I need to run a custom query and I have tried all these methods
$sql = "SELECT acd.*, DATE_FORMAT(acd.cdate, '%d/%m/%Y') cdate_disp,
GROUP_CONCAT(CONCAT_WS(', ', car.type, car.session, crd.name) SEPARATOR '<br />') rd_names,
acb.booking_status my_booking_status
FROM app_data acd
INNER JOIN crmaccounts ca ON ca.id = acd.client_crm_id
LEFT JOIN crmaccount_rdids car ON car.account_id = ca.id
LEFT JOIN crmrd_ids crd ON crd.id = car.rd_id
LEFT JOIN app_bookings acb ON acb.call_ref_id = acd.call_ref AND acb.user_id = 12391
WHERE 1=1
AND acd.client_crm_id NOT IN (select account_id from bstaff WHERE user_id=12391)
GROUP BY acd.id
ORDER BY acd.cdate, ctiming";
DB:select($sql);
throws
Illuminate \ Database \ QueryException (42000)
SQLSTATE[42000]: Syntax error or access violation: 1055 'mydatabase.acd.call_ref' isn't in GROUP BY (SQL: ....)
DB::select($sql);
DB::raw($sql);
DB::select(DB::raw($sql));
//even with pdo
$sth = DB::getPdo()->prepare($sql);
$sth->execute();
$data = $sth->fetchAll(\PDO::FETCH_OBJ);
PDOException (42000)
SQLSTATE[42000]: Syntax error or access violation: 1055 'mydatabase.acd.call_ref' isn't in GROUP BY
It seems like the only way to get working is to list all the table columns in the group by which is doable but not convenient.
I am able to run the same query directly in the phpmyadmin. So I am not sure why when I run it through Laravel it asks me add all columns.
I have MariaDB installed and both Laravel and PhpMyAdmin are connecting to the same instance. Larvel version is 5.8.5.
Please take a look at full query as I asked question here too but couldn't find any answer - https://laracasts.com/discuss/channels/eloquent/query-runs-ok-in-phpmyadmin-but-dbselect-throws-exception
You should try disabling the strict mode in config/database.php, in mysql connection section.
This is not a Laravel issue, but a logical limitation by MySQL. When you add columns to your results set that are not in the GROUP BY clause, different values could be possible for these columns in the result set.
When executing queries in strict mode (which is on by default), MySQL will not allow this because it could lead to unexpected results. If you turn off strict mode MySQL will use the first result it finds for these columns.
You can turn off strict mode in your Laravel config files in config/database.php. However, it is recommended to change your query instead because of the unpredictability.
A full list of checks in strict mode can be found in the MySQL documentation: https://dev.mysql.com/doc/refman/8.0/en/sql-mode.html#sql-mode-strict
I have installed Laravel 4.8 and also configured Auth. further more,I add two tables called as role and states and tried your pdo code - its perfectly works for me! Here, I put my code.
$sql = "SELECT *
FROM users u
INNER JOIN roles r ON r.id = u.role_id
LEFT JOIN states s ON s.id = u.state_id
WHERE 1=1
GROUP BY u.id
ORDER BY u.id ";
$sth = DB::getPdo()->prepare($sql);
$sth->execute();
$data = $sth->fetchAll(\PDO::FETCH_OBJ);
dd($data);
I am eager to know what is your db structure.
Related
I'm running the following query using PHP and I get the error:
1055: Expression #4 of SELECT list is not in GROUP BY clause and contains nonaggregated column '_______.a.name' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by
query
$get_artistid = (int)$_POST['artistid'];
$q = 'SELECT
w.work_id,
w.name,
CONCAT(w.work_id, \'/\', w.seo_url),
a.name,
a.photo_basename
FROM
work_credit wc
JOIN recording r ON wc.work_id = r.work_id
JOIN WORK w ON wc.work_id = w.work_id
JOIN artist a ON wc.artist_id = a.artist_id
WHERE
wc.artist_id = ?
AND r.is_performer = 1
AND r.is_video = 0
GROUP BY
w.work_id';
$conn = fn_connect();
if (!$stmt = $conn->prepare($q)) {
echo 'Error #' . $conn->errno . ': ' . $conn->error;
}
$stmt->bind_param('i', $get_artistid);
$stmt->execute();
$stmt->bind_result($workid, $workname, $workhref, $artistname, $artistbasename);
...
