How to read specific item in a list

[emokoena]

I have created a list and would like to read specific items in it, I only can read 1

import cPickle, sys, shelve

## open file that stores scores
    
choice = None
employees = []
while choice != "0":

    print \
            """
            411 TUCKSHOP

            OPTIONS:
            0 - Exit
            1 - Employee: Add New Employee
            2 - Employee: Display Balance
            3 - Show All Employees and Balances
            4 - Employee: Make deposit
            5 - Purchase items @ Tuck Shop
            6 - Remove all emmployees from database

            """
    choice = raw_input("Choice: ")
    print

    #exit if choice is "0"
    if choice == "0":
        print "Good bye, please call again..."

    #Choice 1 - Add Employees to a file    
    elif choice == "1":
        #Create a sequence
        empnum = int(raw_input("Enter Employee Number: " ))
        name = raw_input("Enter Employee Name:" )
        sname = raw_input("Enter Employee Surname:" )
        dept = raw_input("Enter Department:" )
        bal = float(raw_input("Employee Balance R: "))
        entry = (empnum, name, sname, dept, bal)
        employees.append(entry)
        
        #create pickle file
        pickle_file = open("emp_data.dat", "a+")
        cPickle.dump(entry, pickle_file)
        #close pickle file
        pickle_file.close()
   
        #create a shelve
        #empshelve = shelve.open("emp_data2.da")
        #empshelve ["emp_number"] = [emp_number]
        
        #Ensure sync
        #empshelve.sync()
        #close shelve
        #empshelve.close()

    #Choice 2 - Display Balances
    elif choice == "2":
        
        pickle_file = open("emp_data.dat", "r")
        emps = cPickle.load(pickle_file)
        #Prompt user for search
        for entry in employees:
            empnum, name, sname, dept, bal = entry
        print name, "\t", bal
        empn = int(raw_input("\nChoose employee number: "))
        
        print "\nDisplaying Balance\n"
        if empn in emps:
            print "Balance ==> R ", emps[4] 
        else:
            print "Sorry"
        pickle_file.close()

[milesmajefski]

In Python 3 the print statement is gone, it is replaced by the print function.

http://www.harshj.com/2008/12/09/the-new-print-function-in-python-3/

[dwblas]

You have to check the first element of the tuple against the entry number:

        found = False
        for entry in employees:
            if empn == entry[0]:
                print "Balance ==> R ", entry[4]
                found = True
        if not found:
            print "Not a valid employee number"

Generally, a dictionary is used, with the employee number as the key pointing to a list of items http://greenteapress.com/thinkpython/html/book012.html so you can then use, if empn in employee_dict.

[dwblas]

A couple of hints:
# 6 is easy --> employees = [] ## re-initialize as an empty list
When adding an employee, check if the number is already in the list, since you don't want the same employee number in the list twice.
Use a function to look up an employee number, since you will be doing this for every choice on the menu.
Use a list instead of a tuple so you can change it (make deposits)
Read the pickle file once, before the while loop, instead of every time through the loop, and write/dump/close it once, after the while loop. Adds will be appended to the list that will be pickled.

[emokoena]

All your suggestions have been invaluable and I have gotten someway but when I cPickle the file I fail to read the contents back as a list

#Choice 2 - Display Balances
    elif choice == "2":
        
        lists = []
        infile = open('emp_data.dat', 'r')
        #inlist = cPickle.load(infile)
        while 1:
            try:
                lists.append(cPickle.load(infile))
                
            except (EOFError):
                break
        infile.close()
        #Prompt user for search
        print len(lists)
        empn = int(raw_input("\nChoose employee number: "))
        found = False 
        for entry in lists: 
           if empn == entry[0]: 
                print "Balance ==> R ", entry[4] 
                found = True 
        if not found: 
            print "Not a valid employee number"

[dwblas]

It should be
lists = cPickle.load(in file)

http://wiki.python.org/moin/UsingPickle

[emokoena]

hi dwblas;

I'm really am loosing my mind now. my code is only able to display the first entry, from there my if statement only evaluates to false for all other entries. please correct my code:

#Choice 2 - Display Balances
elif choice == "2":

lists = []
infile = open('emp_data. dat', 'r')
lists = cPickle.load(in file)

empn = int(raw_input(" \nChoose employee number: "))
found = False
for entry in lists:
if empn == entry[0]:
print "Balance ==> R ", entry[4]
found = True
if not found:
print "Not a valid employee number"
infile.close()

[dwblas]

The pickle statement is still wrong. The link in my previous post shows an example of pickling and unpickling. If you are not going to read the post then there is little reason to post. Also, for future problems add a print statement. In this case:

for entry in lists:
    print "testing" empn, entry[0]
    if empn == entry[0]:

please correct my code

That's just rude.