I just want to know one SQL query.
I've got two tables:
id | id_town
1 | 26
id | town
26 | Prague
What's the query if I need to print "Prague"?
Thank you.
PhpMyAdmin prints #1052 - Column 'id' in on clause is ambiguous
I tried this
SELECT town FROM localities JOIN towns ON id = id_town
TRY THIS:
select b.town
from table1 a
inner join table2 b on a.id_town = b.id
I have three tables where table_2 is the middle(connected) between table_1 and table_3
tables
table_id
...
...
table_rest
rest_id
table_id
...
rest
rest_id
...
...
And the query to select I use
SELECT m.table_id, table_name
FROM tables m
JOIN table_rest mr
ON m.table_id = mr.table_id
WHERE rest_id = '$rest_id'
What I need now is to join in this query another table reserv
id
...
status
To check if status is 0, 1,or 2 to not show me anything if there is no status this mean there is no record to show me. In other words this is resserved system where I show on screen few tables. If status is 0,1,2 thats mean the table is taken. If nothing is found for status this mean that there is no record for table and can be shown to user.
EDIT: Sample scenario
tables
table_id
1
2
3
4
5
rest
rest_id
1
2
table_rest
table_id | rest_id
1 2
2 2
3 2
4 2
5 2
So the query that is above will generate 5 tables for rest_id=2 and none for rest_id=1
So now I have another table
reserv
id | status
1 0
2 1
3 2
So in this table reserv currently are saved 3 tables. The idea is to show me other two whit id=4 and id=5 because they are not in table reserv and don't have any status.
Hope is a little bit more clear now.
You have to point from table reserv to which table is beign booked, let's call it reserv.table_id
SELECT m.table_id, table_name
FROM tables m
JOIN table_rest mr
ON m.table_id = mr.table_id
left join reserv
on reserv.table_id = m.id
WHERE rest_id = '$rest_id'
and reserv.status is null (*note)
*note use 'is' or 'is not' depending of your needs, as far as I read, first seems that you want !=, later that what you want is =
It's better if you provide sample data or sqlfiddle. Based on what I realize: Is this what you want:
select tables.table_id, rest.rest_id
from tables
left join table_rest on table_rest.table_id = tables.table_id
left join rest on rest.rest_id = table_rest.rest_id
where rest.rest_id = '$rest_id'
and tables.table_id not in (select id from reserv)
Essentially, I have a table that is like this:
FirstName, LastName, Type
Mark, Jones, A
Jim, Smith, B
Joseph, Miller, A
Jim, Smith, A
Jim, Smith, C
Mark, Jones, C
What I need to do is be able to display these out in PHP/HTML, like:
Name | Total Count Per Name | All Type(s) Per Name
which would look like...
Mark Jones | 2 | A, C
Jim Smith | 3 | B, A, C
Joseph Miller | 1 | A
Jim Smith | 3 | B, A, C
Jim Smith | 3 | B, A, C
Mark Jones | 2 | A, C
I have spent time trying to create a new table based off the initial one, adding these fields, as well as looking at group_concat, array_count_values, COUNT, and DISTINCT, along with other loop/array options, and cannot figure this out.
I've found a number of answers that count and concatenate, but the problem here is I need to display each row with the total count/concatenation on each, instead of shortening it.
How about doing it like this?
SELECT aggregated.* FROM table_name t
LEFT JOIN (
SELECT
CONCAT(FirstName, ' ', LastName) AS Name,
COUNT(Type) AS `Total Count Per Name`,
GROUP_CONCAT(Type SEPARATOR ',') AS `All Type(s) Per Name`
FROM table_name
GROUP BY Name) AS aggregated
ON CONCAT(t.FirstName, ' ', t.LastName) = aggregated.Name
Without an ORDER BY clause, the order the rows will be returned in is indeterminate. Nothing wrong with that, by my personal preference is to have the result to be repeatable.
We can use an "inline view" (MySQL calls it a derived table) to get the count and the concatenation of the Type values for (FirstName,LastName).
And then perform a join operation to match the rows from the inline view to each row in the detail table.
