I can't show in MySQL a specific row that not contains a value.
First of all this is my two tables:
And this is an array with the info of both tables:
Bills table:
[0] => Array
(
[id] => 1
[reference] => #001001
[seat_id] => 101
[client_id] => 10200
)
[1] => Array
(
[id] => 2
[reference] => #001002
[seat_id] => 102
[client_id] => 10400
)
[2] => Array
(
[id] => 3
[reference] => #001003
[seat_id] => 103
[client_id] => 10600
)
Accounting_seats table:
[0] => Array
(
[0] => Array
(
[id] => 1
[seat_id] => 101
[account_id] => taxes_qty
[value] => 0.99
[client_id] => 10200
)
[1] => Array
(
[id] => 2
[seat_id] => 101
[account_id] => tax_base
[value] => 4
[client_id] => 10200
)
[2] => Array
(
[id] => 3
[seat_id] => 101
[account_id] => total
[value] => 4.99
[client_id] => 10200
)
)
[1] => Array
(
[0] => Array
(
[id] => 4
[seat_id] => 102
[account_id] => taxes_qty
[value] => 2.00
[client_id] => 10400
)
[1] => Array
(
[id] => 5
[seat_id] => 102
[account_id] => tax_base
[value] => 8.00
[client_id] => 10400
)
[2] => Array
(
[id] => 6
[seat_id] => 102
[account_id] => shipping_cost
[value] => 2.00
[client_id] => 10400
)
[3] => Array
(
[id] => 7
[seat_id] => 102
[account_id] => total
[value] => 12.00
[client_id] => 10400
)
)
[2] => Array
(
[0] => Array
(
[id] => 8
[seat_id] => 103
[account_id] => taxes_qty
[value] => 3
[client_id] => 10600
)
[1] => Array
(
[id] => 9
[seat_id] => 103
[account_id] => tax_base
[value] => 7
[client_id] => 10600
)
[2] => Array
(
[id] => 10
[seat_id] => 103
[account_id] => shipping_cost
[value] => 3.99
[client_id] => 10600
)
[3] => Array
(
[id] => 11
[seat_id] => 103
[account_id] => total
[value] => 13.99
[client_id] => 10600
)
)
The problem is that I cannot show the bill where the accounting_seats not contain the value "shipping_cost". Only show the 2 bills with shipping_cost or all bills, but I need the bill without shiipping_cost value.
SELECT reference, seat_id FROM bills WHERE seat_id IN(101,102,103) AND seat_id IN (SELECT seat_id FROM accounting_seats WHERE account != 'shipping_cost');
Tons of thanks for any help!
I would reommend not exists:
select reference, seat_id
from bills b
where
b.seat_id in(101,102,103)
and not exists (
select 1
from accounting_seats acs
where acs.account = 'shipping_cost' and acs.seat_id = b.seat_id
)
As compared to not in, not exists usually scales better when the size of the underlying table increases (here, accounting_seats): the above query woulud take advantage of an index on accounting_seats(account, seat_id). Another upside of this approach is that not exists is safe as regard to null values in seat_id, while not in isn't.
Try this:
SELECT reference, seat_id
FROM bills
WHERE seat_id IN(101,102,103)
AND seat_id NOT IN (SELECT seat_id FROM accounting_seats WHERE account = 'shipping_cost');
Related
I want to find users whose usernames start with "developer"; however, when I used the query below, I got 18 results when only two should return, based on the number of users in the database. Is there something wrong with the query that is causing this?
