TIMESTAMP ETEMP# ETIORO PMNAME
1/2/09 8:56 AM 213 I John Deere
1/2/09 1:15 PM 213 O John Deere
1/2/09 1:43 PM 213 I John Deere
1/2/09 6:03 PM 213 O John Deere

I created two queries in access, one with clock in data and one with cloc out data and added the field DateVal for the day. I then created a query with the following SQL language.

SELECT ClockIn.DateVal, ClockIn.[ETEMP#], ClockIn.PMNAME, ClockIn.ETIORO, ClockOut.ETIORO, ClockIn.TIMESTAMP, ClockOut.TIMESTAMP, DateDiff("n",[clockin].[timestamp],[clockout].[timestamp])/60 AS [Hours]

FROM ClockIn INNER JOIN ClockOut ON (ClockIn.[ETEMP#]=ClockOut.[ETEMP#]) AND (ClockIn.DateVal=ClockOut.DateVal)

WHERE (((ClockIn.DateVal)=#1/2/2009#) And ((ClockIn.PMNAME)="John Deere"));

This is the result I get:

DateVal ETEMP# PMNAME ClockIn.ETIORO ClockOut.ETIORO ClockIn.TIMESTAMP ClockOut.TIMESTAMP Hours
1/2/09 213 John Deere I O 1/2/09 1:43 PM 1/2/09 1:15 PM -0.466666667
1/2/09 213 John Deere I O 1/2/09 8:56 AM 1/2/09 1:15 PM 4.316666667
1/2/09 213 John Deere I O 1/2/09 1:43 PM 1/2/09 6:03 PM 4.333333333
1/2/09 213 John Deere I O 1/2/09 8:56 AM 1/2/09 6:03 PM 9.116666667

How do I eliminate the duplication and get true hours worked for the day. I think it can be done with a subquery, but that is out of my league. Any help is greatly appreciated.

I don’t know access syntex, but I can give you my 2 cents how to achive this using one simple query using aggregation.

convert your timestamp to minutes (timestamp – the date), then 8:56 will be 8×60 + 56 = 536, 1:15 => 795
1:43 =>823, 6:03=>1083. If Access can handle case statement, that will be great, if not, add a column called "sign", -1 for clock in, 1 for clock out.

select Sum (theCovertedMinutes * sign)
from your table
group by PMName

You will get -536+795-823+1083 = 519 minute

select
from employees;

the easy way to get into that is to build 2 queries and use the query designer instead of trying to write sql code first.

so, let’s say you have an ‘employees’ table and a table that tracks absences and they’re keyed on an employee id field

in you primary query from the employee table, you want:
emplid empl name and then you want the 3rd data item to be their number of days absent

create a separate totals query that counts the number of days absent, grouped by Emp id. they view that query in sql mode and copy/paste it in your main query in the 2rd column (some editing is required to remove the ‘;’ at the end and other changes depending on your tables)

so now, for every record in your main query, it will run the sub query and display the empid, name, and # days absent.

NOTE: before everyone jumps on it there are much easier way to accomplish this result but it’s a good basic subquery example.

13.Write one query to return the employee_id, job_id, hire_date and department_id of all employees and a second query listing employee_id, job_id, start_date and department_id from the job_hist table and combine the results as one single output. Make sure you suppress duplicates in the output.

14.DJs on Demand just purchased four new CDs. Use an explicit INSERT statement to add each CD to the copy_d_cds table. After completing the entries, execute a SELECT * statement to verify your work.

CD_NUMBERTITLEPRODUCERYEAR
97Celebrate the DayR&B Inc.2003

98Holiday Tunes for All AgesTunes are Us
2004
99Party MusicOld Town Records2004
100Best of Rock and RollOld Town Records2004

15.DJs on Demand has two new events coming up. One event is a fall football party and the other event is a sixties theme party. The DJs on Demand clients requested the songs shown in the table for their events. Add these songs to the copy_d_songs table using an implicit INSERT statement.

IDTITLEDURATIONTYPE_CODE
52Surfing Summer Not known12

53Victory Victory5 MINS
12

16.Monique Tuttle, the manager of Global Fast Foods, sent a memo requesting an immediate change in prices. The price for a strawberry shake will be raised from $3.59 to $3.75, and the price for fries will increase to $1.20. Make these changes to the copy_f_food_items table.

