I am getting an SQL ERROR (1064) Syntax. Is what i am trying to do allowed? As i don't see the syntax error.
`SELECT isc_products.prodname, isc_product_variations.* , isc_product_variation_combinations.vcoptionids,
FROM isc_products
JOIN isc_product_variations
ON isc_products.prodvariationid = isc_product_variations.variationid
JOIN isc_product_variation_combinations
ON isc_product_variation_combinations.vcvariationid = isc_product_variations.variationid`
You have isc_product_variations.variationid twice in your ON statements. Check, if this is what you want, or if there is a second key in you perhaps need isc_product_variations
You have an error on the first line. You have a comma that shouldn't be there:
SELECT isc_products.prodname,
isc_product_variations.* ,
isc_product_variation_combinations.vcoptionids,
-- ^
FROM ...
I'd also advise you not to use SELECT isc_product_variations.* but instead list the columns you want explicitly.
Related
I've been working on a SQL query for a project, and I face an error message when I want to use it.
Here is the query itself :
SELECT COUNT(r) AS auditMade,
SUM(g.nbrMilkingCows) AS cowsAudited,
AVG(r.gainPerCowPerYearTransition) AS averageGainTransition,
AVG(r.gainPerCowPerYearLactation) AS averageGainLactation,
AVG(r.totalGain) AS averageTotalGain,
AVG(r.supplementalCostPerCow) AS averageSuppCost
FROM `smart_calculator_infos` i
INNER JOIN `smart_calculator_result` r ON r.idSmartCalculatorResult = i.idSmartCalculatorResult
INNER JOIN `calculator_general_informations` g ON g.idSmartCalculatorInfo = i.idSmartCalculatorInfo
WHERE i.idUser = 14
MySQL answers me "Unknown column 'r' in field list".
But I dont really understand why I get an error here as I define r in my INNER JOIN.
I'm kinda new at using SQL so maybe there is something pretty obvious I forgot, but I can't seem to understand what.
You can't count an alias itself, so the very first line of your query is what is causing the error:
SELECT COUNT(r)
To remedy this, you could use COUNT(*):
SELECT COUNT(*)
Or, you could count an actual column in the smart_calculator_result table, e.g.
SELECT COUNT(r.idSmartCalculatorResult)
This it the code I am trying to execute:
SELECT ID_K
FROM koncert,
programi
WHERE koncert.ID_K = programi.ID_K
GROUP BY koncert.ID_K
HAVING COUNT (DISTINCT programi.Salla) = 2
It returns this error:
#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 'DISTINCT programi.Salla)=2 LIMIT 0, 25' at line 4.
Tried to change different things but it still won't work .
You should use the count(DISTINCT programi.Salla
) and not count (..) ..remove space between COUNT and (...
SELECT koncert.ID_K
FROM koncert
INNER JOIN programi on koncert.ID_K = programi.ID_K
GROUP BY koncert.ID_K
HAVING COUNT(DISTINCT programi.Salla) = 2
but you need also tablename for avoid ambiguity and use explicit join sintax too
First you should use qualified name for your column, when column name is same in both table
SELECT ID_K FROM
should be
SELECT programi.ID_K FROM
else, you will get ambiguous column error. Otherwise, your query looks fine except removing extra space when calling COUNT (#spencer already mentioned in comment)
Also, it is good practice to join your table using JOIN (or INNER JOIN, LEFT JOIN etc) keyword, which makes your query more clear and readable.
The following query gives me an error in phpmyadmin. It looks syntactically correct to me, and the table/column names match up accordingly. I have tried a number of variations (quoting table names, using as, etc) with no luck.
SELECT *
FROM GROUP
INNER JOIN GROUP_MEMBER ON GROUP.group_id = GROUP_MEMBER.group_id
WHERE group_owner='test';
Error I'm getting:
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 'GROUP INNER JOIN GROUP_MEMBER ON GROUP.group_id = GROUP_MEMBER.group_id WHERE ' at line 2
"group" is a sql-keyword, you need to surround it with backticks if you want to use it as tablename.
GROUP is a reserved word in SQL so it's a bad choice for a table name. If you surround it with backticks it might work but I'd really recommend changing that table name.
SELECT *
FROM `GROUP`
INNER JOIN GROUP_MEMBER ON `GROUP`.group_id = GROUP_MEMBER.group_id
WHERE group_owner='test';
This is not PHPMyAdmin-specfiic error. The problem you have is using a table name GROUP that matches a MySQL reserved word. If you insist on using such a problematic table name, you need to enclose it with backticks anywhere you might use it.
SELECT *
FROM `GROUP`
INNER JOIN GROUP_MEMBER ON `GROUP`.group_id = GROUP_MEMBER.group_id
WHERE group_owner='test';
For some reason I get the following error when running the code below:
#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 'FROM postcodes_demographics INNER JOIN latlon1 on postcodes_demographics.postc' at line 3
I don't understand what I'm doing wrong, thanks for any suggestions!
INSERT INTO pslatlong
SELECT postcodes_demographics.*, latlon1.*,
FROM postcodes_demographics
INNER JOIN latlon1
on postcodes_demographics.postcode = latlon1.postcodens;
You have an errant comma:
SELECT postcodes_demographics.*, latlon1.*, <--- HERE
Remove it.
I would be very surprised if merely removing the comma fixes the problem. When using insert, you should get in the habit of listing all the columns explicitly:
INSERT INTO pslatlong(col1, col2, . . . )
SELECT d.col1, l.col2, . . .
FROM postcodes_demographics d INNER JOIN
latlon1 ll
on d.postcode = ll.postcodens;
You need to do this to be sure that the right column is assigned the right value, to allow auto incrementing columns to be auto-incremented, and to prevent problems based on the number of columns.
This may works:
INSERT INTO pslatlong
SELECT postcodes_demographics.*, latlon1.*
FROM postcodes_demographics
INNER JOIN latlon1
on postcodes_demographics.postcode = latlon1.postcodens;
EDIT: I am using phpMyAdmin interface, and I have been copy/paste the codes from phpMyAdmin to here. The phpMyAdmin seems to run a "different code" as I run the following code, and generating some error message that are referring to that "different code", causing huge confusion.
** Final edit: It seems Safari is causing this: it run the "different query" when I try to run 2nd query below. Use Firefox instead, and it generate correct results. Thanks for the help and sorry for the confusion. **
I have two tables: newsFeeds, comments, where
** newsFeeds contains column PID, comments contains column FID.**
I want to join rows in two tables with matching PID = FID. My code is:
SELECT * FROM newsFeeds
INNER JOIN
(
SELECT * FROM
comments
)
ON comments.FID = newsFeeds.PID
and the error message is "#1248 - Every derived table must have its own alias".
I then add in AS newtb after ) according to other posts here. And the code is then:
SELECT * FROM newsFeeds
INNER JOIN
(
SELECT * FROM
comments
) AS newtb
ON newtb.FID = newsFeeds.PID
But another error shows up: #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 'INNER JOIN( SELECTFID,COUNT(*) AScount FROMcomments
LIMIT 0, 25' at line 8
I wonder how to correctly do this?
You should correct this by removing the derived table:
SELECT *
FROM tb_1 INNER JOIN
tb_2
ON tb_2.FID = tb_1.PID;
MySQL has a tendency to materialize derived tables, which hurts performance.
The answer to your question, though, is to add a name after the parentheses:
SELECT *
FROM tb_1 INNER JOIN
(SELECT *
FROM tb_2
) t2
ON t2.FID = tb_1.PID;