How do i make a script that checks that my input is letters or numbers ?

ministeren · 05-19-2010, 08:30 AM

Hi guys. as mentioned in the headline:

lets say. i would like to know how to make a script that can check that i input 3 letters and 3 numbers in some field.

i am completely noob in this programming scene, but think its quite interesting.

Hope rou gyus can help me.

Thanks in advance.

//Chris

PMP · 05-19-2010, 08:37 AM

What have you tried so far ?
Tried Google ?

grail · 05-19-2010, 08:41 AM

Well you could use straight bash or a command like grep, sed or awk and even other languages such as Perl.

So I would suggest either man of one of the commands or searching google for tutorials on a language.

ministeren · 05-20-2010, 04:01 AM

Hi again.
See below. this is what my script looks like.

Thanks in advance

#!/bin/bash

beertypes=(carlsberg tuborg masterbrew corona)
beertypenames=(car tub mas cor)

beertype=""

findBeerType()
{
num=0
for types in ${beertypenames[@]}; do
if [ $types = ${newbeer:0:3} ]; then
beertype=${beertypes[$num]}
else
num=$(( $num + 1 ))
fi
done
}

checkBeerName()
{

if [ ${#newbeer} -eq 6 ]; then
echo "hej"
if [ $newbeer -eq "[a-z][a-z][a-z][0-9][0-9][0-9]" ] ; then
echo "wowwow"
fi
fi

}

done

findBeerType
checkBeerName

echo $beertype

catkin · 05-20-2010, 04:14 AM

Have a look at bash's [[ ]] test and its =~ regular expression comparison operator. You want a regular expression beginning with ^ and ending with $ (and the regular expression must not be quoted)..

grail · 05-20-2010, 04:18 AM

Right, well firstly CODE tags are your friend, and ours so we can read your script better:

Code:

#!/bin/bash

beertypes=(carlsberg tuborg masterbrew corona)
beertypenames=(car tub mas cor)


beertype=""

findBeerType()
{
    num=0
    for types in ${beertypenames[@]}; do
        if [ $types = ${newbeer:0:3} ]; then
            beertype=${beertypes[$num]}
        else
            num=$(( $num + 1 ))
        fi
    done
}


checkBeerName()
{

    if [ ${#newbeer} -eq 6 ]; then
        echo "hej"
        if [ $newbeer -eq "[a-z][a-z][a-z][0-9][0-9][0-9]" ] ; then
            echo "wowwow"
        fi
    fi

}

done

findBeerType
checkBeerName

echo $beertype

So here are my questions for you:

1. As you are using functions, after you set the 3 variables at the beginning the first command is - done - Please explain??

2. You call the findBeerType function first but then refer to an uninitialised variable - newbeer - How is this supposed to work?

3. checkBeerName also refers to "newbeer" and again I cannot see where it receives values from??

Maybe if you answer some of these we can look at your issue, which I assume is the following:

Quote:

if [ $newbeer -eq "[a-z][a-z][a-z][0-9][0-9][0-9]" ] ; then

ministeren · 05-20-2010, 04:43 AM

Hi again - i thought "done" should be there to tell, that the first part i done

.

i dont know what you mean by uninitialized variable - this is almost the 1 time i tried this scripting .

i cant see it either where checkBeerName gets its value from , This is just how i thought it should be. Im sorry - i'm not skilled at programming

.

i think You are right
this is what i need to check for
if [ $newbeer -eq "[a-z][a-z][a-z][0-9][0-9][0-9]" ] ; then

i need to check , that there are 3 letters and 3 numeric values afterwards.

isnt this the way to do it ?.

Thanks.

I really appreciate your help !!

grail · 05-20-2010, 05:12 AM

Quote:

Im sorry - i'm not skilled at programming

This is not a problem, we all start somewhere

You need to address the issues I have outlined before moving onto the correct way to do the last part.
I would suggest breaking your code down and making sure each function does what it is required to do before moving on.

Code:

findBeerType()
{
    num=0
    for types in ${beertypenames[@]}; do
        if [ $types = ${newbeer:0:3} ]; then
            beertype=${beertypes[$num]}
        else
            num=$(( $num + 1 ))
        fi
    done
}

In this function you have 6 variables:

num - counter initialised to 0

beertypenames - array initialised at the start of the script

types - set to each value in turn in the previous array

newbeer - ???

beertypes - array initialised at the start of the script

beertype - set to one of the beertypes based on comparison between types and newbeer (again this test cannot work as newbeer never set)

Try commenting out the rest of the code and get this function to work correctly first.

catkin · 05-20-2010, 05:15 AM

Quote:

Originally Posted by ministeren

this is what i need to check for
if [ $newbeer -eq "[a-z][a-z][a-z][0-9][0-9][0-9]" ] ; then

i need to check , that there are 3 letters and 3 numeric values afterwards.

isnt this the way to do it ?

It isn't for the following reasons:

-eq is a numeric comparison operator; you want the string comparison operator = (or == if you prefer; they are equivalent).
The double quotes prevent bash expanding what's between them so bash compares $newbeer with exactly what is between the quotes.
The [ ] form of testing does not expand [a-z][a-z][a-z][0-9][0-9][0-9] even without the double quotes. The [[ ]] form of testing is preferred for reasons explained here, which include doing what you want.

Putting that all together, you need if [[ $newbeer = [a-z][a-z][a-z][0-9][0-9][0-9] ]] ; then. This will work, as long as only lowercase letters are used and the letters appear before the numbers -- a refinement of your initial requirement of three letters and three numbers.

catkin · 05-20-2010, 05:23 AM

Quote:

Originally Posted by grail

Code:

            beertype=${beertypes[$num]}
            num=$(( $num + 1 ))

Hello Grail

Minor refinement, ever in search of the perfect code: you don't need the $ in $num in those places. It's not wrong but it's redundant. In both those places, bash expects a numerical expression and variables can be referenced in numerical expressions without a leading $.

Personallly I prefer let num++ over num=$(( $num + 1 )), finding it more legible but that's a matter of personal taste.

grail · 05-20-2010, 05:34 AM

Hi catkin

I didn't put any of them in

I actually just go with ((num++))

catkin · 05-20-2010, 06:35 AM

Quote:

Originally Posted by grail

Hi catkin

I didn't put any of them in

I actually just go with ((num++))

Ah! You are ahead of me already

grail · 05-20-2010, 07:34 AM

nah ... i still read all your posts when i see them

tuxdev · 05-20-2010, 09:36 AM

I personally go with (( ++num )).. something I picked up from too much C++.

Anyhow, catkin's right on the mark. However, I'd use the [[:lower:]] and [[:digit:]] character classes instead of [a-z] and [0-9]. The symbolic character classes respect locale. The manual ones don't, with some pretty spectacular results sometimes.

Also, make sure to quote your array loop, like

Code:

for types in "${beertypenames[@]}" ; do