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Simple negated test condition

 
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# 1  
Old 07-17-2015
Simple negated test condition
Without grep, I'd like to make a test condition so that any a word that does not have the successive letters car in it will be echoed. for example, bluecar will contain "car" so it will show up as a no
Code:
var=bluecar
$ echo $var|if [[ "$var" != "car" ]]; then echo "yes";fi

yes

this variable contains "car" so I want it to print nothing. However, I can do this by using *car* but that appears to grab any letters c a r and not the successive letters car. Is there any way to do this with a test condition?
Last edited by Scrutinizer; 07-17-2015 at 03:37 PM..
# 2  
Old 07-17-2015
Might be something like the following:
Code:
 echo $var | awk '$1 ~ car {print yes}'
or
echo $var | awk '/car/ {print yes}'

# 3  
Old 07-17-2015
Not sure I understand your problem. The *car* pattern seems to do what is requested:
Code:
var=bluecar
echo $var|if [[ "$var" != *car* ]]; then echo "yes";fi
var=blue
echo $var|if [[ "$var" != *car* ]]; then echo "yes";fi
yes

# 4  
Old 07-17-2015
But please omit the echo $var|; the following if statement does not do anything with it.
Unless you eventually want to feed an animal, like the cat in the following:
Code:
echo "$var"|if [[ "$var" != *car* ]]; then cat;fi

This User Gave Thanks to MadeInGermany For This Post:
RudiC
# 5  
Old 07-17-2015
Does this work?
(bash)
Code:
[ -z "${var/car}" ] && echo yes || echo no

EDIT: n/m too hot to think clear, figured after re-reading.
# 6  
Old 07-17-2015
Just using shell variable expansions specified by the standards, you could also use:
Code:
[ "$var" = "${var%car*}" ] && echo yes

or:
Code:
[ -n "${var%%*car*}" ] && echo yes

Note that test expression and [ expression ] are equivalent. The test utility is almost always a shell built-in, but [[ ... ]] (in shells that support it) is part of the grammar of the shell; not a utility. The expressions accepted by test and [[ ... ]] are not the same (although some expressions will be handled the same way by both).
 
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