/exit
From the documentation:
The link to the server will be closed as soon as the execution of the
script ends, unless it's closed earlier by explicitly calling
mysql_close().
https://secure.php.net/function.mysql-connect
Nope, you don't need to call die() or exit(0 if you have another code to run, like you HTML code
I'm not sure if this is a bug, but I don't seem to be able to get 100% coverage with early exit inside a PHP 5.5 generator.
For example:
<?php
function gen($branch = false)
{
yield;
if($branch) {
echo 'branched';
return;
} // This line is never covered.
echo 'did not branch';
}
The closing bracket of the if statement is never covered. It's not considered "dead code", it behaves as though you didn't consider the possibility of exiting the if statement.
I don't want to use an else because, as phpmd would say, "else is never necessary", but I would like to get 100% coverage (as that makes it easy to see if I've missed something in new code without drilling down).
Has anyone else run into this? Is there a solution?
This was a bug in Xdebug, and is resolved in version 2.4.0
Is there a way to immediately stop PHP code execution?
I am aware of exit but it clearly states:
Terminates execution of the script. Shutdown functions and object destructors will always be executed even if exit is called.
So what I want to achieve is to stop the PHP code execution exactly when I call exit or whatever.
Any help?
Edit: After Jenson's answer
Trial 1:
function newExit() {
__halt_compiler();
}
echo "start";
newExit();
echo "you should not see this";
Shows Fatal error: __HALT_COMPILER() can only be used from the outermost scope in which was pretty expected.
Trial 2:
function newExit() {
include 'e.php';
}
echo "start";
newExit();
echo "you should not see this";
e.php just contains __halt_compiler();
This shows startyou should not see this
Edit: Why I want to do this?
I am working on an application that includes a proprietary library (required through virtual host config file to which I don't have access) that comes as encrypted code. Is a sort of monitoring library for security purpose. One of it's behaviours is that it registers some shutdown functions that log the instance status (it saves stats to a database)
What I want to do is to disable this logging for some specific conditions based on (remote IP)
Please see the following information from user Pekka 웃
According to the manual, destructors are executed even if the script gets terminated using die() or exit():
The destructor will be called even if script execution is stopped using exit(). Calling exit() in a destructor will prevent the remaining shutdown routines from executing.
According to this PHP: destructor vs register_shutdown_function, the destructor does not get executed when PHP's execution time limit is reached (Confirmed on Apache 2, PHP 5.2 on Windows 7).
The destructor also does not get executed when the script terminates because the memory limit was reached. (Just tested)
The destructor does get executed on fatal errors (Just tested) Update: The OP can't confirm this - there seem to be fatal errors where things are different
It does not get executed on parse errors (because the whole script won't be interpreted)
The destructor will certainly not be executed if the server process crashes or some other exception out of PHP's control occurs.
Referenced in this question
Are there any instances when the destructor in PHP is NOT called?
whats wrong with return ?
echo "you will see this";
return;
echo "you will not see this";
You can use __halt_compiler function which will Halt the compiler execution
http://www.php.net/manual/en/function.halt-compiler.php
You could try to kill the PHP process:
exec('kill -9 ' . getmypid());
Apart from the obvious die() and exit(), this also works:
<?php
echo "start";
__halt_compiler();
echo "you should not see this";
?>
I'm not sure you understand what "exit" states
Terminates execution of the script. Shutdown functions and object destructors will always be executed even if exit is called.
It's normal to do that, it must clear it's memmory of all the variables and functions you called before. Not doing this would mean your memmory would remain stuck and ocuppied in your RAM, and if this would happen several times you would need to reboot and flush your RAM in order to have any left.
or try
trigger_error('Die', E_ERROR);
In a php script I have some test and after the script the html page.
When a test fail i call die("Test 1 failed");
If no test fail the php script reach the end ?> and then load the html code after the php script.
Is this a good procedure? Or I need to write die() or exit() before the end of php script?
