Php MySQL查询首先返回大多数匹配项

我正在处理一个查询，它应该通过搜索的关键字返回大多数匹配项的数组或标题。甚至不确定查询是否是它应该是的，但它可以工作。别担心打针，我以后再做。这里的目标是返回结果最多的行的标题

   $query = "SELECT * ";
   $query .= "FROM user_rec ";
   $query .= "WHERE sast1 = '" . $mekletie[0] . "'";
   $query .= "OR sast1 = '" . $mekletie[1] . "'";
   $query .= "OR sast1 = '" . $mekletie[2] . "'";

/*NOTE - the query isn't complete yet, it will run sast2 next and so on until sast5 
 and compare all the same searched keywords */

$result = mysqli_query($connection, $query);
    if (!$result) {
       die ("DB query nedarbojas");
    }

---

您似乎想这样做：

$query = 
"SELECT * 
FROM
  (SELECT
    u.*,
    ( sast1 = '" . $mekletie[0] . "' +
      sast1 = '" . $mekletie[1] . "' +
      sast1 = '" . $mekletie[2] . "' +
      sast2 = '" . $mekletie[0] . "' +
      sast2 = '" . $mekletie[1] . "' +
      sast2 = '" . $mekletie[2] . "' ) AS numberofmatches
  FROM user_rec ) m
WHERE
  m.numberofmatches > 0
ORDER BY
  m.numberofmatches DESC";

但是它不起作用，因为如果

sast1

匹配

$mekletie[0]

，那么它就不会匹配

$mekletie[1]

，除非它们是相同的，但从一开始就不是很有用

[编辑] 我看到你的评论，说还有一个stat2。在这种情况下，您可以使用这种方法。但是，如果您为用户存储关键字，您可能需要考虑具有关键字的单独表，您可以这样加入：

$query = 
"SELECT
  u.*
FROM
  user_rec u
  -- Inner join implicitly excludes users that don't have any matching tag.
  INNER JOIN user_tag t ON t.userid = u.userid
WHERE
  t.tag IN ('" .$mekletie[0] . "', '" . $mekletie[1] . "', '" . $mekletie[2] . "')
GROUP BY
  -- Mysql, unlike other databases, allows you to specify only the id 
  -- in group by when actually returning other fields as well.
  u.userid
ORDER BY
  -- You can return the count, or use it just for sorting.
  COUNT(t.userid) DESC"

---

目前还不清楚问题出在哪里，也不清楚您试图实现什么。如果$mekletie中的任何一个匹配sast1，我们将跳转到SAST2并继续查询