比较两个表后如何输出匹配项? [英] How do I output a match after comparing two tables?

问题描述

我想比较两个表. 但是我不知道该怎么办，因为我不够好. 我想知道如何编写查询

I want to compare two tables. But I don't know what to do because I'm not good enough. I want to know how to write query

1号桌

id  chart_num       chart_name            visit            card_amount_received
1        4            user1             2020-04-05               1000                        
2        5            user2             2020-05-05               1000           
3        5            user2             2020-05-05               1000                        
4        5            user2             2020-06-05               1000              
5        6            user3             2020-07-05               1000        
6        6            user3             2020-08-05               1000                 
7        7            user4             2020-09-05               1000                    
8        7            user4             2020-09-05               1000                    

2号表

id           card_date             advenced_amount
1        2020-04-05 17:28:00            1000        
2        2020-05-05 12:12:12            2000   
10       2020-11-05 12:12:12            5000              

想要的结果

条件和顺序如下.

1. 1号表，对每个chart_num具有相同值的列求和并访问.

1. no.1_table, sum a column with the same value of each chart_num and visit.

比较1号结果值的访问和card_count_received与2号表的card_date和advenced_amount值.

Compare the visit and card_count_received of the result value of number 1 with the card_date and the advenced_amount values of no.2_table.

如果chart_num的值和no.1_table的访问值与no.2_table的card_date和no.2_table的added_count值相同，则否则输出错误并进行空处理.

If the value of chart_num and visit of no.1_table is the same as the card_date of no.2_table and the added_count value of no.2_table, ok otherwise output error with null processing.

如何创建查询语句?

数据库技能不足.请给我答复.

Database skills are insufficient. Please give me a reply.

推荐答案

1. 1号表，对每个chart_num具有相同值的列求和并访问.

为此，您要 group by chart_num并访问.任何具有相同chart_num和visit的行将在结果中显示为单行.然后，您可以 sum 收到的金额，这将为一个小组.

For this you want to group by the chart_num and visit. Any rows with the same chart_num and visit will appear as a single row in the result. Then you can sum the amount received, this will add up all the values for a group.

select
  chart_num,
  visit,
  sum(card_amount_received) as card_amount_received
from table1
group by chart_num, visit

chart_name是个问题.您无法显示它，因为它不是group by的一部分.这是一个字符串，因此将其与sum或count之类的函数进行聚合没有任何意义.尽管数据中的chart_num具有相同的chart_name，但这并不能保证.

chart_name is a problem. You can't display it as it is not part of the group by. It's a string so it doesn't make sense to aggregate it with functions like sum or count. Though in the data chart_num has the same chart_name, that is not guaranteed.

一种解决方案是使用 group_concat 将组中的每个名称连接在一起.每个组只能有一个名字.

One solution is to use group_concat to concatenate each name in a group together. There should only be one name per group.

select
  chart_num,
  visit,
  group_concat(chart_name) as chart_name,
  sum(card_amount_received) as card_amount_received
from table1
group by chart_num, visit

但是，正确的解决方案是修复架构. chart_name是重复的，这是要避免的.而是将图表列移动到自己的表中.然后，要获取图表名称，请加入chart_num.

However, the proper solution is fix the schema. chart_name is duplicated, and that's to be avoided. Instead, move the chart columns into their own table. Then to get the chart name, join on the chart_num.

create table charts (
  id serial primary key,
  name varchar(255) not null
);

insert into charts (id, name) values
  (4, 'user1'), (5, 'user2'), (6, 'user3'), (7, 'user4');

alter table table1 drop column chart_name;

select
  charts.id as chart_num,
  visit,
  charts.name as chart_name,
  sum(card_amount_received) as card_amount_received
from table1
join charts on charts.id = chart_num
group by chart_num, visit

1. 将第1个结果值的访问和card_count_receive与card_date和no.2_table的advanced_amount值进行比较.

我们需要左连接，第二个表将card_date与访问进行匹配. 左联接表示左"表中的所有行(即from表)将始终显示，即使联接表中没有匹配项.

We need a left join with the second table matching the card_date with the visit. A left join means all the rows in the "left" table (ie. the from table) will always appear even if there is no match in the join table.

访问是一个日期. card_date不是日期，而是时间戳.为了匹配它们，我们需要将card_date转换为日期.

visit is a date. card_date is not a date but a timestamp. To match them we'll need to convert card_date to a date.

select
  charts.id as chart_num,
  visit,
  charts.name as chart_name,
  sum(card_amount_received) as card_amount_received,
  table2.card_date,
  table2.advanced_amount as amount
from table1
join charts on charts.id = chart_num
left join table2 on date(table2.chart_date) = visit
group by chart_num, visit

如果chart_num和no.1_table的访问值与no.2_table的card_date和no.2_table的added_count值相同，则否则输出错误并进行空处理.

If the value of chart_num and visit of no.1_table is the same as the card_date of no.2_table and the added_count value of no.2_table, ok otherwise output error with null processing.

我们需要将advanced_amount与sum(card_amount_received)进行比较.如果它们相等:好的.如果不是:错误.在标准SQL中，我们将使用 case ，但是MariaDB具有非标准 if ，它更加紧凑.

We need to compare advanced_amount with sum(card_amount_received). If they're equal: ok. If not: error. In standard SQL we'd use a case, but MariaDB has a non-standard if that is much more compact.

select
  charts.id as chart_num,
  visit,
  charts.name as chart_name,
  sum(card_amount_received) as card_amount_received,
  table2.card_date,
  table2.advanced_amount as amount,
  if(table2.advanced_amount = sum(card_amount_received), 'ok', 'error') as result
from table1
join charts on charts.id = chart_num
left join table2 on date(table2.chart_date) = visit
group by chart_num, visit

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