having-clause

SELECT id HAVING maximum count of id

由 China☆狼群 提交于 2019-11-29 11:57:01
问题 Have a products table with item_id and color_id. I'm trying to get the color_id with the most non-null instances. This fails: SELECT color_id FROM products WHERE item_id=1234 GROUP BY item_id HAVING MAX(COUNT(color_id)) with Invalid use of group function This SELECT color_id, COUNT(color_id) FROM products WHERE item_id=1234 GROUP BY item_id Returns color_id count 1, 323 2, 122 3, 554 I am looking for color_id 3, which has the most instances. Is there a quick and easy way of getting what I

What is the correct way to do a HAVING in a MongoDB GROUP BY?

由 随声附和 提交于 2019-11-28 21:11:52
For what would be this query in SQL (to find duplicates): SELECT userId, name FROM col GROUP BY userId, name HAVING COUNT(*)>1 I performed this simple query in MongoDB: res = db.col.group({key:{userId:true,name:true}, reduce: function(obj,prev) {prev.count++;}, initial: {count:0}}) I've added a simple Javascript loop to go over the result set, and performed a filter to find all the fields with a count > 1 there, like so: for (i in res) {if (res[i].count>1) printjson(res[i])}; Is there a better way to do this other than using javascript code in the client? If this is the best/simplest way, say

Unknown column in 'having clause'

由 假装没事ソ 提交于 2019-11-28 11:59:01
i need to find in sakila data base the longest rental period of a movie. i have trie this: SELECT DISTINCT customer.first_name FROM rental, customer WHERE rental.customer_id = customer.customer_id GROUP BY rental.rental_id HAVING ( rental.return_date - rental.rental_date ) =( SELECT MAX(countRental) FROM ( SELECT ( rental.return_date - rental.rental_date ) AS countRental FROM rental, customer GROUP BY rental.rental_id ) AS t1 ) but i am getting the error: 1054 - Unknown column 'rental.return_date' in 'having clause' does anybody know why? i have used a column that's supposed to be the

HOW to use HAVING COUNT(*) with hibernate

由 那年仲夏 提交于 2019-11-28 11:09:07
I need to create a query and I need COUNT(*) and HAVING COUNT(*) = x . I'm using a work around that uses the CustomProjection class, that I downloaded somewhere. This is the SQL that I try to achieve: select count(*) as y0_, this_.ensayo_id as y1_ from Repeticiones this_ inner join Lineas linea1_ on this_.linea_id=linea1_.id where this_.pesoKGHA>0.0 and this_.nroRepeticion=1 and linea1_.id in (18,24) group by this_.ensayo_id having count(*) = 2 This is the code, where I use the Projection Hibernate class: critRepeticion.setProjection(Projections.projectionList() .add( Projections.groupProperty

use mysql SUM() in a WHERE clause

由 五迷三道 提交于 2019-11-28 07:57:55
suppose I have this table id | cash 1 200 2 301 3 101 4 700 and I want to return the first row in which the sum of all the previous cash is greater than a certain value: So for instance, if I want to return the first row in which the sum of all the previous cash is greater than 500, is should return to row 3 How do I do this using mysql statement? using WHERE SUM(cash) > 500 doesn't work You can only use aggregates for comparison in the HAVING clause: GROUP BY ... HAVING SUM(cash) > 500 The HAVING clause requires you to define a GROUP BY clause. To get the first row where the sum of all the

HOW to use HAVING COUNT(*) with hibernate

由 主宰稳场 提交于 2019-11-27 05:58:01
问题 I need to create a query and I need COUNT(*) and HAVING COUNT(*) = x . I'm using a work around that uses the CustomProjection class, that I downloaded somewhere. This is the SQL that I try to achieve: select count(*) as y0_, this_.ensayo_id as y1_ from Repeticiones this_ inner join Lineas linea1_ on this_.linea_id=linea1_.id where this_.pesoKGHA>0.0 and this_.nroRepeticion=1 and linea1_.id in (18,24) group by this_.ensayo_id having count(*) = 2 This is the code, where I use the Projection

use mysql SUM() in a WHERE clause

由 天大地大妈咪最大 提交于 2019-11-27 04:29:29
问题 suppose I have this table id | cash 1 200 2 301 3 101 4 700 and I want to return the first row in which the sum of all the previous cash is greater than a certain value: So for instance, if I want to return the first row in which the sum of all the previous cash is greater than 500, is should return to row 3 How do I do this using mysql statement? using WHERE SUM(cash) > 500 doesn't work 回答1: You can only use aggregates for comparison in the HAVING clause: GROUP BY ... HAVING SUM(cash) > 500

WHERE vs HAVING

由 妖精的绣舞 提交于 2019-11-26 03:14:25
问题 Why do you need to place columns you create yourself (for example select 1 as \"number\" ) after HAVING and not WHERE in MySQL? And are there any downsides instead of doing WHERE 1 (writing the whole definition instead of a column name)? 回答1: Why is it that you need to place columns you create yourself (for example "select 1 as number") after HAVING and not WHERE in MySQL? WHERE is applied before GROUP BY , HAVING is applied after (and can filter on aggregates). In general, you can reference
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