Add command line interface to upload videos

[Yetangitu]

Add command line interface to upload videos. Use php -f upload.php -- -h in view/mini-upload-form for instructions.

The -g parameter takes a comma-separated list of NUMERICAL groups (group IDs, actually) as argument, this due to the current absence of an easy way to get groups by their name, this can be changed when such has been implemented. I have not done this yet as there are things of higher concern, eg. the abundance of SQL injection opportunities in the code...).

Use: php -f upload.php -u <user> -f <file> [-d <description>] [-g "group1[,group2]] [-c] [-h]

    -u user     valid username
    -f file     input filename
    -d descr    description
    -g group(s) comma-separated numerical groups for which this item will be visible,
                item will be public if this is left out
    -c          copy original to destination directory (file will be moved if this is left out)
    -h          this help message

The easiest way to use this is to call the php code from a script, like so:

#!/bin/bash
srcdir="/var/src/YouPHPTube"

pushd "${srcdir}/view/mini-upload-form/" > /dev/null
sudo -u www-data php -f upload.php -- "$@"
popd > /dev/null

(adjust srcdir and www-data to your environment)

[DanielnetoDotCom]

Very good

[DanielnetoDotCom]

I have tested the command line help and didn't show me the help

Show me the PHP help instead upload.php help

[DanielnetoDotCom]

what the -f after php means?
php -f upload.php
i have remove it and works as expected

[Yetangitu]

The '-f' parameter tells php to parse and execute a file:

$ php -h
Usage: php [options] [-f] <file> [--] [args...]
 ...
  -f <file>        Parse and execute <file>.
 ...

Add a double dash after upload.php to tell php it should stop parsing parameters. The remaining parameters will be parsed by upload.php, like so:

$ sudo -u www-data php -f upload.php -- -h

Use: php -f upload.php -u <user> -f <file> [-d <description>] [-g "group1[,group2]] [-c] [-h]

    -u user     valid username
    -f file     input filename
    -d descr    description
    -g group(s) comma-separated numerical groups for which this item will be visible,
                item will be public if this is left out
    -c          copy original to destination directory (file will be moved if this is left out)
    -h          this help message

[DanielnetoDotCom]

With double dash works perfect.
But I think -f after php command is useless (at least for ubuntu)
What OS are u using?

[Yetangitu]

Try again with the double dash to see if it works for you with -f, it should. I'm using Debian, not that far from Ubuntu.

BTW, I notice a lot of files with DOS/Windows line endings, do you edit on Windows?

[DanielnetoDotCom]

Double dash works fine but the -f parameter is unnecessary on ubuntu Em 5 de jun de 2017 15:37, "Frank de Lange" <[email protected]> escreveu: Try again with the double dash to see if it works for you with -f, it should. I'm using Debian, not that far from Ubuntu. — You are receiving this because you modified the open/close state. Reply to this email directly, view it on GitHub <#29 (comment)>, or mute the thread <https://github.com/notifications/unsubscribe-auth/ARLMzTktIyg24ywxwRAguE1JLAqvgy-xks5sBErigaJpZM4NvbzH> .

[Yetangitu]

From man 1 php:

If none of -r -f -B -R -F -E or -S is present but a single parameter is given then this parameter
is taken as the  filename  to  parse  and  execute (same as with -f). If no parameter is present 
then the standard input is read and executed.

In other words, leaving out -f works if there are no other parameters for php but there is a single unnamed parameter.