SELECT SQL_CALC_FOUND_ROWS wp_users.ID,wp_users.user_login
FROM wp_users
INNER JOIN wp_usermeta
ON ( wp_users.ID = wp_usermeta.user_id )
WHERE wp_users.ID !='1' AND wp_users.user_login LIKE 'developer%' OR (wp_usermeta.meta_key = 'nickname' AND wp_usermeta.meta_value LIKE 'developer%') OR (wp_usermeta.meta_key = 'first_name' AND wp_usermeta.meta_value LIKE 'developer%')
ORDER BY user_login ASC
The actual results:
Array (
[0] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[1] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[2] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[3] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[4] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[5] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[6] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[7] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[8] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[9] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[10] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[11] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[12] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[13] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[14] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[15] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[16] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[17] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
[18] => stdClass Object ( [ID] => 6 [user_login] => developer [id] => 6 )
)
Do all users have wp_usermeta entries for nickname and first_name? If not, you may need to change the join to a LEFT JOIN.
SELECT SQL_CALC_FOUND_ROWS DISTINCT wp_users.ID, wp_users.user_login
FROM wp_users
INNER JOIN wp_usermeta
ON wp_users.ID = wp_usermeta.user_id
AND wp_usermeta.meta_key IN ('nickname', 'first_name')
WHERE wp_users.ID != 1
AND ( wp_users.user_login LIKE 'developer%' OR wp_usermeta.meta_value LIKE 'developer%')
ORDER BY user_login ASC
Right now, I have this query:
SELECT COUNT(*) AS Count, SUM(Ask) AS Ask, SUM(Cost) AS Cost, Provider, Factura FROM store_items
WHERE (
Provider NOT IN(SELECT Provider FROM store_provider_invoices)
AND Factura NOT IN(SELECT Factura FROM store_provider_invoices)
)
OR Factura NOT IN(SELECT Factura FROM store_provider_invoices)
GROUP BY Provider, Factura
This is working great, and returns the following array:
Array (
[0] => Array (
[Count] => 1
[ID] => 13
[Ask] => 20.00
[Cost] => 10.00
[Provider] => 5
[Factura] => 8
)
[1] => Array (
[Count] => 1
[ID] => 18
[Ask] => 125.01
[Cost] => 110.01
[Provider] => 5
[Factura] => 34
)
[3] => Array (
[Count] => 3
[ID] => 14
[Ask] => 210.00
[Cost] => 150.00
[Provider] => 6
[Factura] => 5
)
)
What I would like to do is to also return all the ID's that match the query from the store_items table, like:
Array (
[0] => Array (
[ID] => Array (
[0] => 101
)
[Count] => 1
[Ask] => 20.00
[Cost] => 10.00
[Provider] => 5
[Factura] => 8
)
[1] => Array (
[ID] => Array (
[0] => 102
)
[Count] => 1
[Ask] => 125.01
[Cost] => 110.01
[Provider] => 5
[Factura] => 34
)
[3] => Array (
[ID] => Array (
[0] => 103
[1] => 104
[2] => 105
)
[Count] => 3
[Ask] => 210.00
[Cost] => 150.00
[Provider] => 6
[Factura] => 5
)
)
So, for instance, in the last array element above, instead of just returning a Count of 3, also return the ID's of each row that it counted.
You can't really get nested results, but you could use
GROUP_CONCAT(DISTINCT store_items.ID ORDER BY ID) AS siIDs
to get a comma-separated list of id values.
You can also modify the separator
http://dev.mysql.com/doc/refman/5.7/en/group-by-functions.html#function_group-concat
I've got a big query that all contain same types I need for a feed that I list in my app. Now the problem is that this query isn't very fast. I'm thinking that if I limit each individual union it might speed it up a bit but I'm not sure.
So basically my question is how could I optimize this query to execute faster?
