Php as shell script and arguments

[Anon]

Hello.

I have to do some regular jobs on a mysql database (dumping data from specific tables and delete everything from it except the last record).

As this job will be done using a crontable, i m not sure php would be the best solution.

However, that s the way i ll try to do this.

So, by compiling php as an executable and launching my script with php -q it looks like i m able to run php scripts as if they were shell or perl one.

But i m stuck at a simple task: how to get the arguments that were passed to the script ?

Thanks for information.

[Anon]

when compiled as an executable argument on the command line are passed in the $argv[] array and number of args in $argc

ex: testarg.php
-----8<----------8<-----
<?

echo "Number of arguments: $argc\n";

for ($i = 0; $i < $argc ; $argc++)
echo "argv[$i]=".$argv[$i]."\n";

?>
-----8<----------8<-----

when run, this script output is:

php -q testarg.php aaa bbb ccc

Number of arguments: 4
aaa
bbb
ccc

---

argv[0] contains the name of the script

JBL

[Anon]

Thanks a lot, it works ok.

I had some try with $ARGV[0] and $ ARGV[1] without thinking about having a try in lower case.

Have a nice day.