Shell Programming and Scripting

Write down your exact set of requirements as a list of tests that need to be performed. You have the first two done:Do you have any other requirements about what characters are allowed or forbidden, minimum or maximum length restrictions, etc?

With your list of requirements and the code I supplied as a sample, try creating your own script. If it doesn't work, show us the complete set of requirements and show us what you tried. Then we'll try to help you fix the script to meet your requirements.

As you get more experience programming, you'll quickly learn it is MUCH easier to write a program if you write a complete set of requirements first.

PS Note that the script I provided doesn't allow any space characters in the login name.

OK!
My requirement is below:
1. the first and last character must be an underscore or an alphabet;
2. the length of the loginname should be between 8 and 16 characters long;
3. Except for first and last character, the digit or alphabet or 4 kinds of punctuation ( @ . - _ ) is only allowable.

You are off to a great start.

I repeat:

I think I did it.

Code:

grep -n '^[_a-Z][a-Z0-9_@.-]\{6,14\}[_a-Z]$' abc

What confused me just now was why I couldn't replace [a-Z0-9_@.-] with [\w@.-].

---------- Post updated at 03:59 PM ---------- Previous update was at 03:27 PM ----------

I have another question:Why did I get an error with the first code? And it worked well with the second code
first code:

Code:

...
if grep -q "^${loginname}:" /etc/passwd && echo "The loginname already exists." && return 1
elif [ -z "${loginname}" ] && echo "You should enter a nonempty loginname." && return 1
...

Error:
second code:

Code:

...
if grep -q "^${loginname}:" /etc/passwd
then
echo "The loginname already exists."
return 1
elif [ -z "${loginname}" ]
then
echo "You should enter a nonempty loginname." 
return 1
...

According to the standards, \w in a bracket expression in a Basic Regular Expression (as used by grep when the -E and -F options are not active) matches the two characters \ and w.

What I found surprising is that I hadn't seen anything indicating that the login names you wanted to verify were:Note that the a-Z in a range expression is treated as an empty set or as an error on most systems I've seen because "a" comes after "Z" in collating order in the C/POSIX locale. Looking more closely at the standards, it is ambiguous as to whether a backwards range should be accepted. If it is accepted it would contain the characters Z, [, \, ], ^, _, `, and a; not lowercase and uppercase letters.
You also did not mention that your code checking valid login names was going to be included in a function. But, the return statement is only vaiid inside a function.

Whether it is in a function or not, there are two ways to code the equivalent of an if statement (ignoring elif clauses):

Code:

if compound-list
then
compound-list of then clause commands
else
compound-list of else clause commands
fi

which matches what you have shown above as second code, and equivalently

Code:

[B]compound-list &&
compound-list of then clause commands ||
compound-list of else clause commands

What you are showing above as first code is a cross between the two that is a syntax error; the if keyword requires matching then and fi keywords.

I had expected that you were going to try to pattern your updated code based on the code I had supplied earlier. I thought it was working for you, but you had changed the requirements concerning what was to be considered a valid login name.

What you have here is a completely different concept. And, with the short snippets of code you have shown, I no longer have any idea what you're trying to do.

@franksunnn
Is problematic because it assumes ASCII order.
Better is

Code:

grep -n '^[_a-zA-Z][a-zA-Z0-9_@.-]\{6,14\}[_a-zA-Z]$' abc

or

Code:

grep -n '^[_[:alpha:]][[:alnum:]_@.-]\{6,14\}[_[:alpha:]]$' abc

Thank you