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Broken count(*) after adding LEFT JOIN


Broken count(*) after adding LEFT JOIN

By : ncwoehler
Date : November 21 2020, 07:01 PM
To fix the issue you can do If you wonder about a count, just run the query without a group by. You'll see where the rows in the count come from.
In your case, you could probably fix it by changing count(*) to count distinct(n.entry_id).
code :


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Mysql LEFT Join query only returns one row when adding Count() mysql function

Mysql LEFT Join query only returns one row when adding Count() mysql function


By : David Addario
Date : March 29 2020, 07:55 AM
this will help The COUNT() aggregate will return only one row in absence of a GROUP BY clause, and MySQL is lenient about the presence or contents of the GROUP BY (your query would have failed with a syntax error in most other RDBMS).
Since you have multiple columns, you ought to join against a subquery to get the count per Files.GUID. Although MySQL will permit you to GROUP BY Users.GUID without the subquery, which is simpler, you may not get the results you expect from Users.Name or Users.CreationDate. This method is more portable:
code :
SELECT
  Users.GUID,
  Users.Name,
  Users.CreationDate,
  FileCount
FROM
  Users
  /* Subquery returns UserGUID and associated file count */
  LEFT JOIN (
    SELECT UserGUID, COUNT(*) AS FileCount
    FROM Files
    GROUP BY UserGUID
  ) fc ON Users.GUID = fc.UserGuid
WHERE Users.Group = 1
Can anyone help me when my left join statement? I think the mySQL Engine is broken

Can anyone help me when my left join statement? I think the mySQL Engine is broken


By : TomH
Date : March 29 2020, 07:55 AM
will help you When LEFT JOIN, put the right side table's conditions in the ON clause to get true left join behavior! (When in WHERE, you get regular inner join result):
code :
SELECT a.SPERIOD
FROM Payslip a
LEFT JOIN Earndetl b ON a.SPERIOD = b.SPERIOD AND MONTH(b.SPERIOD) = 1
LEFT JOIN Ded_detl c ON a.SPERIOD = c.SPERIOD AND MONTH(c.SPERIOD) = 1
WHERE MONTH(a.SPERIOD) = 1
GROUP BY a.SPERIOD;
Left join on broken?

Left join on broken?


By : kidkarma
Date : March 29 2020, 07:55 AM
it fixes the issue The server time may have become incorrect, so SYSDATE is returning the wrong value. Check the system time and also run a SQL query with SYSDATE function replaced with literals (Just to test). Also related to this check the time zone has not changed. If server time wrong then reset.
count is multiplied after adding left join

count is multiplied after adding left join


By : user2985819
Date : March 29 2020, 07:55 AM
around this issue Your average may also be subtly off -- or at least require additional unnecessarily calculation.
The reason is that the number of rows is being multiplied along two dimensions. The more accurate solution is to aggregate before doing the join:
code :
SELECT f.name, s.num_letters, AVG(t.numbers)
FROM first f INNER JOIN
     (SELECT s.first_id, COUNT(*) as num_letters
      FROM second s 
      WHERE s.letters IN ('aa', 'bb', 'cc', 'dd')  
      GROUP BY s.first_id
     ) s
     ON s.first_id = f.id LEFT JOIN
     third t
     ON t.first_id = f.id
GROUP BY f.id, s.num_letters;
SELECT f.name, s.num_letters, t.avg_numbers
FROM first f INNER JOIN
     (SELECT s.first_id, COUNT(*) as num_letters
      FROM second s 
      WHERE s.letters IN ('aa', 'bb', 'cc', 'dd')  
      GROUP BY s.first_id
     ) s
     ON s.first_id = f.id LEFT JOIN
     (SELECT t.first_id, AVG(t.numbers) as avg_numbers
      FROM third t
      GROUP BY t.first_id
     ) t
     ON t.first_id = f.id;
SQL - LEFT JOIN, but I want COUNT(*) to only count the results from the INNER part of the join

SQL - LEFT JOIN, but I want COUNT(*) to only count the results from the INNER part of the join


By : Sophaneth Khean
Date : March 29 2020, 07:55 AM
wish of those help I want to display the number of purchases each customer has made. If they've made 0 purchases, I want to display 0. , Instead of count(*) use count(purchase_id)
code :
SELECT customer_name, COUNT(purchase_id) AS number_of_purchaes 
FROM customer 
LEFT JOIN purchases ON customer.customer_id = purchases.customer_id 
GROUP BY customer_id,customer_name
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