Phil Ewington - 43 Plc wrote on donderdag 10 juni 2004 14:27:
>> Phil Ewington - 43 Plc wrote on donderdag 10 juni 2004 12:58:
>>
>>> Hi All,
>>>
>>> Can anyone tell me how to pass arguments to a PHP script from the
>>> command line? I know I need to use the -args option but how
>>> exactly, I have tried the following...
>>>
>>> /usr/bin/php myscript.php -args key=value
>>
>> You don't have to call the -args argument. just run the script from
>> command line like this:
>>
>> php myscript.php value
>>
>> Where value is the argument you want to use in the script.
>> These arguments are stored in an arrray ($argv).
>>
>> Do a print_r($argv) and you can see the content.
>
> OK, but if no param name is passed in, I assume that the order they
> are passed in as must be correct or the script will not know what
> values belong to which parameters? This seems a little odd, I assume
> that $argv only exists when accessed from the command line? My script
> needs to be excuted from both command line and from web browsers.
>
How do you mean the order must be correct. Use some code like this to
give it an index value.
for ($counter=0; $counter < $argc; $counter++) {
list($key, $value) = each($argv);
echo ("key ". $key . "has value : " . $value . "\n");
}
>From what i understand the argv & argc are only avaliable when the
script is run from commandline. Just do a check on this and if not
there get some $_POST or $_GET variables.
>>
>>>
>>> but this does not work, the script executes but key is not available
>>> as $_GET["key"] in my script. Any pointers will be much appreciated.
>>>
>>> TIA
>>>
>>> Phil.
>>>
>>> ---
>>> Phil Ewington - Technical Director
>>>
>>> 43 Plc - Ashdale House
>>> 35 Broad Street, Wokingham
>>> Berkshire RG40 1AU
>>>
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