通過一個查詢選擇多個記錄

Question

請給我建議，如何構造選擇查詢。 我有表table與字段type和obj_id 。 我想選擇與下一個數組一致的所有記錄：

$arr = array(
0 => array('type' => 1, 'obj_id' => 5),
1 => array('type' => 3, 'obj_id' => 15),
2 => array('type' => 4, 'obj_id' => 14),
3 => array('type' => 12, 'obj_id' => 17),
);

我想通過一個查詢選擇所需的行，這是真的嗎？ 有點像

select * from `table` where type in (1,3,4,12) and obj_id in (5,15,14,17)

但是此查詢還會返回類型= 3和obj_id = 14的記錄，例如類型= 1和obj_id =17。ps主持人，請修正我的標題，我不知道如何描述我的問題。

update: array $arr could contain more than 500 elems.

Answer 1

據我知道你不能用in了這一點，但你必須退回到這樣的事情

select * from `table` where type=1 and obj_id=5 or type=3 and obj_id=15

Answer 2

如果我理解正確，則可以使用UNION或OR。 使用UNION：

select * from `table` where type = 1 and obj_id = 5
UNION ALL
select * from `table` where type = 3 and obj_id = 15
UNION ALL
select * from `table` where type = 4 and obj_id = 14
UNION ALL
select * from `table` where type = 12 and obj_id = 17

如果需要刪除重復項，請從UNION ALL刪除ALL 。

使用OR：

select * from `table` 
 where (type = 1 and obj_id = 5)
    OR (type = 3 and obj_id = 15)
    OR (type = 4 and obj_id = 14)
    OR (type = 12 and obj_id = 17)

括號很重要 -它們表示需要滿足內部的所有條件。

我建議使用UNION-OR因性能不佳而臭名昭著。

Answer 3

$sql = 'select * from `table` where ';

foreach ($arr as $index => $conditions){
    $sql .= $or.' (';
    foreach($condicions as $key => $value){
       $sql .= $concat. "$key = $value";
       $concat = ' and ';
    }
    $sql .= ')';
    $or = ' or ';
}

應將$ sql形成為：

select * from table where 
  (type = 1 and obj_id = 5) or 
  (type = 3 and obj_id = 15) or 
  (type = 4 and obj_id = 14) or 
  (type = 12 and obj_id = 17); 

Answer 4

如果條件數組變大，則可能需要創建索引臨時表，在table (type, obj_id)上具有復合索引table (type, obj_id)然后執行INNER JOIN 。