SQL计数+左联接+分组依据…缺少行

Question

Trying to list all what's in table 1 and records under it in table 2

Table one each row has an id , and each row in table 2 has idontable1

select table1.*, count(table2.idintable1)as total
from   table1
       left join table2 on table1.id=table2.idintable1

 WHERE table1.deleted='0' AND table2.deleted=0
group by 
table2.idintable1

My current problem is rows from table1 with 0 records in table2 are not displayed I want them to be displayed

Answer 1

The query that you want is:

select t1.*, count(t2.idintable1) as total
from table1 t1 left join
     table2 t2
     on t1.id = t1.idintable1 and t2.deleted = 0
where t1.deleted = 0 
group by t1.id;

Here are the changes:

• The condition on t2.deleted was moved to the on clause. Otherwise, this turns the outer join into an inner join.
• The condition on t1.deleted remains in the where clause, because presumably you really do want this as a filter condition.
• The group by clause is based on t1.id , because t2.idintable1 will be NULL when there are no matches. Just using t1.id is fine, assuming that id is unique (or a primary key) in table1 .
• The table aliases are not strictly necessary, but they make queries easier to write and to read.

Answer 2

You should GROUP BY table1.id .

The LEFT JOIN ensures all the rows from table1 appear in the result set. Those that do not have a pair in table2 will appear with NULL in field table2.idintable1 . Because of that your original GROUP BY clause produces a single row for all the rows from table1 that do not appear in table2 (instead of one row for each row of table1 ).

Answer 3

You have fallen into mysql's non-standard group by support trap.

Change your group by to list all columns of table 1:

group by table1.id, table1.name, etc

or list the column positions of all table1 columns in the select:

group by 1, 2, 3, 4, etc 

Or use a subquery to get the count vs the id, and join table1 to that.