piping data into command line php?

问题:

It is possible to pipe data using unix pipes into a command-line php script? I've tried

$> data | php script.php

But the expected data did not show up in $argv. Is there a way to do this?

回答1:

As I understand it, $argv will show the arguments of the program, in other words:

php script.php arg1 arg2 arg3

But if you pipe data into PHP, you will have to read it from standard input. I've never tried this, but I think it's something like this:

$fp = readfile("php://stdin");
// read $fp as if it were a file

回答2:

PHP can read from standard input, and also provides a nice shortcut for it: STDIN.

With it, you can do things like:

$data = stream_get_contents(STDIN);

This will just dump all the piped data into $data.

If you want to start processing before all data is read, or the input size is too big to fit into a variable, you can use:

while(!feof(STDIN)){
    $line = fgets(STDIN);
}

STDIN is just a shortcut of $fh = fopen("php://stdin", "r"); The same methods can be applied to reading and writing files, and tcp streams.

回答3:

If your data is on one like, you can also use either the -F or -R flag (-F reads & executes the file following it, -R executes it literally) If you use these flags the string that has been piped in will appear in the (regular) global variable $argn

Simple example:

echo "hello world" | php -R 'echo str_replace("world","stackoverflow", $argn);'

回答4:

You can pipe data in, yes. But it won't appear in $argv. It'll go to stdin. You can read this several ways, including fopen('php://stdin','r')

There are good examples in the manual

回答5:

This worked for me:

stream_get_contents(fopen("php://stdin", "r"));

回答6:

Came upon this post looking to make a script that behaves like a shell script, executing another command for each line of the input... ex:

ls -ln | awk '{print $9}'

If you're looking to make a php script that behaves in a similar way, this worked for me:

#!/usr/bin/php
<?php

$input = stream_get_contents(fopen("php://stdin", "r"));

$lines = explode("\n", $input);

foreach($lines as $line) {
    $command = "php next_script.php '" . $line . "'";
    $output = shell_exec($command);
    echo $output;
}

回答7:

If you want it to show up in $argv, try this:

echo "Whatever you want" | xargs php script.php

That would covert whatever goes into standard input into command line arguments.

回答8:

Best option is to use -r option and take the data from the stdin. Ie I use it to easily decode JSON using PHP. This way you don't have to create physical script file.

It goes like this:

docker inspect $1|php -r '$a=json_decode(stream_get_contents(STDIN),true);echo str_replace(["Array",":"],["Shares","  -->  "],print_r($a[0]["HostConfig"]["Binds"],true));'

This piece of code will display shared folders between host & a container. Please replace $1 by the container name or put it in a bash alias like ie displayshares() { ... }