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bash script help

I have a friend who is taking an online course. She is doing very well, but is having trouble with bash scriptng. I have no knowledge of bash scripting syntax so I thought I would pass this along and see if anyone can point us in the right direction.

All this thing does is ask for input and then stops.

Code:

#!/bin/bash

#Individual Project 5 for UNIX CS126

#Create a Menu

echo "Choose an option

a. Create Users

b. Create Groups

c. Create Directories

d. Create Links

e. Set File Permissions

f. exit"

read RESPONSE

if [“$RESPONSE” = "a"]; then

#Create Users

echo "What is your NAME?"

read NAME

if [“$NAME” -!= "Tiara"]; then

echo "Your name is $NAME"

useradd -c $NAME

else

echo "That's my name too."

fi

#Create Groups

if [“$RESPONSE” = "b"]; then

echo “What is the name of the GROUP?”

read GROUP

if test -g $GROUP; then

echo “That group name exists, please try again.”

else

echo “The name of the group is $GROUP.”

groupadd $GROUP

fi

fi

#Create Directories

if [“$RESPONSE” = "c"]; then

echo “What is the name of the DIRECTORY1?”

read DIRECTORY1

if test -d $DIRECTORY1; then

echo “That directory name exists, please try again.”

else

echo “The name of the directory is $DIRECTORY1.”

mkdir $DIRECTORY1

ls -l

echo “What is the name of the DIRECTORY2?”

read DIRECTORY2

if test -d $DIRECTORY2; then

echo “That directory name exists, please try again.”

else

echo “The name of the directory is $DIRECTORY2.”

mkdir $DIRECTORY2

ls -l

echo “What is the name of the DIRECTORY3?”

read DIRECTORY3

if test -d $DIRECTORY3; then

echo “That directory name exists, please try again.”

else

echo “The name of the directory is $DIRECTORY3.”

mkdir $DIRECTORY3

ls -l

echo “What is the name of the DIRECTORY4?”

read DIRECTORY4

if test -d $DIRECTORY4; then

echo “That directory name exists, please try again.”

else

echo “The name of the directory is $DIRECTORY4.”

mkdir $DIRECTORY4

ls -l

fi

fi

fi

fi

fi

#Create Links

if [“$RESPONSE” = "d"]; then

echo “Links all directories together.”

cd $DIRECTORY1

ls $DIRECTORY2_ln $DIRECTORY3_ln $DIRECTORY4_ln

ls

fi

#File Permissions

if [“$RESPONSE” = "e"]; then

echo “ This will modify permissions.”

usermod -g $GROUP $NAME

groups $NAME

chown $NAME $DIRECTORY1

ls -l

chmod u+x, g+w $DIRECTORY1

chmod u+x, g-r, o-r $DIRECTORY2

chmod u+x, g-r, o-r $DIRECTORY3

chmod u+x, g-r, o-r $DIRECTORY4

ls -l

fi

fi

exit 0

Like I said I don't know anythng about that 'read' or the syntax of the if statements. But when I run this, I get the menu, type in 'a' and then it just stops.

Any help?

Thanks...

Change the first line from

Code:

#!/bin/bash

Code:

#!/bin/bash -vxe

That sets debug mode and makes it easier to see where things go wrong. For instance there should be at least a single space between quoted variables and the square bracket, you may reverse meaning by placing an exclamation sign in front of the equal sign -!= but adding a minus in front of that isn't valid and generally speaking it would be much, much easier to use a "case" statement instead of (nested!!!) if-elses.

There's a missing if after

Code:

if [“$RESPONSE” = "a"]; then

#Create Users

echo "What is your NAME?"

read NAME

if [“$NAME” -!= "Tiara"]; then

echo "Your name is $NAME"

useradd -c $NAME

else

echo "That's my name too."

fi

46 of the double quotes are sloping double quotes instead of straight.

As already pointed out, -!= should be !=

In the Create Links and File Permissions sections the variables are not initialised before use.

As already pointed out, a case statement would be much better than all those ifs but maybe this exercise comes in the course before the case statement has been introduced and will be modified to use case later -- cleverly showing the student how much neater a case based solution is.

