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Request #26818 getopt() parses incorrectly
Submitted: 2004-01-06 14:52 UTC Modified: 2010-11-24 12:59 UTC
From: php at richardneill dot org Assigned:
Status: Open Package: *General Issues
PHP Version: * OS: *
Private report: No CVE-ID:
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 [2004-01-06 14:52 UTC] php at richardneill dot org
Description:
------------
When parsing options, if the command line arguments are WRONG, php mangles them in such a way as to give an incorrect, but valid response.

Eg my program normally takes arguments:
      ./getopt.php -a -x foo -y bar
(ie. option a takes no value; options x,y take values)

Therefore, if I run it as
       ./getopt.php -a -x -y bar
this is clearly a mistake: I omitted the value for -x.

HOWEVER, php assumes that I wanted the value for -x to be '-y'

If this is "a feature, not a bug", it might be worth documentig it, or providing a switch to enable the behaviour I'd expect. Thank you.


Reproduce code:
---------------
#!/usr/bin/php
<?
$flags="ax:y:";
$options_array=getopt($flags);

echo "Here are the results of getopt:\n";

foreach ($options_array as $key => $value){
	echo "\tkey:\t $key\t value is:\t $value";
	if ($value===false){
		echo "\t[FALSE]";
	}
	echo "\n";
}
echo "\n";

#N.B. If options -x and -y each expect an argument, then one might expect
# 'getopt.php -x -y foo' to result in $key="x",$value=false; $key="y",$value="foo"
# BUT actually, the result is $key="x",$value="-y"; and foo is left over.


Expected result:
----------------
[rjn]$ ./getopt.php -a -x -y bar
Here are the results of getopt:
        key:     a       value is:              [FALSE]
        key:     x       value is:              [FALSE]      
        key:     y       value is:       bar

Actual result:
--------------
[rjn]$ ./getopt.php -a -x  -y bar
Here are the results of getopt:
        key:     a       value is:              [FALSE]
        key:     x       value is:       -y



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 [2004-01-06 17:30 UTC] sniper@php.net
RTFM: "It does not matter if an argument has leading white space." (it means also: whitespace is omitted)

 [2004-01-06 17:53 UTC] php at richardneill dot org
I agree with you about the manual. BUT, nevertheless, this is not the behaviour that it should have.  Currently, it isn't consistent with the normal (GNU-like) command line behaviour.

If options -x and -y are supposed to take values, then if I enter the command: 
      ./getopt.php -a -x -y bar
it can mean only one of 2 things:

1)I, the user have made a mistake at the command line, and omitted the value for $x. The script should be able to test for this as an error. 

2)I meant to assign a null or false value to $x 
(possible, not likely)

It is *very* unlikely that I meant to assign the value '-y' to $x. (If I had meant to do this, I would have double quoted it).

Please can we at least have an option within getopts to parse options in a standard manner? I'd suggest doing the following:

----------------
If an option expects a value, read along the command line
until reaching the next '-' or the end of line. This value (or false if it is null) is placed in the options_array.
----------------

Thank you.

Richard
 [2004-01-06 18:58 UTC] sniper@php.net
The function uses the GNU getopt function, so the behaviour is exactly same. Moved the FRs.

 [2004-01-06 20:58 UTC] php at richardneill dot org
I do apologise - I thought that GNU's getopts was smarter than that! So, I now agree that PHP is doing the standards-compliant thing. I still think it would be useful if PHP had an option to parse options "my way" :-)

Thanks for your help.

Richard
 [2010-11-24 12:59 UTC] jani@php.net
-Package: Feature/Change Request +Package: *General Issues -Operating System: Linux (Mandrake) +Operating System: * -PHP Version: 4.3.4 +PHP Version: *
 [2010-11-24 12:59 UTC] jani@php.net
-Summary: getopt parses incorrectly +Summary: getopt() parses incorrectly
 
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