grep function not working - bash

Question

what am I doing wrong in the code below?
I am replacing the salary data within the text file, but the Telephone number field (the 3rd column) is being updated instead of the salary field (the 5th column) which is 0.
In the example below, the calculated salary for Ruben is 500.
My Desired output is:
Ruben,1223,97707001,Salaried,500
But instead, I get this (replacing zero between 9770 and 7001 with 535):
Ruben,1223,9775007001,Salaried,0
payroll_employee()
{
echo
echo "[Option: $input]"
echo "Enter Payroll of an employee "
echo
echo -en "Enter employee name: "
read Name
#Retrieve current entry into individual fields
line=`grep -i "$Name" $PAYROLL`
Name=`echo $line | cut -d "," -f1`
EmployeeID=`echo $line | cut -d "," -f2`
EmployeeHP=`echo $line | cut -d "," -f3`
EmployeeType=`echo $line | cut -d "," -f4`
Salary=`echo $line | cut -d "," -f5`
#Check if entry exist in records
if [ `count_lines "^${Name},"` -eq 0 ]
then
echo "Error: This particular record does not exist!!"
else
echo "$Name is ${EmployeeType} employee."
if [ "$EmployeeType" = "Salaried" ]
then
echo $EmployeeType
echo -en "Enter Weekly Salary:"
read swages
if [ -z $swages ]
then
swages=$Salary
else
grep -vi "$Name" $PAYROLL > tmpfile #Perform updating to salary field entry
grep -x "$line" $PAYROLL | sed -e "s/$Salary/$swages/" >> tmpfile
mv tmpfile $PAYROLL
echo "$Name's weekly payroll has been updated to \$$swages!!"
fi
echo
}
Sample code:
update_employee()
{
echo
echo "[Option: $input]"
echo "Updating employee record... "
echo "Please enter the name of the employee to update: "
echo -en "[1]Name: "
read update_name
#Retrieve current entry into individual fields
line=`grep -i "$update_name" $PAYROLL`
oldname=`echo $line | cut -d "," -f1`
oldjob=`echo $line | cut -d "," -f2`
olddept=`echo $line | cut -d "," -f3`
oldsal=`echo $line | cut -d "," -f4`
#Check if entry to update exist in records
if [ `count_lines "^${update_name},"` -eq 0 ]
then
echo "Error: This particular record does not exist!!"
else
while [ "$choice" != "6" ]
do
update_menu #Display update menu for user input,allows update of individual field or all at once
read update_choice
case $update_choice in
"1") echo -en "Please enter employee's new name: "
read new_name
if [ -z $new_name ]
then
new_name=$oldname
elif [ `count_lines "^${new_name},"` -ne 0 ] #Check if name already exist in records
then
echo "Error: Employee [$new_name] already exist in records!"
else
grep -vi "$oldname" $PAYROLL > tmpfile #Perform updating to name field entry
grep -x "$line" $PAYROLL | sed -e "s/$oldname/$new_name/" >> tmpfile
mv tmpfile $PAYROLL
echo "Employee's name $oldname has been updated to [$new_name]!!"
fi
break
;; }
All I changed was add one more column.
Salary=`echo $line | cut -d "," -f5`

Answer 1

payroll_employee()
{
echo
echo "[Option: $input]"
echo "Enter Payroll of an employee "
echo
echo -en "Enter employee name: "
read Name
#Retrieve current entry into individual fields
if [ $(grep -ciw "$Name" $PAYROLL) -eq 0 ]
then
echo "No matches found in $PAYROLL";
else
line=$(grep -iw "$Name" $PAYROLL);
Name=$(echo $line | awk -F "," print $1);
EmployeeID=$(echo $line | awk -F "," print $2);
EmployeeHP=$(echo $line | awk -F "," print $3);
EmployeeType=$(echo $line | awk -F "," print $4);
Salary=$(echo $line | awk -F "," print $5);
fi
if [ "$EmployeeType" == "Salaried" ]
then
echo $EmployeeType
echo -en "Enter Weekly Salary:"
read swages
if [ -z $swages ]
then
swages=$Salary
else
sed "/$Name/d" $PAYROLL
echo "$Name $EmployeeID $EmployeeHP $EmployeeType $swages" >> $PAYROLL;
echo "$Name's weekly payroll has been updated to \$$swages!!"
fi
echo
}
I Prefer use of AWK for delimination
also you can delete line using
sed "/$Name/d" $PAYROLL
Why to store it in other file using grep -vi and rename back and replace with original (Waste of memory & resources)
also "`" backtick should be replaced with new method of using $(..)
Also one make sure when you match strings use == sign
also in grep use -w also just to make sure that you select proper complete string
Ruben baruben rubenner all will be searched in grep if -w is not given

Answer 2

Seems like you are replacing every occurrence of the string $Salary in the payroll line, so if salary is 100 and the employee ID number is 2100, it will be replaced.
Instead of using sed at the end, you would probably be better off generating the output using printf and building the fields up that way.
Something like:
printf "%s,%s,%s,%s,%s\n" $Name $EmployeeID $EmployeeHP $EmployeeType $swages >> tmpfile
EDIT: You should fix the = to == as pointed out in the comments.
Also, to illustrate what I think is happening:
line="ABC,5100,DEF,100"
Salary=100
echo $line | sed -e s/${Salary}/XXX/
ABC,5XXX,DEF,100
If you "anchor" the query by putting a $ at the end of the search string, it will only match the last value.
echo $line | sed -e s/${Salary}$/XXX/
ABC,5100,DEF,XXX
Add some echo statements in your code to check the status of variables....