before i use alias for table i get the error:
: Integrity constraint violation: 1052 Column 'id' in field list is ambiguous
Then i used aliases and i get this error:
unknown index a
I am trying to get a list of category name ( dependant to a translation) and the associated category id which is unique. Since i need to put them in a select, i see that i should use the lists.
$categorie= DB::table('cat as a')
->join('campo_cat as c','c.id_cat','=','a.id')
->join('campo as d','d.id','=','c.id_campo')
->join('cat_nome as nome','nome.id_cat','=','a.id')
->join('lingua','nome.id_lingua','=','lingua.id')
->where('lingua.lingua','=','it-IT')
->groupby('nome.nome')
->lists('nome.nome','a.id');
The best way to debug your query is to look at the raw query Laravel generates and trying to run this raw query in your favorite SQL tool (Navicat, MySQL cli tool...), so you can dump it to log using:
DB::listen(function($sql, $bindings, $time) {
Log::info($sql);
Log::info($bindings);
});
Doing that with yours I could see at least one problem:
->where('lingua.lingua','=','it-IT')
Must be changed to
->where('lingua.lingua','=',"'it-IT'")
As #jmail said, you didn't really describe the problem very well, just what you ended up doing to get around (part of) it. However, if I read your question right you're saying that originally you did it without all the aliases you got the 'ambiguous' error.
So let me explain that first: this would happen, because there are many parts of that query that use id rather than a qualified table`.`id.
if you think about it, without aliases you query looks a bit like this: SELECT * FROM `cat` JOIN `campo_cat` ON `id_cat` = `id` JOIN `campo` ON `id` = `id_campo`; and suddenly, MySQL doesn't know to which table all these id columns refer. So to get around that all you need to do is namespace your fields (i.e. use ... JOIN `campo` ON `campo`.`id` = `campo_cat`.`id_campo`...). In your case you've gone one step further and aliased your tables. This certianly makes the query a little simpler, though you don't need to actually do it.
So on to your next issue - this will be a Laravel error. And presumably happening because your key column from lists($valueColumn, $keyColumn) isn't found in the results. This is because you're referring to the cat.id column (okay in your aliased case a.id) in part of the code that's no longer in MySQL - the lists() method is actually run in PHP after Laravel gets the results from the database. As such, there's no such column called a.id. It's likely it'll be called id, but because you don't request it specifically, you may find that the ambiguous issue is back. My suggestion would be to select it specifically and alias the column. Try something like the below:
$categories = DB::table('cat as a')
->join('campo_cat as c','c.id_cat','=','a.id')
->join('campo as d','d.id','=','c.id_campo')
->join('cat_nome as nome','nome.id_cat','=','a.id')
->join('lingua','nome.id_lingua','=','lingua.id')
->where('lingua.lingua','=','it-IT')
->groupby('nome.nome')
->select('nome.nome as nome_nome','a.id as a_id') // here we alias `.id as a_id
->lists('nome_nome','a_id'); // here we refer to the actual columns
It may not work perfectly (I don't use ->select() so don't know whether you pass an array or multiple parameters, also you may need DB::raw() wrapping each one in order to do the aliasing) but hopefully you get my meaning and can get it working.
Related
I'm working on a MySQL Way of printing an "affiliate tree" and got the thing working with Common Table Expression. I'm using the following code right now:
WITH RECURSIVE recUsers AS
(
SELECT ID, username, sponsorID, 1 AS depth, username AS path
FROM users
WHERE id = 1
UNION ALL
SELECT c.ID, c.username, c.sponsorID, sc.depth + 1, CONCAT(sc.path, ' > ', c.username)
FROM recUsers AS sc
JOIN users AS c ON sc.ID = c.sponsorID
)
SELECT * FROM recUsers;
This selects the tree underneath the user with the id 1.
Now what I'd need to get is a way to pass that id as a parameter, so I don't need to define everything from the beginning every time I want to get the result.. So my idea is to put everything in a stored prodecure and pass the id in as a parameter.. However, so far I didn't get it working and always getting various errors that are very self speaking...
Basically what I've tried was
DELIMITER //
CREATE PROCEDURE getAffiliateTree(IN userid INT())
BEGIN
---my code here, the userid 1 replaced with userid
END//
DELIMITER;
However, this doesn't seem to work.. How can I get this done?