SELECT DISTINCT
alert_id,
uniquekey,
type,
user_id,
date_added
FROM
(
( SELECT
r.alert_id,
r.alert_id AS uniquekey,
'reply' AS `type`,
r.user_id,
r.date_added
FROM
`reply` r
LEFT JOIN `alerts` a
ON r.alert_id = a.alert_content_id
WHERE
r.user_id = :id
AND a.hide = '0'
ORDER BY
date_added DESC )
UNION
( SELECT
r.alert_id,
r.alert_id AS uniquekey,
'replyfromfollowing' AS `type`,
r.user_id,
r.date_added
FROM
`reply` r
LEFT JOIN `alerts` a
ON r.alert_id = a.alert_content_id
WHERE
r.user_id IN( '$followingstring' )
AND a.hide = '0'
ORDER BY date_added DESC )
UNION
( SELECT
i.alert_id,
i.alert_id AS uniquekey,
'liked' AS `type`,
i.user_id,
i.date_added
FROM
`interactions` i
LEFT JOIN `alerts` a
ON i.alert_id = a.alert_content_id
WHERE
i.user_id = :id
AND a.hide = '0'
GROUP BY
alert_id
ORDER BY
date_added DESC )
UNION
( SELECT
i.alert_id,
i.alert_id AS uniquekey,
'likedfromfollowing' AS `type`,
i.user_id,
i.date_added
FROM
`interactions` i
LEFT JOIN `alerts` a
ON i.alert_id = a.alert_content_id
WHERE
i.user_id IN ( '$followingstring' )
AND a.hide = '0'
GROUP BY
alert_id
ORDER BY
date_added DESC )
UNION
( SELECT
alerts as alert_id,
alert_content_id AS uniquekey,
'following' AS `type`,
user_id,
date_added
FROM
alerts a
LEFT JOIN `alerts_content` ac
ON ac.id = a.alert_content_id
WHERE
a.user_id IN ( '$followingstring' )
AND ac.anoniem = '0'
AND a.hide = '0'
GROUP BY
alert_id
ORDER BY
date_added DESC )
) joined
GROUP BY
uniquekey
ORDER BY
date_added DESC
LIMIT
".(int)$start.",20"
table structures
Reply table Structure:
id
user_id
alert_id
description
reply_on_alert
reply_on_reply
date_added
Interaction table Structure:
id
alert_id
action_id
reply_id
user_id
date_added
Alerts table structure(Yes i know BIG mistake naming `id` : `alerts`):
alerts
title
alert_content_id
user_id
cat
lat
lon
state
hide
date_added
alerts_content table structure:
id
alert_id
description
img
results of query:
Array
(
[0] => Array
(
[alert_id] => 173404
[uniquekey] => 173404
[type] => reply
[user_id] => 2
[date_added] => 2015-06-01 16:34:16
)
[1] => Array
(
[alert_id] => 172174
[uniquekey] => 172174
[type] => replyfromfollowing
[user_id] => 1380
[date_added] => 2015-06-01 16:01:04
)
[2] => Array
(
[alert_id] => 171772
[uniquekey] => 171772
[type] => liked
[user_id] => 2
[date_added] => 2015-06-01 15:58:44
)
[3] => Array
(
[alert_id] => 149423
[uniquekey] => 149423
[type] => reply
[user_id] => 2
[date_added] => 2015-06-01 15:25:56
)
[4] => Array
(
[alert_id] => 164742
[uniquekey] => 164742
[type] => reply
[user_id] => 2
[date_added] => 2015-05-12 09:46:39
)
[5] => Array
(
[alert_id] => 163344
[uniquekey] => 163344
[type] => replyfromfollowing
[user_id] => 3
[date_added] => 2015-05-12 09:44:46
)
[6] => Array
(
[alert_id] => 164205
[uniquekey] => 164205
[type] => liked
[user_id] => 2
[date_added] => 2015-05-11 11:06:39
)
[7] => Array
(
[alert_id] => 160890
[uniquekey] => 160890
[type] => replyfromfollowing
[user_id] => 1380
[date_added] => 2015-05-08 14:29:34
)
[8] => Array
(
[alert_id] => 163002
[uniquekey] => 163002
[type] => replyfromfollowing
[user_id] => 1380
[date_added] => 2015-05-08 13:31:12
)
[9] => Array
(
[alert_id] => 159123
[uniquekey] => 159123
[type] => replyfromfollowing
[user_id] => 48
[date_added] => 2015-04-30 15:10:32
)
[10] => Array
(
[alert_id] => 150546
[uniquekey] => 150546
[type] => replyfromfollowing
[user_id] => 16
[date_added] => 2015-04-21 21:52:49
)
[11] => Array
(
[alert_id] => 149497
[uniquekey] => 149497
