Question

我有一份准备好的声明：

if ( $statement = $this->connection->prepare("SELECT question_type, count(*) AS `count` FROM (SELECT question.*, left(question_body, locate('between', question_body)-2) AS question_type FROM question) q WHERE (q.question_type = ? AND q.response_value_id = ?)") ) {
    $statement->bind_param("si", $question_type, $response_value_id);
    $statement->execute();
    return $statement->get_result()->fetch_assoc();
} else {
    var_dump($this->db->error);
}

以下是查询：

SELECT question_type, count(*) AS `count` FROM 
(SELECT question.*, left(question_body, locate('between', question_body) - 2) 
AS question_type FROM question) 
q WHERE q.question_type = 'Did you return home' AND q.response_value_id = 4

问题：由于某种原因，虽然我已经尝试在phpMyAdmin上运行查询并且它完美地工作，但是准备好的语句返回false。如果我执行准备好的语句而没有警告错误，我会收到错误：bind_param() on boolean。

如果我添加到查询的末尾：

GROUP BY q.question_type

然后一切正常。但是，这不是我想要的，因为它为计数而不是0返回null，而且我也不明白它如何在没有GROUP BY的情况下工作。

Answer 1

不使用group by的聚合函数的使用，并且不允许在最新版本的mysql（5.7）中使用检查你是否正确版本，无论如何尝试使用集合中没有提到的列的agregated函数例如：

  "SELECT question_type, count(*) AS `count` 
    FROM (
      SELECT question.*, left(question_body, locate('between', question_body)-2) AS question_type 
      FROM question ) q 
  WHERE q.question_type = ? AND q.response_value_id = ?
  GROUP BY question_type)"

或

  "SELECT min(question_type), count(*) AS `count` 
    FROM (
      SELECT question.*, left(question_body, locate('between', question_body)-2) AS question_type 
      FROM question ) q 
  WHERE (q.question_type = ? AND q.response_value_id = ?)"

Answer 2

如果您想按question_type计算，则需要添加GROUP BY q.question_type。否则，在select语句中取count（*）之前取出question_type。试试这个：

SELECT count(*) AS `count` FROM 
(SELECT question.*, left(question_body, locate('between', question_body)-2) 
AS question_type FROM question) 
q WHERE q.question_type = 'Did you return home' AND q.response_value_id = 4

Answer 3

除了显示其他字段（在您的示例中为question_type）之外，每当使用聚合函数时，必须使用GROUP BY。

您可以在使用GROUP BY时计算的字段上使用ISNULL。

准备好的PHP SQL语句在没有GROUP BY子句的情况下返回false

3 个答案: