can php take command line arguments when run from the shell?
I need to use it to send a file to an e-mail address. I am using it to send a backup of my database which is dumped using mysqldump . Each table is then checked against the old version to see if there were any changes, and then the ones which are changed I will pass to the php scrip to send
Full Version: PHP
Parameters are handled by PHP through the use of two predefined global variables $argc and $argv representing the number of parameters and their values as shown below:
Assuming the following command line:
...And the PHP script being executed:
The following output will be produced:
As you can see, the $argc parameter is an integer representing the total number of parameters passed on the command line. However, notice that the second variable $argv has an extra value! To assist the developer, the first value of the $argv array will be the name of the script executed followed by any command line parameters. Note that regardless of the presence of command line parameters, $argv will always contain the name of the script file being executed.
-------------------------------------------------------------
You can use a foreach:
Assuming the following command line:
CODE
$ ./myscript.php param1 param2
...And the PHP script being executed:
CODE
<?php
print "The total parameters count was: $argc\n";
print "The parameters were: ";
foreach ($argv as $val){
print "$val ";
}
print "\n";
?>
print "The total parameters count was: $argc\n";
print "The parameters were: ";
foreach ($argv as $val){
print "$val ";
}
print "\n";
?>
The following output will be produced:
CODE
$ ./myscript.php param1 param2
The total parameters count was: 2
The parameters were: myscript.php param1 param2
The total parameters count was: 2
The parameters were: myscript.php param1 param2
As you can see, the $argc parameter is an integer representing the total number of parameters passed on the command line. However, notice that the second variable $argv has an extra value! To assist the developer, the first value of the $argv array will be the name of the script executed followed by any command line parameters. Note that regardless of the presence of command line parameters, $argv will always contain the name of the script file being executed.
-------------------------------------------------------------
You can use a foreach:
CODE
foreach ($argv as $option)
{
switch ($option)
{
case 'one':
echo '1';
break;
case 'two':
echo '2';
break;
}
echo "\t";
}
{
switch ($option)
{
case 'one':
echo '1';
break;
case 'two':
echo '2';
break;
}
echo "\t";
}
thanks. that looks good. 
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