比较两个表后如何输出匹配?
我想比较两张表.但我不知道该怎么办，因为我不够好.我想知道怎么写查询
no.1_table
id chart_num chart_name 访问 card_amount_received1 4 用户1 2020-04-05 10002 5 用户2 2020-05-05 10003 5 用户2 2020-05-05 10004 5 用户2 2020-06-05 10005 6 用户3 2020-07-05 10006 6 用户3 2020-08-05 10007 7 用户4 2020-09-05 10008 7 用户4 2020-09-05 1000no.2_table
id card_date advenced_amount1 2020-04-05 17:28:00 10002 2020-05-05 12:12:12 200010 2020-11-05 12:12:12 5000想要结果
条件和顺序如下.
no.1_table，将每个chart_num的值相同的列相加，访问.
将数字1的结果值的visit和card_count_received与no.2_table的card_date和advenced_amount值进行比较.
如果no.1_table的chart_num和visit的值与no.2_table的card_date和no.2_table的added_count值相同，则ok，否则输出错误，处理为null.
如何创建查询语句?
数据库技能不足.请给我答复.
解决方案
- no.1_table，将每个chart_num的值相同的一列相加，访问.
为此，您要group bychart_num 并访问.任何具有相同chart_num 和访问的行都将在结果中显示为单行.然后您可以sum 收到的金额，这将添加将一个组的所有值加起来.
选择图表编号，访问，总和(card_amount_received)作为 card_amount_received从表 1按chart_num分组，访问chart_name 是个问题.您无法显示它，因为它不是 group by 的一部分.它是一个字符串，因此将它与 sum 或 count 等函数聚合是没有意义的.虽然在数据中chart_num 具有相同的chart_name，但不能保证.
一种解决方案是使用 group_concat 连接每个在一个组中一起命名.每个组只能有一个名字.
选择图表编号，访问，group_concat(chart_name) 作为chart_name，总和(card_amount_received)作为 card_amount_received从表 1按chart_num分组，访问但是，正确的解决方案是修复架构.chart_name 是重复的，这是要避免的.相反，将图表列移动到它们自己的表中.然后要获取图表名称，请加入 chart_num.
创建表格图表(id 序列主键，名称 varchar(255) 不为空);插入图表(id，name)值(4, 'user1'), (5, 'user2'), (6, 'user3'), (7, 'user4');alter table table1 drop column chart_name;选择chart.id 作为chart_num，访问，chart.name 作为chart_name，总和(card_amount_received)作为 card_amount_received从表 1在charts.id = chart_num 上加入图表按chart_num分组，访问<块引用>
- 将数字1的结果值的visit和card_count_received与no.2_table的card_date和advanced_amount值进行比较.
我们需要一个左连接，第二个表匹配 card_date 和访问.左连接表示左"表中的所有行(即 from 表)将始终出现，即使连接表中没有匹配项.
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.
选择chart.id 作为chart_num，访问，chart.name 作为chart_name，总和(card_amount_received)作为 card_amount_received，table2.card_date,table2.advanced_amount 作为金额从表 1在charts.id = chart_num 上加入图表在日期(table2.chart_date)= 访问左连接表 2按chart_num分组，访问<块引用>
如果no.1_table的chart_num和visit的值与no.2_table的card_date和no.2_table的added_count值相同，则ok，否则输出错误，处理为null.
我们需要比较advanced_amount 和sum(card_amount_received).如果它们相等:好的.如果不是:错误.在标准 SQL 中，我们会使用 case，但是MariaDB 有一个非标准的 if更紧凑.
选择chart.id 作为chart_num，访问，chart.name 作为chart_name，总和(card_amount_received)作为 card_amount_received，table2.card_date,table2.advanced_amount 作为金额，if(table2.advanced_amount = sum(card_amount_received), 'ok', 'error') 结果从表 1在charts.id = chart_num 上加入图表在日期(table2.chart_date)= 访问左连接表 2按chart_num分组，访问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
no.1_table
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
no.2_table
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
want result
The conditions and order are as follows.
no.1_table, sum a column with the same value of each chart_num and visit.
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.
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.
How do I create a query statement?
Database skills are insufficient. Please give me a reply.
解决方案
- no.1_table, sum a column with the same value of each chart_num and visit.
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 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.
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
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
- Compare the visit and card_count_received of the result value of number 1 with the card_date and the advanced_amount values of no.2_table.
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.
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
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.
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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