opelraja
asked on
add data from two tables into one table
I have a table called employees: within it I have various fields two of which being:
emp_uid, and emp_shift_duration.
emp_uid is unique.
emp_shift_duration varies a bit becasue most emps work either 4 hours, 5 hours, 6 hours or 8 hours.
Sometimes people are out and I want to use their hours to do special projects so I want to keep tabs on when they miss hours. The shift sup. can keep track of this but in order to save time I only want him to 'edit' the missing hours when needed, not add hours eveyday so I want the table populated ahead of time.
so I created another table which is called :
emp_hours_worked: fields are 3 : emp_uid, projected_hours,and date.
I wish to populate emp_hours_worked with employee data from employees table.
emp_uid, and emp_shift_duration; and also populate date with the days of the year for 'each' employee: this will be a very long table because the result will be that each employee will have a record for each date of the year.: therefore using a simple php script the sup. can go in and edit any necessary dates and then I will run a query to sum up the hours for the current pay weeks.
emp_hours_worked will look something like this.
emp_uid,projected_hours,da
3432,4,01-01-2013
3432,4,01-02-2013
3432,4,01-03-2013...etc.
7715,8,01-01-2013
7715,8,01-02-2013...etc...
3432,4,01-01-2013
7715,8,01-01-2013
9824,5,01-01-2013
9954,8.01-01-2013... etc. depending how its shown but did this to get the idea through.
So I am looking for sql statement to 'insert' all this data into the emp_hours_worked table.
Well anyways this is my idea any better solutions would be welcome.
emp_uid, and emp_shift_duration.
emp_uid is unique.
emp_shift_duration varies a bit becasue most emps work either 4 hours, 5 hours, 6 hours or 8 hours.
Sometimes people are out and I want to use their hours to do special projects so I want to keep tabs on when they miss hours. The shift sup. can keep track of this but in order to save time I only want him to 'edit' the missing hours when needed, not add hours eveyday so I want the table populated ahead of time.
so I created another table which is called :
emp_hours_worked: fields are 3 : emp_uid, projected_hours,and date.
I wish to populate emp_hours_worked with employee data from employees table.
emp_uid, and emp_shift_duration; and also populate date with the days of the year for 'each' employee: this will be a very long table because the result will be that each employee will have a record for each date of the year.: therefore using a simple php script the sup. can go in and edit any necessary dates and then I will run a query to sum up the hours for the current pay weeks.
emp_hours_worked will look something like this.
emp_uid,projected_hours,da
3432,4,01-01-2013
3432,4,01-02-2013
3432,4,01-03-2013...etc.
7715,8,01-01-2013
7715,8,01-02-2013...etc...
3432,4,01-01-2013
7715,8,01-01-2013
9824,5,01-01-2013
9954,8.01-01-2013... etc. depending how its shown but did this to get the idea through.
So I am looking for sql statement to 'insert' all this data into the emp_hours_worked table.
Well anyways this is my idea any better solutions would be welcome.
ASKER CERTIFIED SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
ASKER
I figured it out like this:
I created a table 'dates with just '_dates' as a field : being the whole year day to day of 2013.
then I used this query:
Insert into emp_hours_worked(emp_uid,p
SELECT emp_uid, emp_shift_duration,_date
FROM employees_table AS a, date AS b;
So it populated a line for each employee for each day. Now I just need a faster laptop for this huge table.
I created a table 'dates with just '_dates' as a field : being the whole year day to day of 2013.
then I used this query:
Insert into emp_hours_worked(emp_uid,p
SELECT emp_uid, emp_shift_duration,_date
FROM employees_table AS a, date AS b;
So it populated a line for each employee for each day. Now I just need a faster laptop for this huge table.
SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
'employees e left join <other tab t> on field'