Two things I would suggest:
Use INT, not INT(). The optional length argument to integer types is deprecated in MySQL 8.0 (which I know you're using, because you're using CTE syntax). Even if you did use the length argument, using an empty argument is not legal syntax.
Make sure that the userid input parameter name is distinct from all of the columns in the tables you reference. That is, if the table has a column named userid (any capitalization), then change the name of your input parameter. Otherwise you may make ambiguous expressions like:
... WHERE userid = userid
Even though you intend one of these to be the column and the other to be the parameter, the SQL parser has no way of knowing that. It ends up treating both as the column name, so it's trivially true on all rows of the table.
Actually, a third thing I would suggest: when you ask questions, "it doesn't seem to work" isn't clear enough. Did it produce an error? If so, what was the full error message? Did it produce no error, but didn't give you the result you wanted? If so, show a mocked-up example of what you expected, and what the query produced that didn't match. It helps to be as clear as you can when you post questions, so readers don't have to guess what trouble you need help with.
I just discovered that if you use the same name in bindparams twice in the same query, the second value clobbers the first one. For a contrived example:
db.session.query(MyTable).filter(
db.or_(
db.text("my_table.field = :value").bindparams(value=value1),
db.text("my_table.field = :value").bindparams(value=value2),
)
)
Here you would only get things with value2. value1 would not appear in the query.
Is there a general purpose way to fix this?
Btw, db.text() bits in my real query access nested jsonb properties, so please don't answer telling me to just use the column objects in this query in place of db.text().
I have a table in my MySQL database, compatibility_core_rules, which essentially stores pairs of ids which represent compatibility between parts which have fields with those corresponding ids. Now, my aim is to get all possible compatibility pairs by following the transitivity of the pairs (e.g. so if the table has (1,2) and (2,4), then add the pair (1,4)). So, mathematically speaking, I'm trying to find the transitive closure of the compatibility_core_rules table.
E.g. if compatibility_core_rules contains (1,2), (2,4) and (4,9), then initially we can see that (1,2) and (2,4) gives a new pair (1,4). I then iterate over the updated pairs and find that (4,9) with the newly added (1,4) gives me (1,9). At this point, iterating again would add no more pairs.
So my approach is to create a view with the initial pairs from compatibility_core_rules, like so:
CREATE VIEW compatibility_core_rules_closure
AS
SELECT part_type_field_values_id_a,
part_type_field_values_id_b,
custom_builder_id
FROM compatibility_core_rules;
Then, in order to iteratively discover all pairs, I need to keep replacing that view with an updated version of itself that has additional pairs each time. However, I found MySQL doesn't like me referencing the view in its own definition, so I make a temporary view (with or replace, since this will be inside a loop):
CREATE OR REPLACE VIEW compatibility_core_rules_closure_temp
AS
SELECT part_type_field_values_id_a,
part_type_field_values_id_b,
custom_builder_id
FROM compatibility_core_rules_closure;
No problems here. I then reference this temporary view in the following CREATE OR REPLACE VIEW statement to update the compatibility_core_rules_closure view with one iteration's worth of additional pairs:
CREATE OR REPLACE VIEW compatibility_core_rules_closure
AS
SELECT
CASE WHEN ccr1.part_type_field_values_id_a = ccr2.part_type_field_values_id_a THEN ccr1.part_type_field_values_id_b
WHEN ccr1.part_type_field_values_id_a = ccr2.part_type_field_values_id_b THEN ccr1.part_type_field_values_id_b
END ccrA,
CASE WHEN ccr1.part_type_field_values_id_a = ccr2.part_type_field_values_id_a THEN ccr2.part_type_field_values_id_b
WHEN ccr1.part_type_field_values_id_a = ccr2.part_type_field_values_id_b THEN ccr2.part_type_field_values_id_a
END ccrB,
ccr1.custom_builder_id custom_builder_id
FROM compatibility_core_rules_closure_temp ccr1
INNER JOIN compatibility_core_rules_closure_temp ccr2
ON (
ccr1.part_type_field_values_id_a = ccr2.part_type_field_values_id_a OR
ccr1.part_type_field_values_id_a = ccr2.part_type_field_values_id_b
)
GROUP BY ccrA,
ccrB
HAVING -- ccrA and ccrB are in fact not the same
ccrA != ccrB
-- ccrA and ccrB do not belong to the same part type
AND (
SELECT ptf.part_type_id
FROM part_type_field_values ptfv
INNER JOIN part_type_fields ptf
ON ptfv.part_type_field_id = ptf.id
WHERE ptfv.id = ccrA
LIMIT 1
) !=
(
SELECT ptf.part_type_id
FROM part_type_field_values ptfv
INNER JOIN part_type_fields ptf
ON ptfv.part_type_field_id = ptf.id
WHERE ptfv.id = ccrB
LIMIT 1
)
Now this is where things go wrong. I get the following error:
#1146 - Table 'db509574872.compatibility_core_rules_closure' doesn't exist
I'm very confused by this error message. I literally just created the view/table only two statements ago. I'm sure the SELECT query itself is correct since if I try it by itself and it runs fine. If I change the first line to use compatibility_core_rules_closure2 instead of compatibility_core_rules_closure then it runs fine (however, that's not much use since I need to be re-updating the same view again and again). I've looked into the SQL SECURITY clauses but have not had any success. Also been researching online but not getting anywhere.