[type] => reply
[user_id] => 2
[date_added] => 2015-04-10 15:19:06
)
[12] => Array
(
[alert_id] => 141078
[uniquekey] => 141078
[type] => liked
[user_id] => 2
[date_added] => 2015-04-10 15:15:32
)
[13] => Array
(
[alert_id] => 125466
[uniquekey] => 125466
[type] => replyfromfollowing
[user_id] => 3
[date_added] => 2015-04-09 00:15:22
)
[14] => Array
(
[alert_id] => 134592
[uniquekey] => 134592
[type] => replyfromfollowing
[user_id] => 3
[date_added] => 2015-04-09 00:11:04
)
[15] => Array
(
[alert_id] => 124194
[uniquekey] => 124194
[type] => likedfromfollowing
[user_id] => 3
[date_added] => 2015-04-09 00:08:35
)
[16] => Array
(
[alert_id] => 128645
[uniquekey] => 128645
[type] => likedfromfollowing
[user_id] => 3
[date_added] => 2015-04-09 00:07:29
)
[17] => Array
(
[alert_id] => 144867
[uniquekey] => 144867
[type] => replyfromfollowing
[user_id] => 3
[date_added] => 2015-04-06 13:59:19
)
[18] => Array
(
[alert_id] => 133355
[uniquekey] => 133355
[type] => liked
[user_id] => 2
[date_added] => 2015-03-31 16:16:15
)
[19] => Array
(
[alert_id] => 141075
[uniquekey] => 141075
[type] => liked
[user_id] => 2
[date_added] => 2015-03-30 15:17:01
)
)
Some possibilities, in no particular order:
Optimization #1:
Use LIMIT in the subqueries, too. However, since you are using OFFSET, it may not be obvious how to do it.
Before the query, calculate $start+20 and put it into, say, $limit. Then use LIMIT $limit for the inner queries. No, don't use an OFFSET on them. This guarantees that you get enough rows from each query to satisfy the outer OFFSET $start LIMIT 20.
Optimization #2:
Restructure the tables so that you don't need to JOIN to another table (alerts) to find out whether to show the record. That is, having hide prevents a number of potential optimizations. Before advising further, we need to understand the need for LEFT. Are there rows in reply, etc, that are not in alerts? If not, get rid of the LEFT and look into searching alerts with theOFFSETandLIMIT`, then joining to the 4 other tables.
Optimization #3:
Restructure the data so that there is one core table, with alerts and the 4 other tables hanging off it. Be sure to have most (all?) of the fields needed for this query in this new table.
Optimization #4:
The current structure requires the full scan of each of the 4 tables before even thinking about OFFSET and LIMIT. This smells like "pagination"; is it? For optimizing "pagination", the goals are to avoid the table scan and OFFSET; instead remember where you "left off" so that the query can be
WHERE ... AND x < $left_off
ORDER by x DESC
LIMIT 20
This should make it possible to read only 20 rows, not the entire table(s). That would make the query much faster, especially for later pages. (A bigger OFFSET costs more time`)
I discuss pagination optimization in my blog.
i am trying to access the id column of one of my table but i get "Undefined property" error but i can access the other columns with no problems,here is my code
in the controller i get all records as follows
$employees=Employee::all();
return View::make('admin.employee',compact('tittle','heading','employees'));
and i my view
#foreach($employees as $employee)
<tr class="success">
<td>{{$employee->first_name}}</td>
<td>{{$employee->last_name}}</td>
<td>{{$employee->phone}}</td>
<td>{{$employee->salary}}</td>
<td>{{$employee->created_at}}</td>
<td>Edit</td>
<td>Resign</td>
<td>View</td>
</tr>
#endforeach
How can i access the id column from my table?