The script should exit after the "please try again" errors or the user does ot have an opportunity to try again.

Error messages are conventionally written to stderr, not echo's default stdout. Maybe this convention has not been introduced yet.

A function to test directory existence and create a directory would be much neater. Maybe functions have not been introduced yet.

EDIT:

The Create Links section does not have a ln command (the first ls should be ln -s ?)

Debugging would be easier if the commands were only displayed, not executed. This could be done by using a $debug variable as shown below (change to debug= when debugged).

Code:

#!/bin/bash

#Individual Project 5 for UNIX CS126

#Create a Menu

echo "Choose an option

a. Create Users

b. Create Groups

c. Create Directories

d. Create Links

e. Set File Permissions

f. exit"

read RESPONSE

echo "DEBUG: RESPONSE is '$RESPONSE'"

debug=echo

if [ "$RESPONSE" = "a" ]; then

    #Create Users

    echo "What is your NAME?"

    read NAME

    if [ "$NAME" != "Tiara" ]; then

        echo "Your name is $NAME"

        $debug useradd -c $NAME

    else

        echo "That's my name too."

    fi

fi

#Create Groups

if [ "$RESPONSE" = "b" ]; then

    echo "What is the name of the GROUP?"

    read GROUP

    if test -g $GROUP; then

        echo "That group name exists, please try again."

    else

        echo "The name of the group is $GROUP."

        $debug groupadd $GROUP

    fi

fi

#Create Directories

if [ "$RESPONSE" = "c" ]; then

    echo "What is the name of the DIRECTORY1?"

    read DIRECTORY1

    if test -d $DIRECTORY1; then

        echo "That directory name exists, please try again."

    else

        echo "The name of the directory is $DIRECTORY1."

        $debug mkdir $DIRECTORY1

        ls -l

        echo "What is the name of the DIRECTORY2?"

        read DIRECTORY2

        if test -d $DIRECTORY2; then

            echo "That directory name exists, please try again."

        else

            echo "The name of the directory is $DIRECTORY2."

            $debug mkdir $DIRECTORY2

            ls -l

            echo "What is the name of the DIRECTORY3?"

            read DIRECTORY3

            if test -d $DIRECTORY3; then

                echo "That directory name exists, please try again."

            else

                echo "The name of the directory is $DIRECTORY3."

                $debug mkdir $DIRECTORY3

                ls -l

                echo "What is the name of the DIRECTORY4?"

                read DIRECTORY4

                if test -d $DIRECTORY4; then

                    echo "That directory name exists, please try again."

                else

                    echo "The name of the directory is $DIRECTORY4."

                    $debug mkdir $DIRECTORY4

                    ls -l

                fi

            fi

        fi

    fi

fi

#Create Links

if [ "$RESPONSE" = "d" ]; then

    echo "Links all directories together."

    cd $DIRECTORY1

    ls $DIRECTORY2_ln $DIRECTORY3_ln $DIRECTORY4_ln

    ls

fi

#File Permissions

if [ "$RESPONSE" = "e" ]; then

    echo " This will modify permissions."

    $debug usermod -g $GROUP $NAME

    groups $NAME

    $debug chown $NAME $DIRECTORY1

    ls -l

    $debug chmod u+x, g+w $DIRECTORY1

    $debug chmod u+x, g-r, o-r $DIRECTORY2

    $debug chmod u+x, g-r, o-r $DIRECTORY3

    $debug chmod u+x, g-r, o-r $DIRECTORY4

    ls -l

fi

I told her I thought the -!= was wrong. :-D I had never seen that, but like I said, I'm not familiar with bash script syntax.

After I posted this I went downstairs and had dinner and she called me and said she had changed it to use a case statement and said it was working! She's pretty smart.

She is going to send me the latest code. I'll post it up when I see it.

I'll also point her back over to this thread and look at your suggestions, etc.

Thanks for your input!

Agree with all above comments, its mostly small details; nonetheless critical for proper functioning... :)

You/she may find these worth bookmarking if you/she haven't already
http://rute.2038bug.com/index.html.gz
http://tldp.org/LDP/Bash-Beginners-G...tml/index.html
http://www.tldp.org/LDP/abs/html/