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Problem with Looping

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# 8  
Old 06-11-2013
Quote:
Originally Posted by RudiC
Be careful - if you don't modify the creation of the variable, you're stuck in an infinite loop. So you better not define i to be a constant. However, depending on your shell (bash in my case), this may work for you:
Code:
until typeset i=J; [ "$i" cond1 -a "$i" cond2 -a "$i" cond3 ]; do :; done

I understand the usage of until. But it doesn't work well here.
I'm so sad because if I use your code, I will not get the error indication when the variable i doesn't meet any condition.
And thanks for your advice. In fact, the variable i will be assigned the value by the command read.
e.g.
Code:
read -p "Please enter something: " i

and what I want is to check user's input whether to meet all the three conditions. If not, reenter again!
Thank you.

---------- Post updated at 05:39 PM ---------- Previous update was at 05:37 PM ----------

Quote:
Originally Posted by Corona688
This is shell, you don't "create" a variable per se. If you use a variable you haven't used before, it's just blank, not an error.
I don't create a variable. I just want to assign the value to the variable by using command read.
like this:
Code:
read -p "Please enter something: " i

THX
# 9  
Old 06-11-2013
Quote:
Originally Posted by franksunnn
I'm so sad because if I use your code, I will not get the error indication when the variable i doesn't meet any condition.
You can avoid printing the first time.

Code:
ERR=0
until [ ... ]
do
        [ "$ERR" -eq 1 ] && echo "error"
        ...
        ERR=1
done

The && is a short-form if/then. If the first command succeeds, the second command will be executed.
# 10  
Old 06-11-2013
Also Rudi's solution should work
Code:
until
 read -p "Please enter something: " i &&
 [ "$i" != 1 ] &&
 [ "$i" != 2 ] &&
 [ "$i" != 3 ]
do
  echo "error $i is invalid"
done
echo "fini with $i"

Quote:
Please enter something: 1
error 1 is invalid
Please enter something: 2
error 2 is invalid
Please enter something: 3
error 3 is invalid
Please enter something: 4
fini with 4
Last edited by MadeInGermany; 06-11-2013 at 07:19 PM.. Reason: better example
# 11  
Old 06-11-2013
Quote:
Originally Posted by MadeInGermany
Also Rudi's solution should work
Code:
until
 read -p "Please enter something: " i &&
 [ "$i" != 1 ] &&
 [ "$i" != 2 ] &&
 [ "$i" != 3 ]
do
  echo "error $i is invalid"
done
echo "fini with $i"

what if I want to print different indications?
e.g.:
if variable i equals to 1, I will get the standard output "hello!"
if variable i equals to 2, I will get the standard output "nice!"
if variable i equals to 3, I will get the standard output "good!"
# 12  
Old 06-11-2013
A simple way to do that is case.

Code:
case "$VAR" in
1)
        echo "a"
        ;;
2)
        echo "b"
        ;;
3)
        echo "c"
        ;;
esac

# 13  
Old 06-12-2013
Going back to the original posting in this thread, the following might do what you were trying to do:
Code:
#!/bin/bash
# name: loginname_validate
# purpose: to validate the loginname which is used to add a new user
i=0
while [ $i -ne 1 ]
do
        #for ksh: if ! IFS="" read -r loginname?"Please enter a new loginname: "
        #for bash:if ! IFS="" read -r -p"Please enter a new loginname: " loginname
        if printf "Please enter a new loginname: " &&
                ! IFS="" read -r loginname
        then    printf '\nUnexpected EOF or error reading login name.\n' >&2
                exit 2
        fi
        if [ -z "$loginname" ]
        then    echo 'Loginname must not be empty.'
        elif [ ${#loginname} -lt 8 -o ${#loginname} -gt 16 ]
        then    echo 'Loginname must be between 8 and 16 characters long.'
        elif [ "$loginname" == "${loginname#[_[:alpha:]]}" ]
        then    echo 'Loginname 1st character must be underscore or alphabetic.'
        elif [ "$loginname" == "${loginname#*[[:lower:]]}" ]
        then    echo 'Loginname must contain a lowercase alphabetic character.'
        elif [ "$loginname" == "${loginname#*[[:upper:]]}" ]
        then    echo 'Loginname must contain an uppercase alphabetic character.'
        elif [ "$loginname" == "${loginname#*[[:digit:]]}" ]
        then    echo 'Loginname must contain a numeric character.'
        elif [ "$loginname" == "${loginname#*[[:punct:]]}" ]
        then    echo 'Loginname must contain a punctuation character.'
        elif [ "$loginname" != "${loginname#*[![:alnum:][:punct:]]}" ]
        then    echo "Loginname can only contain alphanumeric and punctuation characters."
        elif [ -d ~"$loginname" ]
        then    printf "Loginname already exists: %s\n" "$loginname"
        else    echo 'You have entered a valid loginname.'
                i=1
                continue
        fi
        echo 'Loginname must meet following criteria:
1.  8 <= number of characters in name <= 16,
2.  1st character in name must be an alphabetic character or an underscore,
3.  name must contain at least one lowercase letter,
4.  name must contain at least one uppercase letter,
5.  name must contain at least one numeric character,
6.  name must contain at least one punctuation character, and
7.  name must not already be assigned.'
done
printf "Loop terminated with loginname set to \"%s\"\n" "$loginname"

However, I find some things VERY strange about this problem. I've never seen a Linux or UNIX system that wanted login names to contain punctuation characters like `, ~, ', ", !, #, $, ^, &, *, (, ), {, }, \, /, <, and >. The requirement to have punctuation characters sounds more like a validation check for passwords than for login names.