when i print_r($employees);
Illuminate\Database\Eloquent\Collection Object ( [items:protected] => Array ( [0] => Employee Object ( [fillable:protected] => Array ( [0] => profile_photo_id [1] => first_name [2] => last_name [3] => gender [4] => phone [5] => address [6] => age [7] => salary ) [connection:protected] => [table:protected] => [primaryKey:protected] => id [perPage:protected] => 15 [incrementing] => 1 [timestamps] => 1 [attributes:protected] => Array ( [id] => 4 [profile_photo_id] => 4 [first_name] => Jason [last_name] => cbvc [gender] => Female [phone] => 2147483647 [address] => jhjhjh [age] => 12 [salary] => 5566556 [created_at] => 2014-12-20 11:29:41 [updated_at] => 2014-12-20 11:29:41 ) [original:protected] => Array ( [id] => 4 [profile_photo_id] => 4 [first_name] => Jason [last_name] => cbvc [gender] => Female [phone] => 2147483647 [address] => jhjhjh [age] => 12 [salary] => 5566556 [created_at] => 2014-12-20 11:29:41 [updated_at] => 2014-12-20 11:29:41 ) [relations:protected] => Array ( ) [hidden:protected] => Array ( ) [visible:protected] => Array ( ) [appends:protected] => Array ( ) [guarded:protected] => Array ( [0] => * ) [dates:protected] => Array ( ) [touches:protected] => Array ( ) [observables:protected] => Array ( ) [with:protected] => Array ( ) [morphClass:protected] => [exists] => 1 ) [1] => Employee Object ( [fillable:protected] => Array ( [0] => profile_photo_id [1] => first_name [2] => last_name [3] => gender [4] => phone [5] => address [6] => age [7] => salary ) [connection:protected] => [table:protected] => [primaryKey:protected] => id [perPage:protected] => 15 [incrementing] => 1 [timestamps] => 1 [attributes:protected] => Array ( [id] => 5 [profile_photo_id] => 5 [first_name] => gugu [last_name] => cbvc [gender] => Female [phone] => 2147483647 [address] => jhjhjh [age] => 12 [salary] => 5566556 [created_at] => 2014-12-20 11:40:44 [updated_at] => 2014-12-20 11:40:44 ) [original:protected] => Array ( [id] => 5 [profile_photo_id] => 5 [first_name] => gugu [last_name] => cbvc [gender] => Female [phone] => 2147483647 [address] => jhjhjh [age] => 12 [salary] => 5566556 [created_at] => 2014-12-20 11:40:44 [updated_at] => 2014-12-20 11:40:44 ) [relations:protected] => Array ( ) [hidden:protected] => Array ( ) [visible:protected] => Array ( ) [appends:protected] => Array ( ) [guarded:protected] => Array ( [0] => * ) [dates:protected] => Array ( ) [touches:protected] => Array ( ) [observables:protected] => Array ( ) [with:protected] => Array ( ) [morphClass:protected] => [exists] => 1 ) ) )
Give us your $employees object , edit question with
print_r($employees).
Also there may be a chance the $id is protected, in that case use.
$employee->getOriginal('id');
$sql=sprintf("SELECT * FROM stitch);
This my array.
[stitch] => Array
(
[0] => Array
(
[id] => 7,
[name] => Sew buttonhole to front fly
)
[1] => Array
(
[id] => 8,
[name] => Sleeve hem
)
)
)
I need this result
[stitch] => Array
(
[0] => Array
(
[id] => 7,
[name] => Sew buttonhole to front fly
[number_stitch] => 1
)
[1] => Array
(
[id] => 8,
[name] => Sleeve hem
[number_stitch] => 2
)
)
)
how i do these [number_stitch] ?
Try
select s.*, #rowcount:=#rowcount+1 ‘number_stitch’ from stitch s, (SELECT #rowcount:=0) r order by id;