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PHP's white screen of death [duplicate]
Closed 7 years ago.
I'm calling a php script with shell_exec that runs fully operational WHEN called from the terminal (php somescript.php). However, using shell_exec I am able to check the "echo test" I'm issuing inside the script but the database queries with $mysqli are not being commited. Here is the script (again, works well when called from terminal):
//CALLED PHP SCRIPT
include_once("../mysqlconnection.php");
$query = "some query";
echo "this gets printed via shell_exec"
$mysqli->query($query); //this does not execute...
and mysqlconnection.php:
//connects well
$mysqli = new mysqli($db_ip, $db_user, $db_password, $db_database);
What exactly could be the problem? I call the script with shell_exec('php some_script.php');. Thank you very much for your help.
EDIT:
As I said, everything works well via terminal. A couple of things:
I have a debug function that is NOT being called although it is in mysqlconnection.php. Worst of it, if I completely mess up the code trying to provoke variable erros (like removing the include) it still "runs". Does nothing.
If I echo in the original php script it gets printed (as in the first code example echo "this gets printed via shell_exec" EVEN if the rest of the code is messed, semicolons missing and etc. I don't really know what might be off, it's odd. It only points all the flaws if ran by terminal although it also echoes the message if called from a php script
root
/folder/called php script
/script that calls the other php script
/mysqlconnection.php
My code is basically exactly how it is here, no need to change anything...
What exactly could be the problem?
Lack of error reporting.
Having errors properly reported, you will see what is wrong with your include and mysqli query.
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 am currently working from home (on a Sunday!) and I am trying to figure out why my Perl script is returning NULL to PHP from where it is called. However, I don't see how I can debug the Perl script itself. The PHP file returns a warning that I am trying to do an array operation on a non-array object (because the expected array is actually NULL returned by PHP). The logs of the webserver have only logged this warning as well - no Perl errors.
Is there a place where specific 'external' logs are stored on a server? Or, is there a better way to debug a Perl file that is been run from a PHP file that is required in a main PHP file? Debugging isn't necessary (I don't need a debug mode) but I'd like to see the errors or warnings at least.
You can add following code at the top of your Perl script:
sub debug_log
{
open my $log_fh, ">>", "/tmp/debug.log";
print $log_fh $_[0];
warn $_[0];
close $log_fh;
}
$SIG{__WARN__} = \&debug_log;
$SIG{__DIE__} = \&debug_log;
This way all the warnings and die messages should end up in /tmp/debug.log.
In a php script I have some test and after the script the html page.
When a test fail i call die("Test 1 failed");
If no test fail the php script reach the end ?> and then load the html code after the php script.
Is this a good procedure? Or I need to write die() or exit() before the end of php script?
No you don't have to write that and this is not best practice. If the script reaches the end without fatal errros it will exit.
If this means "testing" for you, you're wrong. Testing should be done using unit tests. For php there is phpunit. Give it a try, that's the proper way of testing your code.
Edit: As CompuChip says in a comment, the only useful use case for exit is when you're writing a php based shell script that should return an error code. See the parameter section of the documentation for the exit() function.
You should never be using die() or exit in your production PHP scripts except in very specific cases. Instead, re-work your code paths to simply show an error message to the user rather than exiting the script early.
No you don't need that, but when writing console PHP scripts, you might want to check with for example Bash if the script completed everything in the right way. That's when you use exit() or die()
Is the die() or exit() function needed in the end of a php script?
No, PHP will end the script itself. If the script is an included file (called from another file) then it will end script in the included file and then continue with any code in the original file after where you included (if there is any code).
So you put die() or exit() where ever you want or need it.
For testing, put it after each block of code you test. I use them in some parts of testing if I just want PHP to show me something then stop, such as print out an array to make sure it's being constructed correctly etc.
eg:
print_r($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 $?
