"Read" command in bash script is being skipped

Question

I'm creating a script to update my linux distribution if I need to wipe the HD or I need to install Linux on another machine. So this script basically install all the programs I usually need. At the beginning a have a "read" command that asks if I want to install all the packages automatically or not. If I choose not, for each program not found it should ask me I want it to be installed and I use this code

if [[ $installall == "yes" ]]; then
    echo " Installing $sciprog..."
    sudo apt-get install -y $sciprog >/dev/null
    {
        scitest=`dpkg -s $sciprog | grep Status`
    } 2>${HOME}/musthave.errorlog
    if [[ $scitest != "Status: install ok installed" ]]; then
        echo " I've encountered problems installing $sciprog that I can't resolve. "
        echo " Consider installing $sciprog manually. "
        {
            echo "=========="
            echo " $sciprog"
        } >>${HOME}/musthave.notinstalled
    else
        echo " $sciprog installed correctly!"
        {
            echo "=========="
            echo " $sciprog"
        } >>${HOME}/musthave.installed
    fi
else
    echo " Seems like $sciprog is not installed... Do you want to download it?"
    echo " Type 'y' for yes."

    read secondyn ### THIS IS THE GUILTY COMMAND ###

    if [[ $secondyn == "y" ]]; then
        echo " Installing $sciprog ..."
        sudo apt-get install -y $sciprog >/dev/null
        {
            checkinstall=`dpkg -s $sciprog | grep Status`
        } 2>>${HOME}/musthave.errorlog
        if [[ $checkinstall != "Status: install ok installed" ]]; then
            echo " I've encountered problems installing $sciprog that I can't resolve. "
            echo " Consider installing $sciprog manually. "
            {
                echo "=========="
                echo " $sciprog"
            } >>${HOME}/musthave.notinstalled
        else
            echo " $sciprog installed correctly!"
            {
                echo "=========="
                echo " $sciprog"
            } >>${HOME}/musthave.installed
        fi
    else
        echo " Skipping $sciprog ..."
        {
            echo "=========="
            echo " $sciprog"
        } >>${HOME}/musthave.notinstalled
    fi
### some more code which works as expected. All the code above is inside a 
### while...do...done loop which reads line by line the file at the end
done <${HOME}/file.list

But if I run the script, it skips the "read" command in the else clause and assumes it to be "n"...

I can't figure out why, there are other read function also inside if...then...else...fi loops and they work as expected...

Any ideas?

It's hard to tell because it looks like your indenting got mangled horribly in posting, but you seem to be missing a final `fi`. Also, the `read` works for me. — jw013, Aug 02 '11 at 11:59
Also, what is the shebang at the top of your script? `#!/bin/bash` or other? AND, just to be sure, you **are** running this from a command-line in a terminal session, yes? If you are crontabing it, the script will read a '' as the result and keep going. Good luck. — shellter, Aug 02 '11 at 15:01
@jw013 well, maybe it has been cut while I copied/pasted it but the fis are correct, otherwise the script would raise an error — , Aug 02 '11 at 15:30
@shellter it's `#!/bin/bash` yes. And yes, I run the script from the command line using `./script.sh` — , Aug 02 '11 at 15:31
@Vieler : Forgot to ask, do you see the messages above the `read` but inside the `else`? (It seems like you know what you're doing, so sorry for asking what is probably too obvious a question) ; Good luck! — shellter, Aug 02 '11 at 17:59
@shellter Don't worry, actually I'm not that good in bash scripting, it's rather new for me. Anyway, yes, the echo commands work fine and I see the strings printed on stdout. — , Aug 03 '11 at 09:12

score 16 · Accepted Answer · answered Aug 03 '11 at 09:55

The relevant portions of the code are still not complete but based on the comments I'm going to guess that your while loop looks like

while read -r ... ; do 
    # do stuff ...

    # read user input
    read -r var

done < file

From this the problem is immediately apparent: the inner read is getting its input from the same place as the outer loop, namely stdin which has been redirected from file, and not the user. For a slightly more portable alternative that does not depend on kernel-level support for /dev/tty, just use a different file descriptor other than stdin for the while loop.

while read -r ... <&9; do
    # loop stuff

    # stdin still attached to the terminal untouched, 
    # so this reads from the terminal as expected
    read -r var 

done 9< file

Notice that this example uses fd 9 for the file, leaving fd 0 (stdin) alone. Take a look at the BashFAQ 089 for more details.

Yes, that was the problem. Also I guess this solution is better for portability so thanks for putting me on the right track. — , Aug 03 '11 at 10:04

score 9 · Answer 2 · answered Aug 02 '11 at 19:08

9

Try reading from the controlling terminal device:

read secondyn </dev/tty

answered Aug 02 '11 at 19:08

jon

91
1

score 1 · Answer 3 · answered Aug 03 '11 at 05:15

1

read secondyn < /proc/${PPID}/fd/0

That will look to the parent's input, which should still be stdin.

answered Aug 03 '11 at 05:15

brightlancer

2,059
15
7

"Read" command in bash script is being skipped

3 Answers3