compare variable against regular expression?


 
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# 1  
Old 05-22-2008
compare variable against regular expression?

is it possible? if so, how?

i want to check a variable whether is it a number or letter in an if-else statement
# 2  
Old 05-22-2008
Code:
echo $var | grep "^[0-9][0-9]*$"

if [ $? -eq "0" ]
then
  statement when number
else 
  statement when char
fi

# 3  
Old 05-22-2008
can explain to me why you do two [0-9][0-9]?
# 4  
Old 05-22-2008
* in regexp mean 1 or more occurances of preceeding char. just to make sure that it should have aleast one num char
# 5  
Old 05-22-2008
is it possible to just directly compare regex in the if-else statement?
# 6  
Old 05-22-2008
u can use typeset to set variable type as int. like:-

Code:
typeset -i var

it will give error while assigned non-int value to it

Code:
$ var=sdf
$ ksh: sdf: bad number

after assignment u can chk for command status

if [ $? -eq "0" ]
then
statement when number
else
statement when non -int
fi
# 7  
Old 05-22-2008
The if command simply evaluates another command, and examines its exit code to decide whether to take the then branch or the else branch (if present). You can use the expr command to match a string against a regular expression, or echo | grep. I would recommend case for this, though, even if it actually uses glob expressions, not true regular expressions.

Some shells also have built-in regex operators with the [[ conditional ]] syntax.

Code:
if echo "$var" | grep '[^0-9]'; then
  echo not a number  
fi
if expr "$var" : '[^0-9]' >/dev/null; then
  echo not a number
fi
case $var in
  *[!0-9]*) echo not a number;;    # not a regex proper
esac
if [[ "$var" =~ [^0-9] ]] ; then
  echo not a number
fi

All of these are more or less functionally equivalent. (They will not reject an empty string as not a number; implementing that is left as an exercise.)

Last edited by era; 05-22-2008 at 07:23 AM.. Reason: Will accept empty string as well as numbers, actually
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