PHP is right, i'm supposed to have the error. The query should end with
GROUP BY
w.work_id, a.name, a.photo_basename
But, when i'm running the same query using a database management tool like TablePlus or SQLPro or else, the query executes correctly without the error. The reason why i'm using an external app is to create, edit, update, optimize the query. Having an app that doesn't pick up the error defeats the purpose.
When I run the following on any external db management app
SELECT ##sql_mode
I get
ONLY_FULL_GROUP_BY,STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION
ONLY_FULL_GROUP_BY looks like it's enable. Still, it doesn't pick up the error.
Any ideas why all of the external apps (all of them) are doing this? Do I need to configure something.
I have: Apache 4.4.41, PHP 7.3.11, mySQL 5.7.28 using Catalina 10.15.3
I need to write a query builder from sql which has a join within a left join.
This is the SQL which has a join within another left join.
select v_dts.* from v_dts
left join(select in_d2.ref_book from in_h
inner join in_d2 on(in_h.doc_code=in_d2.doc_code and in_h.book=in_d2.book)
where in_h.doc_code='IN' group by in_d2.ref_book) as i
on(v_dts.book=i.ref_book)
(Problem)This is the Query Builder that I try to convert from the SQL above.
$order_progress = DB::table('v_dts')
->select('v_dts.*')
->leftJoin('in_h',function($join_in_h){
$join_in_h->select('in_d2.ref_book');
$join_in_h->join('in_d2',function($join_in_d2){
$join_in_d2->on('in_h.doc_code','=','in_d2.doc_code');
$join_in_d2->on('in_h.book','=','in_d2.book');
$join_in_d2->where('in_h.doc_code','=','IN');
$join_in_d2->groupBy('in_d2.ref_book');
});
})
->get();
However, my query builder is wrong.
It show an error message
SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax
and generated SQL
SQL: select v_dts.* from v_dts left join (in_h inner join in_d2 on in_h.doc_code = in_d2.doc_code and in_h.book = in_d2.book and in_h.doc_code = IN)
Anyone can help me to figure out my problem?
Thanks.
You need a subquery left join. ->leftJoin('in_h',function($join_in_h){...} does not do a subquery join; that's a regular left join with a fancy join clause.
Instead use ->leftJoinSub.
You may use the joinSub, leftJoinSub, and rightJoinSub methods to join a query to a sub-query. Each of these methods receive three arguments: the sub-query, its table alias, and a Closure that defines the related columns.
Write the subquery.
$ref_books = DB::table('in_h')
->select('in_d2.ref_book')
->join('in_d2', function($join){
$join->on('in_h.doc_code','=','in_d2.doc_code')
->where('in_h.book','=','in_d2.book');
})
->where('in_h.doc_code','=','IN')
->groupBy('in_d2.ref_book');
And use it in another query.
$order_progress = DB::table('v_dts')
->select('v_dts.*')
->leftJoinSub($ref_books, 'i',function($join){
$join->on('v_dts.book', '=', 'i.ref_book');
})
->get();
You can use ->dd() and ->dump() to dump the generated SQL for debugging.
See "Sub-Query Joins" in the Laravel docs.
Note: I don't have Laravel so I can't check this is 100% correct, but it should get you on the right path.
I have a MYSQL query that i need to run on Laravel 5.6 query builder. my query is
SELECT paper_id,user_id,COUNT(payments.user_id),users.district
FROM payments
LEFT JOIN users ON payments.user_id = users.id
WHERE payments.paper_id=3
GROUP BY users.district HAVING COUNT(payments.user_id)>=0;
I have tried running this on Laravel DB::Raw with this code
$data=DB::Raw('SELECT paper_id,user_id,COUNT(payments.user_id),users.district
FROM payments
LEFT JOIN users ON payments.user_id = users.id
WHERE payments.paper_id='.$paper_id.'
GROUP BY users.district HAVING
COUNT(payments.user_id)>=0
');
as well as through this code
$data2=DB::table('payments')
->leftJoin('users','payments.user_id','users.id')
->select('paper_id','user_id','users.district',
DB::Raw('COUNT(payments.user_id)'))
->where('payments.paper_id',$paper_id)
->groupBy('users.district')
->select(DB::Raw('HAVING COUNT(payments.user_id)>=0'))
->get();
Error I get on $data (first query) query
[{}]
i get an empty response
Error I get on `$data2 (second query) query
SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error
in your SQL syntax; check the manual that corresponds to your MariaDB server
version for the right syntax to use near 'HAVING COUNT(payments.user_id)>=0 from
payments left join users on payments at line 1 (SQL: select HAVING
COUNT(payments.user_id)>=0 from payments left join users on
payments.user_id = users.id where payments.paper_id = 1 group by
users.district)
But i get SQL syntax error can someone please let me know how to convert SQL to Laravel query builder format.
