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how to check if a string consist of any space?

 
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# 1  
Old 07-29-2008
how to check if a string consist of any space?
hi all
how do i check if a string consist of any space by using shell script.
i have the following code
while test -z $string
do
//prompting for another string if string is length 0
done

when i type "a b" it give me an error

test: a: binary operator expected

thanks
# 2  
Old 07-29-2008
You need to quote "$string" inside double quotes if it contains any whitespace. Aside from that, it might be more elegant to use case instead of if, especially for coping with all-whitespace input as well as blank inputs. Also, with case, there is no need to quote the variable you are testing:

Code:
while true; do
  read string
  case $string in *[^\ ]*) break;; *) echo try again;; esac
done

The above will pass if there is any non-whitespace character in the input.
Last edited by era; 07-29-2008 at 04:12 AM.. Reason: Elaborate with an explicit while loop
# 3  
Old 07-29-2008
-z tests, if the string is empty, not if it contains a blank.

This checks for any blanks and/or tabs
Code:
egrep -q "[[:space:]]" infile
echo $?

You can check $? if it found something.
# 4  
Old 07-29-2008
if i am taking user input and the input contain space??
# 5  
Old 07-29-2008
Just as era showed already. You should try it a bit yourself, example on the prompt:

Code:
root@isau02:/data/tmp/testfeld> while true; do read a; echo ${a}| egrep -q "[[:space:]]"; if [[ $? = 0 ]]; then echo blank found; fi; done     jajaja
jaja jaja
blank found
lala            lala
blank found

# 6  
Old 07-29-2008
The following will fail if there is a space anywhere in the input.

Code:
case $string in *\ *) echo try again;;  *) break;; esac

If you want to use egrep, you can avoid the Useless Use of Test $? with this:

Code:
if echo "$a" | egrep -q "[[:space:]]" ; then ...

The code above fails to quote $a correctly, but in this particular case, it's harmless. Still, a single case is cheaper and simpler than echo + grep and IMHO more elegant.
# 7  
Old 07-29-2008
sorry..i tried using all these code..but it dun work
i have the following code

while test -z $username
do
echo -n "enter username: " ;read username
done

this code gives me an error if i enter something like this
a b
test: a: binary operator expected

thanks a lot
 
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