display ghostscript version number through PHP in ubuntu

Question

I have written the following code for showing the version number of ghostscript:

<html>
<head>
<title></title>
</head>
<body>
<?

$ver = shell_exec("/usr/bin/gs --version");
//$ver = exec(GS_BIN . " --version");
print "$ver";
print "A";

?>
</body>
</html>

I can get the A printed, but not the version number why?

Thanks.

@lokheart: If you execute that command manually in the shell, what does it print on screen? @Rocket: sorry. — hakre
– hakre, Commented Jul 8, 2011 at 19:02

Marc B · Accepted Answer · 2011-07-08 15:53:56Z

2

Possibly ghostscrsipt is writing the data out to STDERR instead of STDOUT. Try doing

/usr/bin/gs --version 2>&1

to redirect stderr to stdout and try again

answered Jul 8, 2011 at 15:53

Marc B

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4 Comments

lokheart Over a year ago

I need to try the gs version number because I need to use php to use gs to capture some image, but I can't run it using exec, so I try a smaller task of showing the version number, but still fail

hakre Over a year ago

@lokheart: Have you tried to redirect STDERR to STDOUT like Marc B suggested?

lokheart Over a year ago

i tried $ver = shell_exec('/usr/bin/gs --version 2>&1'); and the output is string(111) "/usr/bin/gs: /opt/lampp/lib/libgcc_s.so.1: version GCC_4.2.0' not found (required by /usr/lib/libstdc++.so.6) "`, what has gone wrong?

Marc B Over a year ago

If gs works from the command line, then whatever shell is being invoked by PHP doesn't have the proper library path, and can't find the gcc library that GS was compiled against.

Community · Accepted Answer · 2017-05-23 10:34:20Z

1

You should use var_dump($ver); for debugging purposes, because your code just works:

$ php -r "echo shell_exec('/usr/bin/gs --version');"
8.71

I just had it run on my linux box and according to shell_exec() Docs, it should be fine.

Things to look for:

Safe Mode enabled?
exec() can return the exit code / return status of the command.
if it returns NULL, see this answer.

STDERR and shell_exec()

shell_exec() will only return the commands output written to STDOUT. In case the command can not be invoked by the shell, this function will return NULL and it will hide away what has been reported as error.

To include errors as well in the return value, STDERR needs to be redirect to STDOUT. This is done by adding 2>&1 to the end of the command. Here is the same example code with a wrong command for demonstration:

$ php -r "var_dump(shell_exec('/usr/bin/gs2 --version 2>&1'));"
string(44) "sh: /usr/bin/gs2: No such file or directory
"

edited May 23, 2017 at 10:34

CommunityBot

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answered Jul 8, 2011 at 15:53

hakre

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2 Comments

lokheart Over a year ago

and I have checked phpinfo(), safe mode is disabled

hakre Over a year ago

I added a reference to the question. Check if it helps.

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