Question

我有candidates，candidate-skills和skills的MySql表。哪种方法是选择具备所有技能的候选人

我尝试使用以下查询。但这不准确。

Select `t`.*, GROUP_CONCAT(DISTINCT(s.name)) as skills, 
GROUP_CONCAT(DISTINCT(s.id)) as skill_ids 
FROM `candidates` `t` 
LEFT JOIN `candidate-skills` `cs` ON `t`.`id` = `cs`.`can_id` 
LEFT JOIN `skills` `s` ON `cs`.`skill_id` = `s`.`id` 
where s.id in ('8','10') 
GROUP BY `t`.`id` 
ORDER BY `t`.`id` desc

我想要的两点是：

应该显示所有技能（当评论其中的条件时）
将显示所有技能的记录。（具有一种技能的记录也显示为我使用数组中的位置）

我正在使用codeigniter框架。

http://sqlfiddle.com/#!9/b75c3/49

Answer 1

而不是使用having子句。

select `t`.*, GROUP_CONCAT(DISTINCT(s.name)) as skills, 
GROUP_CONCAT(DISTINCT(s.id)) as skill_ids
FROM `candidates` `t` 
LEFT JOIN `candidate-skills` `cs` ON `t`.`id` = `cs`.`can_id` 
LEFT JOIN `skills` `s` ON `cs`.`skill_id` = `s`.`id` 
GROUP BY `t`.`id` 
having find_in_set ('8', skill_ids) and find_in_set ('10', skill_ids) 
ORDER BY `t`.`id` desc

Codeigniter中的

//take all skill ids in array
$ids=['8','10'];
$this->db->select("t.*");
$this->db->select("GROUP_CONCAT(DISTINCT(s.name)) as skills");
$this->db->select("GROUP_CONCAT(DISTINCT(s.id)) as skill_ids");
$this->db->from("candidates t");
$this->db->join("candidate-skills cs","t.id = cs.can_id");
$this->db->join("skills s","cs.skill_id = s.id");
$this->db->group_by("t.id");
foreach ($ids as $id) {
$this->db->having("find_in_set ('$id', skill_ids)");
}
$this->db->order_by("t.id","desc");
$query=$this->db->get();
$candidates=$query->result();

Answer 2

最灵活的方式是使用多个JOIN; GROUP_CONCAT和逗号分隔的列表被视为反模式，如果连接不是以正确的顺序完成，则它可能不起作用（技能组1,2,5被认为与1,5,2不同）。 / p>

SELECT c.* FROM candidates AS c
    JOIN candidateskills AS cs ON (cs.cand_id = c.id)
    JOIN skills AS sk1 ON (cs.skill_id = sk1.id)
    JOIN skills AS sk2 ON (cs.skill_id = sk2.id)
    ...other sk(N)...
    WHERE (sk1.skill = 'waterskiing')
      AND (sk2.skill = 'snowboarding')
      ...
    ;

这可以轻松定制技能，例如，如果每种技能都具备技能水平，并且您需要滑雪技能达到或超过5级。这种灵活性与GROUP_CONCAT有关。

但是对于简单的匹配，你可以通过选择你想要的技能并计算它们来更快地完成：

SELECT c.* FROM candidates AS c
    JOIN candidateskills AS cs ON (cs.cand_id = c.id)
    WHERE cs.skill_id IN (1, 7, 24, 19, 115)
GROUP BY c.id
HAVING COUNT(1) = 5;

（在更合适的SQL中，您需要明确指出c的所有字段而不是＆＃34; c。*＆＃34;，并在GROUP BY子句中重复它们。更聪明的RDBMS服务器不会只要你按照c的主键进行分组，MySQL 当前无论如何都不关心，但在严格模式下，它会这样做。

对于每项技能，您都会对技能进行单一，快速的查询，以检索其ID并汇总上述查询。

或者，只要您具有完全匹配的技能，您就可以在单个更大的查询中执行此操作：

SELECT c.* FROM candidates AS c
    JOIN candidateskills AS cs ON (cs.cand_id = c.id)
    JOIN skills AS s ON (cs.skill_id = s.id)
    WHERE s.skill IN ('javascript', 'html5', 'php')
GROUP BY c.id
HAVING COUNT(1) = 3;

因为你想在PHP中使用它：

$skills = array('javascript', 'html5', 'php');

$skno   = count($skills);
$set    = implode(',', array_fill('?', $skno));
$params = $skills;
$params[] = $skno;

$query = "SELECT c.* FROM candidates AS c
    JOIN candidateskills AS cs ON (cs.cand_id = c.id)
    JOIN skills AS s ON (cs.skill_id = s.id)
    WHERE s.skill IN ({$set})
GROUP BY c.id 
HAVING COUNT(1) = ?";

$stmt = $db->prepare($query);
$stmt->execute($params);
while ($candidate = $stmt->fetch(PDO::FETCH_ASSOC)) {
    ...
}

Answer 3

也许这个

        Console.WriteLine("Define Array Size? ");
        int number = Convert.ToInt32(Console.ReadLine());

        Console.WriteLine("Enter numbers:\n");
        int[] arr = new int[number];

        for (int i = 0; i < number; i++)
        {
            arr[i] = Convert.ToInt32(Console.ReadLine());
        }

        for (int i = 0; i < arr.Length; i++ )
        {
            Console.WriteLine("Array Index: "+i + " AND Array Item: " + arr[i].ToString());
        }

        Console.ReadKey();

使用GROUP_CONCAT

3 个答案: