Hi. I've been confused by this the last two days.
I am using Flynax for my Smarty project, but inside I can't update the admin app->common->configuration
I am getting this error:
MYSQL ERROR
Error: Unknown column 'Array' in 'having clause'
Query: SELECT `IP` AS `Count` FROM `fl_login_attempts` WHERE `IP` = '123.236.230.132' AND `Status` = 'fail' AND `Interface` = 'admin' GROUP BY `Date` HAVING TIMESTAMPDIFF(HOUR, `Date`, NOW()) < Array
Function: error
Class: rlDb
File: /home/autotrad/public_html/includes/classes/rlDb.class.php (line# 116)
Does anyone know how to solve it - is anyone getting the same error?
Yea Sundar disabling "register_globals" will solve the problem for sure!
Related
I am facing this error for the past 2-3 days. I hold this task to check this later in my free time. But still, I am getting the same error. Please help me to find the error. I am tired to check all solutions on google. But nothing works.
Error
Illuminate\Database\QueryException: SQLSTATE[42000]: Syntax error or access violation: 1055 'dating-app.messages.id' isn't in GROUP BY (SQL: select * from messages group by room_id having is_read = 0) in file C:\xampp\htdocs\dating-app\vendor\laravel\framework\src\Illuminate\Database\Connection.php on line 692
Here is my query
$messages = Message::groupBy('room_id')->having('is_read', 0)->get();
return response()->json(['status' => true, 'message' => $messages]);`
Any solution appreciated!
That's because any column you use in a select statement must appear in the group by clause. So select * and group by don't get along. Remove groupBy('room_id') in your relationship and see how that goes.
Alternatively, disable MySQL strict mode by going into config/database.php and setting strict => false for MySQL, if you are using MySQL version >5.7.
source
I have just upgraded Magento from 1.5.1.0 to 1.9.2.1 and I am stuck at this error when I try to view a product:
a:5:{i:0;s:1427:"SQLSTATE[42S22]: Column not found: 1054 Unknown column 'rule.calculate_subtotal' in 'field list', query was: SELECT `main_table`.`tax_calculation_rate_id`, `main_table`.`tax_calculation_rule_id`, `main_table`.`customer_tax_class_id`, `main_table`.`product_tax_class_id`, `rule`.`priority`, `rule`.`position`, `rule`.`calculate_subtotal`, `rate`.`rate` AS `value`, `rate`.`tax_country_id`, `rate`.`tax_region_id`, `rate`.`tax_postcode`, `rate`.`tax_calculation_rate_id`, `rate`.`code`, IF(title_table.value IS NULL, rate.code, title_table.value) AS `title` FROM `tax_calculation` AS `main_table` INNER JOIN `tax_calculation_rule` AS `rule` ON `rule`.`tax_calculation_rule_id` = main_table.tax_calculation_rule_id INNER JOIN `tax_calculation_rate` AS `rate` ON rate.tax_calculation_rate_id = main_table.tax_calculation_rate_id LEFT JOIN `tax_calculation_rate_title` AS `title_table` ON rate.tax_calculation_rate_id = title_table.tax_calculation_rate_id AND title_table.store_id = '1' WHERE (customer_tax_class_id = 3) AND (product_tax_class_id IN ('2')) AND (rate.tax_country_id = 'GB') AND (rate.tax_region_id IN(0, 0)) AND (rate.zip_is_range IS NULL) AND (rate.tax_postcode IS NULL OR rate.tax_postcode IN('*', '', 'ig9 5bw', 'ig9 5bw*', 'ig9 5b*', 'ig9 5*', 'ig9 *', 'ig9*', 'ig*', 'i*')) ORDER BY `priority` ASC, `tax_calculation_rule_id` ASC, `tax_country_id` DESC, `tax_region_id` DESC, `tax_postcode` DESC, `value` DESC";i:1;s:6703:"#0 /home/wwwarabesquecost/public_html/lib/Varien/Db/Statement/Pdo/Mysql.php(110): Zend_Db_Statement_Pdo->_execute(Array)
I have tried to manually add it using this link : https://sarfarazlaghari.wordpress.com/2015/06/06/magento-sqlstate42s22-column-not-found-1054-unknown-column-rule-calculate_subtotal-in-field-list/ but this doesn't make a difference.
Has anybody had this issue before?
Regarding
I have tried to manually add it using this link : https://sarfarazlaghari.wordpress.com/2015/06/06/magento-sqlstate42s22-column-not-found-1054-unknown-column-rule-calculate_subtotal-in-field-list/ but this doesn't make a difference
Are you saying you've added the column to the tax_calculation_rule table but you're still getting the above error? If that's the case then you've updated the wrong database table, or a configuration file in app/etc is pointed at a different database from the one you think it is.
If that's not what you're saying then be clearer in your question.