123
http://php.net/exit
From the documentation:
The link to the server will be closed as soon as the execution of the
script ends, unless it's closed earlier by explicitly calling
mysql_close().
https://secure.php.net/function.mysql-connect
Nope, you don't need to call die() or exit(0 if you have another code to run, like you HTML code
After so much trouble I find out that when I use the flush function in my PHP mail script then I get garbage or dump characters on browser like below.
The code is below
if ($mail->Send()) {
echo "<br><font color=darkgreen>[$num successful send to $to]</font> ";
// flush();
return true;
}
If I comment that flush line then out is simple English but I uncomment that the whole page the text looks like garbage.
Now is that a PHP problem, browser problem or server problem?
If I use the same script from the shell, I mean execute inside the shell terminal then I can see the HTML output. But it does not work in browsers.
I found the answer to my own question. I had to turn
zlib_compression off
in my php.ini settings file.
(What does that mean and why did it work?. I had been trying this for 1 year but could not resolve the problem but now it worked.)
I have a PHP script that creates other PHP files based on user input. Basically, there are files containing language specific constants (define) that can be translated by the user. In order to avoid runtime errors, I want to test newly written files for parse errors (due to "unusual" character sequences). I have read several posts here on SO (like PHP include files with parse errors) and tried a function that uses
$output = exec("php -l $filename");
to determine whether a file parses correctly. This works perfectly on my local machine, but at on the provider's machine, the output of calls to exec("php ...") seems to be always empty. I tried a call to ls and it gives me output, leading me to the assumption that PHP is somehow configured to not react to command line invocations or so. Does anyone know a way around this?
EDIT: I forgot to mention, I had already tried shell_exec and it gives no result, either. In response to sganesh's answer: I had tried that too, sorry I forgot to mention. However, the output (second argument) will always be an empty array, and the return value will always be 127, no matter if the PHP file to test has syntax errors or not.
I had the same problem. The solution that worked for me was found in running-at-from-php-gives-no-output. I needed to add output redirection.
$output = exec("php -l $filename 2>&1");
You can try with exec second and third arguments.
second argument will have the output of the command.
third argument will have the return value.
And exec will return only last line of the command.
$filename = "a.php";
$output = exec("php -l $filename",$op,$ret_val);
print $output."\n";
print $ret_val."\n";
var_dump($op);
By executing shell_exec(), you can see the output as if you executed that file via command line. You can just see if there is an error right here.
<?php
if (strpos(shell_exec('php -l file.php'), 'Syntax Error')) {
die('An error!');
}
There may also be a possibility that shell_exec() or exec() may be disable by your host.
Nice idea to check the file validity :-)!
Now, from the PHP manual for exec():
Note: When safe mode is enabled, you can only execute files within the safe_mode_exec_dir. For practical reasons, it is currently not allowed to have components in the path to the executable.
Can you check if this is not the case for you?
Also, can you check by providing the full path of the PHP interpreter in the exec() instead of only php. Let me know how you fare.
Pinaki
the correct way is to add >2&1 as tested on a windows system using imagemagick!
I worked around my original problem by using a different method. Here is what I do now:
Write a temporary file with contents <?php include "< File to test >"; echo "OK"; ?>
Generate the correct URL for the temporary file
Perform HTTP request with this URL
Check if result equals "OK". If yes, the file to test parses without errors.
Delete temporary file
Maybe this could be done without the temporary file by issuing an HTTP request to the file to test directly. However, if there is a parse error and errors are suppressed, the output will be empty and not discernible from the output in the case of a file that gives no parse errors. This method is risky because the file is actually executed instead of just checked. In my case, there is only a limited number of users who have access to this functionality in the first place. Still, I'm naturally not entirely happy with it.
Why the exec() approach did not work, I still do not know exactly. pinaki might be right by suggesting to provide the full path to the PHP executable, but I cannot find out the full path.
Thank you everyone for answering, I upvoted you all. However, I cannot accept any of your answers as none of your suggestions really solved my problem.