I have the following
#!/bin/bash
aprograms=`pgrep a`
echo $aprograms
which outputs:
alejandro@ubuntu:~$ bash test.sh
2 6 7 8 12 13 16 17 20 27 ...
I want to control if there is a value inside $aprograms. I tryed the following (Which I dont know if its even a valid approach):
if [ $value in $aprograms ];then
echo "found"
But doesnt work. Is there a correct way to control if there is a value is inside $aprograms?
you could use bash's parameter expansion.
shopt -s extglob
var="2 6 7 8 12 13 16 17 20 27"
if [ "${var/17?( )/}" != "$var" ] ; then echo "match"; fi
#!/bin/bash
function foo () {
echo 2 6 7 8 12 13 16 17 20 27
}
function search_value () {
for i in $2; do {
if [ $i -eq $1 ]; then {
echo "found"
} fi
} done
}
search_value 13 "$( foo )"
Just change foo with your input program:
search_value 13 "$( pgrep a )"
search_value a "a b c" will throw an error.
" around variables and using == as operator instead of -eq should work even for strings.search_value 12 "1 ; do; end ; ls ;" :)if [[ " $aprograms " =~ \ $value\ ]] ; then
echo "found"
fi
Also, with grep
if pgrep a | grep -q -w "$value" ; then
echo "found:
fi
-eq
is equal to
if [ "$a" -eq "$b" ]
Since pgrep outputs PIDs one per line, you can use grep to find the one you're looking for.
#!/bin/bash
aprograms=$(pgrep a)
echo "$aprograms"
if grep -Fxqs "$value" <<< "$aprograms"
then
echo "found"
fi
Always quote variables when they are expanded in order to preserve whitespace.
The -F option to grep uses the pattern as a fixed string instead of a regex. The -x option matches the whole line so smaller strings or numbers don't match parts of larger ones. The options -q and -s suppress output and error output. If your version of grep doesn't have these, then use:
if grep -Fx "$value" <<< "$aprograms" > /dev/null >&2