The query that you found at that link has an obvious error: it create a new field in the table with the same name of the table.
The right query, which solve the error, is the following:
ALTER TABLE tax_calculation_rule
ADD calculate_subtotal TINYINT(1) NOT NULL DEFAULT 0
Hope it helps.
You shouldn't run sql alternations just like that. The OP mentioned this has happened after an upgrade to 1.9x.
Having such errors means that your codebase is upgraded but your database is still outdated.
The best way for this would be to alter the core_resource table for the module (e.g. tax_setup) to the version that his database was aware previously, and let magento handle the database updates.
I'm running the following SQL in Laravel:
$sql = 'SELECT university.id, university.name, MAX(uni_score) AS score
FROM (SELECT uni_id, place AS uni_score FROM ranking) AS tmp
LEFT JOIN university ON university.id = tmp.uni_id
ORDER BY score';
$result = DB::select(DB::raw($sql));
However the code throws this error:
Column not found: 1054 Unknown column 'uni_score' in 'field list'.
uni_score is an alias for place field in ranking table. The query above works fine when ran directly in phpMyAdmin.
What am I doing wrong?
Try using the DB::statement($query); method.
I have the following query:
SELECT * FROM `alerts` WHERE `title` LIKE `%template%`
This should return at least 3 results with titles that includes the word 'template' but I'm getting the following error: -
1054 - Unknown column '%template%' in 'where clause'
As far as I can tell its syntactically correct and calling the correct column names. What am I missing?
Use single quotes for the %template%:
SELECT * FROM `alerts` WHERE `title` LIKE '%template%'
Few weeks ago my code was all fine, but recently I've been unable to load my test database on my web app (built using CakePhp).
This is the error :
Warning (512): SQL Error: 1064: You have an error in your SQL syntax;
check the manual that corresponds to your MySQL server version for the
right syntax to use near ') LEFT JOIN blogs AS Blog ON
(Song.blog_id = Blog.id) WHERE 1 = 1' at line 1
[CORE/cake/libs/model/datasources/dbo_source.php, line 684]
SELECT COUNT(*) AS `count`
FROM `songs` AS `Song`
LEFT JOIN libraries AS `Library` ON (`Song`.`id` = `Library`.`song_id` AND `Library`.`user_id` =)
LEFT JOIN `blogs` AS `Blog` ON (`Song`.`blog_id` = `Blog`.`id`)
WHERE 1 = 1
Warning (512): SQL Error: 1064: You have an error in your SQL syntax;
check the manual that corresponds to your MySQL server version for the
right syntax to use near ') WHERE 1 = 1 ORDER BY Song.postdate
desc LIMIT 100' at line 1
[CORE/cake/libs/model/datasources/dbo_source.php, line 684]
SELECT *, Blog.id
FROM `songs` AS `Song`
LEFT JOIN libraries AS `Library` ON (`Song`.`id` = `Library`.`song_id` AND `Library`.`user_id` =)
WHERE 1 = 1
ORDER BY `Song`.`postdate` DESC
LIMIT 100
You don't have a condition for the Library.userid column:
Library.user_id =)
You need to specify a value on the right side of the equals sign!
your error is here:
Library.user_id =)
maybe you want to change it to:
Library.user_id ='$user_id')
It appears that some value for user_id is not being passed into the query:
`Library`.`user_id` =)
//-----------------^^^^
Likely, there is some PHP variable missing here, or a PDO parameter is not being correctly bound.
SELECT COUNT(*) AS `count` FROM `songs` AS `Song`
left JOIN libraries AS `Library`
ON (`Song`.`id` = `Library`.`song_id` AND `Library`.`user_id` =) <-- HERE's something missing
LEFT JOIN `blogs` AS `Blog` ON (`Song`.`blog_id` = `Blog`.`id`) WHERE 1 = 1
SELECT *,
Blog.id FROM `songs` AS `Song` left JOIN libraries AS `Library`
ON (`Song`.`id` = `Library`.`song_id` AND `Library`.`user_id` =) <-- HERE's something missing
WHERE 1 = 1 ORDER BY `Song`.`postdate` desc LIMIT 100
Since this is a CakePHP question, my answer will approach from the paradigms of the framework. However, given your SQL output, you are either NOT using CakePHP or you are doing something that is definitely a no-no.
Please post the Find statement that you are using the build this SQL statement and your Library's hasMany relationship to Blog. Unless you are using the query method, I think you would get a PHP error before you could build this type of SQL statement using the CakePHP find conventions. If you are using the query method to find your data, then don't. This is a simple query that can be done with CakePHP find easily.
Your library model should like like:
public $hasMany => array('Blog');
And the find:
$this->Library->find(
'all',
array(
'conditions' => array(
'1' => 1 //Not sure why you have Where 1=1
)
'order' => "Library.name DESC",
'limit' => 100
)
);