php - SQL:选择没有任何行具有特定列值的 ID

Question

我想选择不包含 VAL = 'current' 的任何行的不同 ID(与多行关联)。

例如，在这样的表格中:

PK | ID | VAL 
-------------
 1 | 23 | deleted
 2 | 23 | deleted
 3 | 23 | deleted
 4 | 45 | current
 5 | 45 | deleted
 6 | 45 | deleted
...|................

我希望它返回 ID 23，因为它没有 VAL='current' 的行。请注意，在此表中，主键 (PK) 是唯一的，但 ID 不是(因此需要使用 DISTINCT 或 GROUP BY)。

这是我在 PHP 中的内容:

$conn = someConnect("");

// returns ids associated with the number of rows they have where VAL != 'current'
$sql = "SELECT id, count(*) FROM table WHERE val != 'current' GROUP BY id"

$stid = oci_parse($conn, $sql);
oci_execute($stid);

oci_fetch_all($stid, $arr, OCI_FETCHSTATEMENT_BY_ROW);

foreach ($arr as $elm) {
   $id = key($elm);
   $non_current_count = $elm[$id];

   // counts the number of rows associated with the id, which includes VAL = 'current' rows
   $sql2 = "SELECT count(*) FROM table WHERE id = $id";

   $stid2 = oci_parse($conn, $sql2);
   oci_execute($stid2);
   $total_count = oci_fetch_array...
   if ($total_count != $non_current_count) {
      $output[] = $id;
   } 
   ...
}

oci_close($conn);

这就是它的一般要点。如您所见，需要两条 SQL 语句才能完成此任务。有更短的方法吗？

Answer 1

SELECT DISTINCT id
FROM table
WHERE id NOT IN (SELECT id
                 FROM table
                 WHERE val = 'current')

或:

SELECT a.id
FROM table a
LEFT JOIN table b ON a.id = b.id AND b.val = 'current'
WHERE b.id IS NULL