SELECT CONCAT(d.FirstName,' ',d.LastName) AS name
, c.total_coount_per_name
, c.all_types_per_name
FROM mytable d
JOIN ( SELECT b.FirstName
, b.LastName
, GROUP_CONCAT(DISTINCT b.Type ORDER BY b.Type) AS all_types_per_name
, COUNT(*) AS total_count_per_name
FROM mytable b
GROUP
BY b.FirstName
, b.LastName
) c
ON c.FirstName = d.FirstName
AND c.Last_name = d.LastName
ORDER BY d.FirstName, d.LastName
If you have an id column or some other "sequence" column, you can use that to specify the order the rows are to be returned; same thing in the GROUP_CONCAT function. You can omit the DISTINCT keyword from the GROUP_CONCAT if you want repeated values... 'B,A,B,B,C',
I have 4 tables as below
Table1: restos
Structure:
resu_id resu_name resu_address
-------------------------------------
1 ABC Exapmple
2 DEF Example
3 GHD Example
Table2:foodtype
Structure:
id typename
---------------
12 Indian
23 Punjabi
Table3: resto_foodtypes
Structure:
resu_id foodty_id
--------------------
1 12
2 23
3 12
Table4: discnts
Structure:
id resu_id amt_dscPer(%age discount)
---------------------------
19 1 15
20 2 25
Now i want to display the restaurant along with discounts available for the restauarant.
Currently restaurants are getting displayed but for the restaurant not present in discnts table are returning null values from discnts table.
below is the query that m using
SELECT * from `restos` r join resto_foodtypes rf on rf.resu_id = r.resu_id
join foodtype f on rf.foodty_id = f.id left join discnts dcfm on
r.resu_id= dcfm.resu_id where true;
I want that the restaurants that are not present in discnts table should not be included in resultset. For e.g. resu_id=3 is not present in discnts table.
To exclude results that have no entry in discnts table, use a INNER JOIN instead of a LEFT JOIN.
To include these results, but have a 0 displayed (instead of the defaults NULL), you can use IFNULL(expr, 0) function:
SELECT
r.resu_name AS Name,
f.typename AS Food,
IFNULL(dcfm.amt_dscPer, 0) AS Discount
FROM `restos` r
JOIN resto_foodtypes rf ON rf.resu_id = r.resu_id
JOIN foodtype f ON rf.foodty_id = f.id
LEFT JOIN discnts dcfm ON r.resu_id= dcfm.resu_id;
IFNULL returns the first parameter if it is not null, the second if expr is indeed null.
I have these tables:
table 1 : attendance
-------------------------------
ID | DATE | EMPLOYEE_ID |
-------------------------------
1 2013-09-10 1
2 2013-09-10 2
3 2013-09-10 3
-------------------------------
table 2: employee
---------------
ID | NAME |
---------------
1 Smith
2 John
3 Mark
4 Kyle
5 Susan
6 Jim
---------------
My actual code to show employee option.
while($row=mysql_fetch_array($query)){
echo "<option value='$row[employee_id]'>$row[first_name] $row[last_name]</option>";
}
?>
How can i show the list of employee that not registered in table 1?
The condition is if an employee already registered in table 1, they won't appear on the option.
I want to show the list in <option>'s of <select> element. So it will return: kyle, susan, jim.
Please tell me the correct query or if there is any better option, it'll be good too. Please give some solution and explain. Thank you very much
UPDATE / EDIT:
it also based on current date, if in table 1 have no latest date e.g. today it's 2013-09-15. It will show all of employee.
You can do this with a left join and then checking for no matches:
select e.*
from employee e left outer join
attendance a
on e.id = a.employee_id
where a.employee_id is null;
This is probably the most efficient option in MySQL.
EDIT:
To include a particular date, add the condition to the on clause:
select e.*
from employee e left outer join
attendance a
on e.id = a.employee_id and a.date = date('2013-09-20')
where a.employee_id is null;
If I understood correctly this should work, get all employees whose ID is not in the attendance table.
SELECT * FROM employee WHERE employee.ID NOT IN
(SELECT EMPLOYEE_ID FROM attendance)