以下场景如何获取记录

Question

我有一个如下表：

node_name              id            term_name
----------------------------------------------
test1                  001           physics 
test1                  001           maths    
test1                  001           chemistry    
test2                  002           physics    
test2                  002           maths

给定术语名称的组合，我想找到 id 集仅包含给定术语名称的所有行。

例如，给定术语名称physical & maths 我的输出应该如下所示

node_name              id            term_name
----------------------------------------------
test2                  002           physics   
test2                  002           maths

Id set 001 还包含 chemistry 这就是为什么不应该包含它。

Answer 1

您的问题：获取不存在具有相同 id 但存在其他 term_names 的其他行的所有行

SELECT * FROM <table> x WHERE
  term_name IN ('physics','maths') AND
  NOT EXISTS (SELECT * FROM <table> WHERE id=x.id AND term_name NOT IN ('physics','maths'))

Answer 2

首先，您需要解析查询以转换“&”到 SQL 'OR' 运算符
在 PHP 中：

//Parse the query
        $arr = explode('&',$query);
    $where = '';
//get the term count
    $count = count($arr);
    foreach($arr as $value){
    $where .= "term_name = '" . $value . "' OR";
    }
    //Remove last or
    $where = rtrim($where,'OR');

然后：
最后使用 L

"select node_name ,count(1) as Total from my table where $where
group by node_name
having Total =" . $count

：

您的查询必须采用以下格式：

select x,count(1) as total from mytable where field1 = 'term1' or field1 = 'term2' having total = 2

Answer 3

一种可能的方法是：

select id, node_name 
from nodes join 
  (select id, 
       count(*) 
from nodes
  where node_name in ('physics','math')
group by id
having count(*) = 2 // this is set when generating the query ) as eligible_nodes
  on nodes.id = eligible_nodes.id