blocks|key|328152|text|小组成员将确保汽车具有所需的两种功能。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|328153|select+c.id,+c.name
++++from+cars+c
++++++++inner+join+options+o
++++++++++++on+c.id+=+o.id_car
++++++++++++++++and+o.option+in+('ABS','ESP')
++++where+c.age+=+2
++++group+by+c.id,+c.name
++++having+count(distinct+o.option)+=+2|code-block|syntax|javascript|328154|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The group by/having will ensure that the car has both of the desired features.

<pre><code>select c.id, c.name
 from cars c
 inner join options o
 on c.id = o.id_car
 and o.option in ('ABS','ESP')
 where c.age = 2
 group by c.id, c.name
 having count(distinct o.option) = 2
</code></pre>

blocks|key|349532|text|SELECT+*+FROM+CARS+WHERE+id+IN+
(SELECT+id_car+FROM+OPTIONS+WHERE+GROUP_CONCAT(option)+='ABS,ESP'+
GROUP+BY+id_car)+
WHERE+age+='2+years'+GROUP+BY+CARS.name|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|349533|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>SELECT * FROM CARS WHERE id IN 
(SELECT id_car FROM OPTIONS WHERE GROUP_CONCAT(option) ='ABS,ESP' 
GROUP BY id_car) 
WHERE age ='2 years' GROUP BY CARS.name
</code></pre>

blocks|key|328085|text|您需要在OPTIONS表上使用一个联接：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|328086|SELECT+c.*+FROM+CARS+c
JOIN+OPTIONS+o+ON+o.id_car=c.id+AND+o.option='ABS'
JOIN+OPTIONS+o2+ON+o2.id_car=c.id+AND+o2.option='ESP'
WHERE+c.age+>=+2|code-block|syntax|javascript|328087|或子查询：|328088|SELECT+c.*+FROM+CARS+c
WHERE+c.age+>=+2
AND+c.id+IN+(+SELECT+o.id_car+FROM+OPTIONS+o+WHERE+o.id_car=c.id+AND+o.option='ABS')
AND+c.id+IN+(+SELECT+o.id_car+FROM+OPTIONS+o+WHERE+o.id_car=c.id+AND+o.option='ESP')|328089|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You need to use a join on the OPTIONS table:

<pre><code>SELECT c.* FROM CARS c
JOIN OPTIONS o ON o.id_car=c.id AND o.option='ABS'
JOIN OPTIONS o2 ON o2.id_car=c.id AND o2.option='ESP'
WHERE c.age &gt;= 2
</code></pre>

Or subqueries:

<pre><code>SELECT c.* FROM CARS c
WHERE c.age &gt;= 2
AND c.id IN ( SELECT o.id_car FROM OPTIONS o WHERE o.id_car=c.id AND o.option='ABS')
AND c.id IN ( SELECT o.id_car FROM OPTIONS o WHERE o.id_car=c.id AND o.option='ESP')
</code></pre>

blocks|key|349518|text|SELECT+*+FROM+CARS+LEFT+JOIN+`OPTIONS`+ON+`OPTIONS`.`id_car`+=+`CARS`.`id`+WHERE+`OPTIONS`.`option`+LIKE+'%25ABS%25'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|349519|如果你想要更多，你必须让PHP编写每个选项的SQL添加，这样你的最终结果将是|unstyled|349520|SELECT+*+FROM+CARS+LEFT+JOIN+`OPTIONS`+ON+`OPTIONS`.`id_car`+=+`CARS`.`id`+WHERE+(`OPTIONS`.`option`+LIKE+'%25ABS%25'+OR++`OPTIONS`.`option`+LIKE+'%25ESP%25')|349521|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>SELECT * FROM CARS LEFT JOIN `OPTIONS` ON `OPTIONS`.`id_car` = `CARS`.`id` WHERE `OPTIONS`.`option` LIKE '%ABS%'
</code></pre>

Will do ABS if you wanting more you will have to have PHP write the SQL additions per Options so your End result would be

<pre><code>SELECT * FROM CARS LEFT JOIN `OPTIONS` ON `OPTIONS`.`id_car` = `CARS`.`id` WHERE (`OPTIONS`.`option` LIKE '%ABS%' OR `OPTIONS`.`option` LIKE '%ESP%')
</code></pre>

blocks|key|328128|text|我建议您将'+age‘列更改为'date_created'，或者将其保留为age，只使用一个整数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|328129|如果将其用作整数，只需执行以下操作：|328130|SELECT+c.id,+c.car
+++FROM+cars+c
+++INNER+JOIN+options+o
+++ON+o.id_car+=+c.id
+++WHERE+c.age+>=+2|code-block|syntax|javascript|328131|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I suggest you change the 'age' column, either to 'date_created', or just leave it as age and only use an integer.

If you use as an integer, just do this:

<pre><code>SELECT c.id, c.car
 FROM cars c
 INNER JOIN options o
 ON o.id_car = c.id
 WHERE c.age &gt;= 2
</code></pre>

blocks|key|349553|text|我也试着用这种方法把福特1和3的车还给我.|type|unstyled|depth|inlineStyleRanges|entityRanges|data|349554|我只想要3辆带腹肌和(尤指)的福特。|349555|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

I also tried this way and returns me both of ford 1 and 3....

I only want 3 Ford the with abs and esp

blocks|key|2467803|text|试一试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2467804|++Select+new_car.name++
++++From+(select+*+from+cars+where+year=2)+as+new_car+
++++inner+join+option+as+op+
++++on+(op.id_car+=+new_car.id+AND+(op.option+=+`ABS`+OR+op.option+=+`ESP`))|code-block|syntax|javascript|2467805|:)|2467806|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Try it:: 

<pre><code> Select new_car.name 
 From (select * from cars where year=2) as new_car 
 inner join option as op 
 on (op.id_car = new_car.id AND (op.option = `ABS` OR op.option = `ESP`))
</code></pre>

:)

I have 2 MySQL tables like in the example below:
CARS
<pre><code>Id CAR NAME AGE

1 Ford 2 years
2 AUDI 1 years 
3 Ford 2 years
</code></pre>
OPTIONS
<pre><code>Id id_car option

1 1 ESP
2 2 ABS
3 3 ABS
4 3 ESP
</code></pre>
And I need to select all cars of 2 years old which have ABS AND ESP.
So it should return in this example: 3 Ford

Select MySQL data with where clause in 2 tables

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有两个MySQL表，如下例所示：汽车Id     CAR NAME          AGE1   Ford        2 years2   AUDI        1 years 3   Ford        2 years选项Id  id_car   option1    1      ESP2    2          ABS3    3          ABS4    3   

问在两个表中选择具有where子句的MySQL数据
EN

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问在两个表中选择具有where子句的MySQL数据EN

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问在两个表中选择具有where子句的MySQL数据
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