Hello,

I'm not too good with bash scripting. How do I get input from the user? I've used the read -p before but that won't suffice this time because I need to be able to run the script on one command.

This is my task in detail:

I'm writing a scrip to convert any video file to a FLV type. The script should run from a web browser on the users command.

So the PHP needs to be something like:
How do I get the contents after -i and -o into my shell script as separate variables?

EDIT: I've been googling this with no luck, is this even possible? It has to be, otherwise how do those config scripts work?!?!


Thanks,
vxc

If you omit the "-i" and "-o" its actually a bit easier.

e.g.
/var/www/myfolder/scripts/vid2flv.sh input_file output_file

In your script input_file would be $1 and output_file would be $2.

If you're doing any functions or sets in your script you probably want to make them named variables so that you don't later use the wrong "positional parameters".

e.g.
INFILE=$1
OUTFILE=$2
echo "The input file was $INFILE and the output file was $OUTFILE."

Note: The above all applies to shell scripting. Haven't done any PHP myself so don't know if your exec syntax is correct for that.

Type "man bash" for more details. Look for positional parameters. Also look for other information such as the meaning of $*, $@ etc... which are built in mechanisms to deal with all the arguments passed in from the command line.

This works just fine. Thanks. And also thanks for giving me the right keyword, positional parameters and the man tips.