How to run PHP script (with parameters) under the php command line-PHP Tutorial-php.cn

How to run PHP script (with parameters) under the php command line

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Release： 2016-07-25 09:03:04

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echo "Hello from the CLI";
?>

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Now, try running this program in the command line prompt by calling the CLI executable file and provide the file name of the script: #php phphello.php Output Hello from the CLI

Use standard input and output You can use these three constants in your PHP scripts to accept user input or display the results of processing and calculations. To understand this better, take a look at the following code.

// ask for input
fwrite(STDOUT, "Enter your name: ");
// get input
$ name = trim(fgets(STDIN));
// write input back
fwrite(STDOUT, "Hello, $name!");
?>

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Look what happens when you run it: shell> php hello.php Enter your name: Joe Hello, Joe! ?>

In this script, the fwrite() function will first write a message to the standard output device, asking for the user's name. It then reads the user input information obtained from the standard input device into a PHP variable and combines it into a string. Then use fwrite() to print the string to the standard output device.

#---Use command line arguments It is common to enter arguments to a program at the command line to change the way it runs. You can do this with CLI programs as well. The PHP CLI comes with two special variables specifically designed for this purpose: one is the $argv variable, which saves the parameters passed to the PHP script through the command line as separate array elements; the other is the $argc variable, which is used To save the number of elements in the $argv array.

It is very simple to write a piece of code in PHP script that reads $argv and processes the parameters it contains. Test the following sample script to see how it works:

print_r($argv);
?>

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Run this script by passing it some arbitrary values, and check the output:

shell> php phptest.php chocolate 276 "killer tie, dude!" Array ( [0] => test.php [1] => chocolate [2] => 276 [3] => killer tie, dude! )

As you can see from the output, the value passed to test.php will automatically appear as an array element in $argv. Note that the first argument to $argvis is always the name of the script itself.

Here is a more complex example:

// check for all required arguments
// first argument is always name of script!
if ($argc != 4) {
die("Usage: book.php ");
}
// remove first argument
array_shift($argv);
// get and use remaining arguments
$checkin = $argv[0];
$nights = $argv[1];
$type = $argv[2];
echo " You have requested a $type room for $nights nights, checking in on $checkin. Thank you for your order! ";
?>

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Here are examples of its usage : shell> php phpbook.php 21/05/2005 7 single You have requested a single room for 7 nights, checking in on 21/05/2005. Thank you for your order!

Here, the script will first check $argc to ensure that the number of independent variables meets the requirements. It then extracts each argument from $argv and prints them to standard output.