17.Add the new customers shown below to the copy_f_customers table. You may already have added Katie Hernandez. Will you be able to add all these records successfully?

ID FIRST_
NAME LAST_
NAME ADDRESS CITY STATE ZIP PHONE_NUMBER
145 Katie Hernandez 92 Chico Way Los Angeles CA 98008 8586667641
225 Daniel Spode 1923 Silverado Denver CO 80219 7193343523
230 230 Adam Zurn 5 Admiral Way Seattle WA 4258879009

18.Sue Doe has been an outstanding Global Foods staff member and has been given a salary raise. She will now be paid the same as Bob Miller. Update her record in copy_f_staffs.

19.When would you want a DEFAULT value?

20.Complete the GRADUATE CANDIDATE table instance chart. Credits is a foreign key column referencing the requirements table.

21.Write the syntax to create the grad_candidates table. Confirm creation of the table using DESCRIBE.

Column
Name
student_idlast_namefirst_namecreditsgraduation_date

Key Type

Nulls/Unique

FK Column

DatatypeNUMBERVARCHAR2VARCHAR2NUMBERDATE

Length63

22.In your o_employees table, enter a new column called “Termination.” The datatype for the new column should be VARCHAR2. Set the DEFAULT for this column as SYSDATE to appear as character data in the format: February 20th, 2003.

23.Create a new column in the o_employees table called start_date. Use the TIMESTAMP WITH LOCAL TIME ZONE as the datatype.

24.Using the column information for the animals table below, name constraints where applicable at the table level, otherwise name them at the column level. Define the primary key (animal_id). The license_tag_number must be unique. The admit_date and vaccination_date columns cannot contain null values.
animal_id NUMBER(6)
name VARCHAR2(25)
license_tag_number NUMBER(10)
admit_date DATE
adoption_id NUMBER(5),
vaccination_date DATE

25.Create the animals table. Write the syntax you will use to create the table.

the answer is 2

The bonus code is a number and it varies from 100 to 114. After having numerous records, I now want to do a query that will give me all my bonus code entries ( all 12 fields) in ascending order. For example I dont want

100
111
112

100
113
114

I want

100, 100, 111, 112, 113, 114.

Any idea how I can overcome this?? Im think ill have to do a subquery…but im not sure.

SELECT blah
FROM blah
WHERE blah
SORT BY ascending <- this does the trick

3-Key
ABC

4-Key
ABCD

I need to get D, put it into 3-key, when there’s a match based on ABC. Here’s the catch. ABC are allowed to be equal on both tables, and D is the only real designator or uniqueness on the 4-key. Examples:

4-Key
A B C D
ALGEBRA 12349999 X7
ALGEBRA 12349999 A2
ALGEBRA 12349999 P6
ALGEBRA 12349996 F8
ALGEBRA 12349199 K1
ALGEBRA 12349299 E2
ALGEBRA 12349013 C9
ALGEBRA 12349988 Z5

3 Key
A B C
ALGEBRA 12349999
ALGEBRA 12349999
ALGEBRA 12349999
ALGEBRA 12349996
ALGEBRA 12349199
ALGEBRA 12349299
ALGEBRA 12349013
ALGEBRA 12349988

The problem – when I try joining the two tables I get the "ORA-01427: single-row subquery returns more than one row" because it doesn’t know how to handle the first three rows in 3-Key (which D do I take?)

How can I make it so that it fills in all the rows after the first three, in other words, fill in when the query returns one row, and do nothing but don’t fail if it returns more than one?

