Error while displaying data from MySql using bash script - mysql

Question

Here I have a database msmdb which contains a table emp.
I am trying to display the values in that table using a bash script but there comes an error saying command not found (at line 8 after the echo).
USER_NAME=root
USER_PWD=prj
DB_NAME=msmdb
flag=0
delimiter="a3f4g52"
echo "select id ,'"$delimiter"', name,'"$delimiter"', city, '"$delimiter"', salary FROM emp" | \
mysql -u $USER_NAME --password=$USER_PWD --database=$DB_NAME | \
while read row ;
do
if [ $flag == 0 ]; then
flag=1
else
name=` echo $row | \
awk -F "$delimiter" '{print $1}' | \
sed -e 's/ *$//' | sed 's/^ *//'`
age=` echo $row | \
awk -F "$delimiter" '{print $2}' | \
sed -e 's/ *$//' | sed 's/^ *//'`
bornPlace=`echo $row | \
awk -F "$delimiter" '{print $3}' | \
sed -e 's/ *$//' | sed 's/^ *//'`
echo ">$name<"
echo ">$age<"
echo ">$bornPlace<"
fi
done
Any idea about what might be the reason for error ??

Answer 1

In line 8 you have a space following the linebreak (\) therefore the linebreak doesn't work.

Answer 2

Got The Script working properly ......
Just had to remove ' \ ' from the program and it worked.
The Correct bash script is
USER_NAME=root
USER_PWD=prj
DB_NAME=msmdb
flag=0
delimiter="a3f4g52"
echo "select id ,'"$delimiter"', name,'"$delimiter"', city, '"$delimiter"', salary FROM emp" |
mysql -u $USER_NAME --password=$USER_PWD --database=$DB_NAME |
while read row ;
do
if [ $flag == 0 ]; then
flag=1
else
name=` echo $row | awk -F "$delimiter" '{print $1}' | sed -e 's/ *$//' | sed 's/^ *//'`
age=` echo $row | awk -F "$delimiter" '{print $2}' | sed -e 's/ *$//' | sed 's/^ *//'`
bornPlace=` echo $row | awk -F "$delimiter" '{print $3}' | sed -e 's/ *$//' | sed 's/^ *//'`
echo ">$name<"
echo ">$age<"
echo ">$bornPlace<"
fi
done