hide output with errors from stdout

Question

need hide error from the script

for s in $s_list; do
    if [ "${s}" = "test" ]; then
        db_status=$(mysql -h localhost -P 3306 -u test -ptest -e "show create database test;"  | awk {'print $1'} | tail -n 1 )
        db_status_error=$(mysql -h localhost -P 3306 -u test -ptest -e "show create database test;" 2>&1 | awk {'print $1'} | tail -n 1 )
        # echo $db_status_error   
        if [ "$db_status" == "test" ]; then  
            echo "Database exist, need wait..." 
            sleep 2;
        elif [ "$db_status_error" == "ERROR" ] < /dev/null > /dev/null 2>&1 ; then
            echo "Database does not exist" 
            sleep 2;
            exit 0
        fi
    fi
done

result is

ERROR 1049 (42000) at line 1: Unknown database 'test'
Database does not exist

i need just line with Database does not exist

Hint: the message is probably coming from your db_status=$(...) line, not from your elif [ ... ] line... — twalberg
– twalberg, Commented May 17, 2016 at 17:50

user3657941 · Accepted Answer · 2016-05-17 18:08:28Z

1

You need to redirect stderr to /dev/null when you set db_status:

db_status=$(mysql -h localhost -P 3306 -u test -ptest -e "show create database test;" 2> /dev/null | awk {'print $1'} | tail -n 1 )

answered May 17, 2016 at 18:08

user3657941

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