Running Raw SQL Queries
$data = DB::select('SELECT paper_id,user_id,COUNT(payments.user_id),users.district
FROM payments
LEFT JOIN users ON payments.user_id = users.id
WHERE payments.paper_id= ?
GROUP BY users.district HAVING
COUNT(payments.user_id)>=0', [$paper_id]);
Parameter binding provides protection against SQL injection.
https://laravel.com/docs/5.6/database#running-queries
groupBy / having
->select(DB::Raw('HAVING COUNT(payments.user_id)>=0'))
=>
->having(DB::Raw('COUNT(payments.user_id)'), '>=', 0)
https://laravel.com/docs/5.6/queries#ordering-grouping-limit-and-offset
Try This:
When use Raw query
$data = DB::select('SELECT paper_id,user_id,COUNT(payments.user_id),users.district
FROM payments
LEFT JOIN users ON payments.user_id = users.id
WHERE payments.paper_id='.$paper_id.'
GROUP BY users.district HAVING
COUNT(payments.user_id)>=0
');
OR
$data2= DB::table('payments')->select([
'payments.paper_id','payments.user_id','users.district',
DB::Raw('COUNT(payments.user_id)')
])
->leftJoin('users','payments.user_id','users.id')
->where('payments.paper_id',$paper_id)
->havingRaw("COUNT(payments.user_id)>=0")
->groupBy('users.district')
->get();
I think select function should use one time only in a query, No need to add multiple time. And before add any questin alway search about your problem. Hope it will work..
For your error
SQLSTATE[42000]: Syntax error or access violation: 1055
In this file
vendor\laravel\framework\src\Illuminate\Database\Connectors\Connector.php
at this location in your laravel replace this file this for laravel5.4
PDO::ATTR_EMULATE_PREPARES => true,
I'm trying to use a left join to bridge two tables and force index an index that only exists on the joined table, but I get the following error:
You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use
near 'FORCE INDEX (l.sfdcId) WHERE l.sfdcId = '003A000001eR0HsIAK'
ORDER BY a.activity' at line 3
Here's an output of the query being run (works fine if I remove the FORCE INDEX):
SELECT a.activityDate,a.primaryAttributeValue,a.attributeDescription,l.firstName,l.lastName,l.title,l.email
FROM activities AS a LEFT JOIN
leads AS l
ON a.leadId = l.leadId FORCE INDEX (l.sfdcId)
WHERE l.sfdcId = '003A000001eR0HsIAK'
ORDER BY a.activityDate DESC
Any idea why this would fail?
The FORCE INDEX goes after the table definition:
SELECT a.activityDate,a.primaryAttributeValue,a.attributeDescription,
l.firstName,l.lastName,l.title,l .email
FROM activities a LEFT JOIN
leads l FORCE INDEX (sfdcId)
ON a.leadId = l.leadId
WHERE l.sfdcId = '003A000001eR0HsIAK'
ORDER BY a.activityDate DESC ;
I am getting error on pagination
$writers = $writer::join('categories','categories.id','=','writers.category_id')
->where([['categories.category_name', '=', $category]])->paginate(1);
Can anyone tell how to paginate an inner join query.
This is the error message I'm getting:
QueryException in Connection.php line 770: SQLSTATE[42000]: Syntax error or access violation: 1140 Mixing of GROUP columns (MIN(),MAX(),COUNT(),...) with no GROUP columns is illegal if there is no GROUP BY clause (SQL: select count(*) as aggregate from writers` inner join categories on categories.id = writers.category_id where (categories.category_name = Comic))
You need to specify GROUP_BY clause on your query
$writers = Writer::join('categories','categories.id','=','writers.category_id')
->where('categories.category_name', '=', $category)
->groupBy('writers.id')
->paginate(1);
UPDATE
If you still get an error, check your config/database.php. Be sure that in mysql settingsstrict = false
UPDATE 2
strict mode works on mysql starting from 5.7. If you have mysql under 5.7, set strict => false. You can check this link for more information:
Strict mode