My code:
update grade_table b
set b.D =
(select distinct
a.D from local_college_course a
where
trim(b.A||b.B||b.C)=trim(a.A||a.B||a.C)
and a.D is not null
group by a.D
having count(distinct a.D) = 1
)
where exists (select ‘x’ from local_college_course a
where
trim(b.A||b.B||b.C)=trim(a.A||a.B||a.C)
and a.D is not null)

I replaced the actual column names with ABCD to save space. I need it to look like this

3-Key
A B C D
ALGEBRA 12349999
ALGEBRA 12349999
ALGEBRA 12349999
ALGEBRA 12349996 F8
ALGEBRA 12349199 K1
ALGEBRA 12349299 E2
ALGEBRA 12349013 C9
ALGEBRA 12349988 Z5

I can’t just dedupe the 4-key table because there are lot of fields I need to maintain on it.
tried your suggestion. Its not identifying the columns in the second subquery –

select * from local_college_course l, ( <insert Query 1> ) k
where k.A=l.A and k.B=l.B and k.C=l.C

ERROR at line 5:
ORA-00904: "L"."C": invalid identifier
both excellent answers, but I’m having better luck with techieguys. I’m reduced to this error:

ORA-00904: "T1"."C": invalid identifier

its in ref to line 15: WHERE t.a = t1.a AND t.b = t1.b AND t.c = t1.c)

any thoughts on what I’m doing wrong?

Oy gevalt!

Ok, you can do this with multiple subqueries:

First you’d trim 4 Key to the lines with one D value per unique ABC:
(I’m not checkign syntax on this stuff):
Query1
—————–
select A, B,C , count (*)
from local_college_course
group by A,B,C
having count (*) =1

You actually have ‘distinct D’ in your ‘having’. Who designed this table and why do they hate you?

Next you’d rejoin that with original local_college_course:

Query 2
——————-
select * from local_college_course l, ( <insert Query 1> ) k
where k.A=l.A and k.B=l.B and k.C=l.C

and then use THAT to join with grade_table:

update grade_table b
set b.D= ( select D from ( <insert QUERY 2> ) where trim ( yadda yadda yadda ….

You get the idea.

Good luck.

This is wrong but its all i can think of and everything needed should be there.

SELECT fname, surname
FROM actor
WHERE a_id=(SELECT actor_id FROM acts WHERE movie_id=(SELECT Count(*) FROM acts GROUP BY actor_id HAVING (COUNT(*)>3)));

Its wrong in the first subquery but i dont know what to do?
SELECT fname, surname
FROM actor
WHERE a_id=(SELECT actor_id FROM acts GROUP BY actor_id HAVING (COUNT(*)>3));

This is better but wont work because the sub querie returns 2 answers and it only allows 1 for sub queries?

SELECT fname, surname
FROM actor
WHERE a_id IN (SELECT actor_id FROM acts GROUP BY actor_id HAVING (COUNT(*)>3));

select *, (select value from Sub_Data, claimant where Sub_Data.claimant_id = Claimant.claimant_id and Sub_Data.id = ‘72′) as B from Claimant

The above returns multiple rows on the subquery which obviously wont work,
Any ideas? this was kind of hard to explain so it may be confusing…

Thnks Keg, I will try that tomorrow and see if that is what I am looking for
ok, same question but I need to add something, what if I want to also get the result where the id is say ‘22′, ‘72′, and ‘600′ for each claimant id; how would I get a select statement where it returns all the results for claimant along with columns of the result of value with the above 3 ids from the sub_data table?

You can do this by referencing the sub_data table under 3 aliases.

select
c.*,
s72.value as value_72,
s22.value as value_22,
s600.value as value_600
from
claimant c
left outer join sub_data s72
on c.claimant_id = s72.claimant_id and
s72.id = ‘72′
left outer join sub_data s22
on c.claimant_id = s22.claimant_id and
s22.id = ‘22′
left outer join sub_data s600
on c.claimant_id = s600.claimant_id and
s600.id = ‘600′

those statements need to be 1 row.
just put a conditional like where whatever = whatever

this query is executed in mysql but I don’t understand how to display this sql output in php page. I am able to display first value i.e assign_name,credit value but can’t display counted value which is given subquery.

UPDATE: Seems I misread your question.

What you need to do is set a "column name" for the computed value, for example:

SELECT count(ColumnID) AS RowCount FROM TableName

The "AS RowCount " is the key to giving it a column name, this way you will be able to reference it much easier with in PHP, just like your other columns.

I get a list of 5 items.

Now I want to exclude those 5 ids from another query, with something like this:

select content from sentences where id NOT IN ( (select id from table1 where user="me") )

But that gives me an error of "Subquery returns more than 1 row". Anyone have any suggestions to make this work?

u cannot use sub queries in mysql 4.1

u need to hava 5.0