No you don't have to write that and this is not best practice. If the script reaches the end without fatal errros it will exit.
If this means "testing" for you, you're wrong. Testing should be done using unit tests. For php there is phpunit. Give it a try, that's the proper way of testing your code.
Edit: As CompuChip says in a comment, the only useful use case for exit is when you're writing a php based shell script that should return an error code. See the parameter section of the documentation for the exit() function.
You should never be using die() or exit in your production PHP scripts except in very specific cases. Instead, re-work your code paths to simply show an error message to the user rather than exiting the script early.
No you don't need that, but when writing console PHP scripts, you might want to check with for example Bash if the script completed everything in the right way. That's when you use exit() or die()
Is the die() or exit() function needed in the end of a php script?
No, PHP will end the script itself. If the script is an included file (called from another file) then it will end script in the included file and then continue with any code in the original file after where you included (if there is any code).
So you put die() or exit() where ever you want or need it.
For testing, put it after each block of code you test. I use them in some parts of testing if I just want PHP to show me something then stop, such as print out an array to make sure it's being constructed correctly etc.
eg:
print_r($array);
exit();
For other code tests, I sometimes just echo "Section A worked", etc, such as within if/else. If I want to know if a particular part of code is working or if some criteria is being met or not (basically, it lets you trace where PHP itself is going within your code).
All that said, don't use die() or exit() in production code. You should use a more friendly and controlled messaging setup. For security reasons and visual, as you could potentially give them some info like "ERROR Failed to load SomethingSecret". Also it doesn't look pretty when you page only half loads and then puts out an on screen error message which likely means nothing to the end user.
Have a read through this:
PHP Error handling: die() Vs trigger_error() Vs throw Exception
No !
This is not recommanded to use it
Use trigger_error or error_log to log the tests in your error.log. Then check it.
No you don't have to use these functions at the end of the script, because it exists anyway at the end of the script.
No need to put a die or an exit at the end of the scipt.
But you may use exit to terminate your script with a specific exit code (by default it's 0).
E.g
$ php -r "/* does nothing */;"
$ echo $?
0
$ php -r "exit(123);"
$ echo $?
123
http://php.net/exit
From the documentation:
The link to the server will be closed as soon as the execution of the
script ends, unless it's closed earlier by explicitly calling
mysql_close().
https://secure.php.net/function.mysql-connect
Nope, you don't need to call die() or exit(0 if you have another code to run, like you HTML code
I was working on this book and I found an unexpected situation.
When working in a PHP/HTML hybrid file, the book says, if I use PHP try/catch without the ob buffer thing, and if there's an error somewhere in the middle of the file, the PHP engine won't be able to reach the catch{} line since some output has already been sent to the browser. The book then says this situation could be fixed by using ob_start(), ob_end_clean() and ob_end_flush()
However, when I was playing with the book's sample code, the try/catch worked just fine without the ob buffer stuff. By saying just fine I mean if there's an exception the catch{} line could be reached and executed without a problem.
I used a Linode VPS for testing, the PHP version is 5.3.2. I set up the VPS with some ordinary Linode script.
Why is that? :)
That statement is nonsense. PHP's try..catch works just as you would expect, regardless of output that has or hasn't already been sent to the browser.
Seeing that even the errata to that book contains errors, I'd say it's simply not a very good book and am not surprised at all that it contains wrong statements. Just a very quick glance reveals:
In step 4, the third line of code should read as follows:
if ($_POST && isset($missing) && !empty($missing)) {
No, it should actually read:
if ($_POST && !empty($missing)) {
or
if ($_POST && $missing) {
The author apparently does not understand how to use empty.
Further:
An opening curly brace is missing at the end of line 4 of the code in step 2. It should read:
if (!#include('includes/connection.inc.php')) {
That reeks of bad practices and should be rewritten to:
if (!file_exists('includes/connection.inc.php')) {
or:
require_once 'includes/connection.inc.php';
I'm sure there's more... :)