Face Problems in Scripting

[james99]

LIMIT=20
for((a=0;a < LIMIT; a++))
do
echo "Adding User$a..."
useradd -m -u 19$a -g test -d /home/test$a -s /bin/sh -c "Test$a" tester$a
done
echo;
exit 0

the above is the code i figure out after reading some online BASH guide
but everytime i add the first few digits from 0 - 9 [190 - 199] is added to the
System users and 10 -19 [1910 - 1919] is to the Normal user

how can i add another 0 infront of those Single Digit number 0-9 to prevent them from adding into the System Users? 0 - 9 -> 00 - 09

Thanks James.

[Franklin52]

Hi James,

You can format the numbers with printf, try this:

LIMIT=20
for((a=0;a < LIMIT; a++))
do
  echo "Adding User $a ..."
  user=`printf "%02d" $a`
  useradd -m -u 19$user -g test -d /home/test$user -s /bin/sh -c "Test$user" tester$user
done
echo;
exit 0

Regards

[james99]

Great it works but all numbers have a 0 infront[after some tweaking]
i just need those 0 - 9 to have a 0 infront

IF i use a if else condition i have think of

if [ a < 10 ]
then
user=`printf "&#37;01d"$a`
else
user=`printf $a`
fi

when i try to mkdir /home/test$user
it did make some directories but they have no 0 infront for all folders as in test0 - test99 after executing
but it said no such file or directory for the line {if [ a < 10 ]}

any ideas?

[Franklin52]

In your printf statement there must be a space between the arguments.
This should work for the digits 0-99:

user=`printf "%02d" $a`

You don't have to use an if else statement.

Regards

[james99]

In your printf statement there must be a space between the arguments.
This should work for the digits 0-99:

user=`printf "%02d" $a`

You don't have to use an if else statement.

Regards

Thanks but if i want to apply it to 0 - 9 only and leave the rest untouch?

[Franklin52]

Thanks but if i want to apply it to 0 - 9 only and leave the rest untouch?

Right, try this to see what you get:

LIMIT=20
for((a=0;a < LIMIT; a++))
do
  user=`printf "%02d" $a`
  echo $user
done

Regards

[james99]

Right, try this to see what you get:

LIMIT=20
for((a=0;a < LIMIT; a++))
do
  user=`printf "&#37;02d" $a`
  echo $user
done

Regards

Thanks.
I solve it by using 2 FOR loops
1 with user=`printf "%01d" $a` for 0 - 9 Results : 00-09
another is user=`printf $a` for 10 - 99 Results : 10-99

[Cabhan]

printf "%01d" 9

that produces "09"? It shouldn't...and you shouldn't need two different for loops. The "%02d" pattern means "Treat the argument as a number. The number should be 2 characters long, and if it is less than 2 characters, then left-pad it with 0s.

In fact, here's my script:

#!/bin/bash

# @(#)  printf-2-digits         Demonstrate printf's formatting abilities

LIMIT=20

for((a=0; a < LIMIT; a++)); do
        user=$(printf '%02d' $a)
        echo $user
done

It produces the following output:

alex@danu ~/test/bash $ ./printf-2-digits 
00
01
02
03
04
05
06
07
08
09
10
11
12
13
14
15
16
17
18
19

[Franklin52]

printf "%01d" 9

that produces "09"? It shouldn't...and you shouldn't need two different for loops. The "%02d" pattern means "Treat the argument as a number. The number should be 2 characters long, and if it is less than 2 characters, then left-pad it with 0s.

In fact, here's my script:

#!/bin/bash

# @(#)  printf-2-digits         Demonstrate printf's formatting abilities

LIMIT=20

for((a=0; a < LIMIT; a++)); do
        user=$(printf '%02d' $a)
        echo $user
done

[/code]

That's what I've early mentioned:

Right, try this to see what you get:

LIMIT=20 
for((a=0;a < LIMIT; a++)) 
do 
  user=`printf "%02d" $a` 
  echo $user 
done

But I think he didn't try it

Regards