content instead of path in shell

问题

Pngquant has the following example for php

// '-' makes it use stdout, required to save to $compressed_png_content variable
    // '<' makes it read from the given file path
    // escapeshellarg() makes this safe to use with any path
    $compressed_png_content = shell_exec("pngquant --quality=$min_quality-$max_quality - < ".escapeshellarg(    $path_to_png_file));

I want to replace $path_of_file with the actual content.

This will avoid wasting I/O when converting a file from one format to png and then optimize it

What will be the new shell_exec() command in that situation

回答1:

I am no PHP expert, but I believe you are looking for a 2-way pipe (write and read) to another process, so that you can write data to its stdin and read data from its stdout. So, I think that means you need proc_open() which is described here.

It will look something like this (untested):

$cmd = 'pngquant --quality ... -';

$spec = array(array("pipe", "r"), array("pipe", "w"), array("pipe", "w"));

$process = proc_open($cmd, $spec, $pipes);

if (is_resource($process)) 
{

    // write your data to $pipes[0] so that "pngquant" gets it
    fclose($pipes[0]);

    $result=stream_get_contents($pipes[1]);
    fclose($pipes[1]);

    proc_close($process);
}

来源：https://stackoverflow.com/questions/37838561/content-instead-of-path-in-shell