> 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.

>
> >
> > 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
> >
> > T: +44 (0)1189 789 500
> > F: +44 (0)1189 784 994
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> >
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