[SOLVED] Opening files as input in bashj

barry07 · 07-24-2012, 05:52 AM

I want to use the text of a file as the input for a bash read command.

to use the -u option I need a file descriptor, presumbly this means I open the file and get a file descriptor. I have tried the reference manual, scripting guide and advance scripting guide, but have found no help.
So, how do I open a file?

barry

pan64 · 07-24-2012, 06:15 AM

is this what you are looking for: http://www.linuxquestions.org/questi...y-line-136784/ ?

414N · 07-24-2012, 06:19 AM

Maybe

Code:

while read VAR
do
	#play with $VAR
done < /path/to/file

is all you need...

barry07 · 07-24-2012, 06:28 AM

Thanks for the replies. That would solve the immediate problem, but I want to know how to attach files to channels other than 0,1,2 (stdin, stdout, stderr) I can find everythin but how to open the files and attach them to a channel number.

Am I barking up the wrong tree?

Barry

pan64 · 07-24-2012, 06:45 AM

what do you mean by channel? how do you want to use those channels? What is your original goal with it?

barry07 · 07-24-2012, 07:41 AM

Thanks all. I have found the answer. Syntax is

3< filename

then read -u 3 line

that does me. Thanks everyone

Barry