Does anyone have any ideas what is happening and how to solve it?
MySQL doesn't support sub-queries in views.
You'll have to separate them... ie. using another view containing the sub-query inside you main view.
Running the create statement for that view will render an error, not creating it, hence the doesn't exist error you're getting when querying it.
hi i am executing nested "select" query in mysql .
the query is
SELECT `btitle` FROM `backlog` WHERE `bid` in (SELECT `abacklog_id` FROM `asprint` WHERE `aid`=184 )
I am not getting expected answer by the above query. If I execute:
SELECT abacklog_id FROM asprint WHERE aid=184
separately
I will get abacklog_id as 42,43,44,45;
So if again I execute:
SELECT `btitle` FROM `backlog` WHERE `bid` in(42,43,44,45)
I will get btitle as scrum1 scrum2 scrum3 msoffice
But if I combine those queries I will get only scrum1 remaining 3 atitle will not get.
You Can Try As Like Following...
SELECT `age_backlog`.`ab_title` FROM `age_backlog` LEFT JOIN `age_sprint` ON `age_backlog`.`ab_id` = `age_sprint`.`as_backlog_id` WHERE `age_sprint`.`as_id` = 184
By using this query you will get result with loop . You will be able to get all result with same by place with comma separated by using IMPLODE function ..
May it will be helpful for you... If you get any error , Please inform me...
What you did is to store comma separated values in age_sprint.as_backlog_id, right?
Your query actually becomes
SELECT `ab_title` FROM `age_backlog` WHERE `ab_id` IN ('42,43,44,45')
Note the ' in the IN() function. You don't get separate numbers, you get one string.
Now, when you do
SELECT CAST('42,43,44,45' AS SIGNED)
which basically is the implicit cast MySQL does, the result is 42. That's why you just get scrum1 as result.
You can search for dozens of answers to this problem here on SO.
You should never ever store comma separated values in a database. It violates the first normal form. In most cases databases are in third normal form or BCNF or even higher. Lower normal forms are just used in some special cases to get the most performance, usually for reporting issues. Not for actually working with data. You want 1 row for every as_backlog_id.
Again, your primary goal should be to get a better database design, not to write some crazy functions to get each comma separated number out of the field.
In my Rails testing environment, I have a user_id that looks like 1234-567abc89. I'm getting inconsistent behaviour by querying this user in different tables. Most of the queries are working, but running one particular query fails:
ActiveRecord::StatementInvalid (Mysql::Error: Unknown column '1234' in
'where clause': SELECT * FROM `point_allocations` WHERE (user_id = 1234-567abc89) ):
So for some reason, everything beyond the hyphen is getting cut off. I realized that for the queries that work, it is looking up user 1234 instead of 1234-567abc89, but if all the others work, any idea why only this one would return an error?
You need to include quotations.
SELECT * FROM `point_allocations` WHERE (user_id = '1234-567abc89')
Because the user_id column expects character-typed data, it will take your value (1234-567abc89) and parse it as an integer, truncating the content after the hyphen. If you include it in quotations, it will accept it as a string and transfer properly.
Enjoy and good luck!