Notes:
  1. This script is written to be portable for use with any POSIX conforming shell. The way the ksh and bash read built-ins specify prompts are incompatible with each other. The portable if statement in brown above can be removed along with the comment at the start of one of the two lines above it for use with ksh only or with bash only.
  2. The code in orange above prevents characters in the space class and the control character class from being accepted. For login names, I would think this would be a useful addition. For passwords, I would think this code should be removed.
  3. Your original script didn't mention it, but I would think any validation program for a new user name would need to verify that the name being checked is not already in use as a login name. Unfortunately, I don't know of any portable way to test it. (A grep on /etc/passwd would work on some systems, but certainly not all.) The code in blue above tries to check this by looking for the home directory for given login name. If the shell on your system finds the home directory for the given name, it seems safe to assume that the name is already in use. If for some reason some users' directories are not accessible, some duplicate names might not be caught by this test. If this test isn't sufficient on your system, I assume that you'll be able to replace it with something that works in your environment. If you don't want a test like this at all, just remove the code in blue.
Hope this helps...
This User Gave Thanks to Don Cragun For This Post:
Neo
# 14  
Old 06-12-2013
Quote:
Originally Posted by Don Cragun
Going back to the original posting in this thread, the following might do what you were trying to do:
Code:
#!/bin/bash
# name: loginname_validate
# purpose: to validate the loginname which is used to add a new user
i=0
while [ $i -ne 1 ]
do
        #for ksh: if ! IFS="" read -r loginname?"Please enter a new loginname: "
        #for bash:if ! IFS="" read -r -p"Please enter a new loginname: " loginname
        if printf "Please enter a new loginname: " &&
                ! IFS="" read -r loginname
        then    printf '\nUnexpected EOF or error reading login name.\n' >&2
                exit 2
        fi
        if [ -z "$loginname" ]
        then    echo 'Loginname must not be empty.'
        elif [ ${#loginname} -lt 8 -o ${#loginname} -gt 16 ]
        then    echo 'Loginname must be between 8 and 16 characters long.'
        elif [ "$loginname" == "${loginname#[_[:alpha:]]}" ]
        then    echo 'Loginname 1st character must be underscore or alphabetic.'
        elif [ "$loginname" == "${loginname#*[[:lower:]]}" ]
        then    echo 'Loginname must contain a lowercase alphabetic character.'
        elif [ "$loginname" == "${loginname#*[[:upper:]]}" ]
        then    echo 'Loginname must contain an uppercase alphabetic character.'
        elif [ "$loginname" == "${loginname#*[[:digit:]]}" ]
        then    echo 'Loginname must contain a numeric character.'
        elif [ "$loginname" == "${loginname#*[[:punct:]]}" ]
        then    echo 'Loginname must contain a punctuation character.'
        elif [ "$loginname" != "${loginname#*[![:alnum:][:punct:]]}" ]
        then    echo "Loginname can only contain alphanumeric and punctuation characters."
        elif [ -d ~"$loginname" ]
        then    printf "Loginname already exists: %s\n" "$loginname"
        else    echo 'You have entered a valid loginname.'
                i=1
                continue
        fi
        echo 'Loginname must meet following criteria:
1.  8 <= number of characters in name <= 16,
2.  1st character in name must be an alphabetic character or an underscore,
3.  name must contain at least one lowercase letter,
4.  name must contain at least one uppercase letter,
5.  name must contain at least one numeric character,
6.  name must contain at least one punctuation character, and
7.  name must not already be assigned.'
done
printf "Loop terminated with loginname set to \"%s\"\n" "$loginname"

However, I find some things VERY strange about this problem. I've never seen a Linux or UNIX system that wanted login names to contain punctuation characters like `, ~, ', ", !, #, $, ^, &, *, (, ), {, }, \, /, <, and >. The requirement to have punctuation characters sounds more like a validation check for passwords than for login names.

Notes:
  1. This script is written to be portable for use with any POSIX conforming shell. The way the ksh and bash read built-ins specify prompts are incompatible with each other. The portable if statement in brown above can be removed along with the comment at the start of one of the two lines above it for use with ksh only or with bash only.
  2. The code in orange above prevents characters in the space class and the control character class from being accepted. For login names, I would think this would be a useful addition. For passwords, I would think this code should be removed.
  3. Your original script didn't mention it, but I would think any validation program for a new user name would need to verify that the name being checked is not already in use as a login name. Unfortunately, I don't know of any portable way to test it. (A grep on /etc/passwd would work on some systems, but certainly not all.) The code in blue above tries to check this by looking for the home directory for given login name. If the shell on your system finds the home directory for the given name, it seems safe to assume that the name is already in use. If for some reason some users' directories are not accessible, some duplicate names might not be caught by this test. If this test isn't sufficient on your system, I assume that you'll be able to replace it with something that works in your environment. If you don't want a test like this at all, just remove the code in blue.
Hope this helps...
You are right! Loginname doesn't look so complex. I just took a little practice.
I have a question: what if I want the loginname to start with an underscore or an alphabet and not to contain any spaces? What should I do?
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