php - 准备好的 PHP SQL 语句在没有 GROUP BY 子句的情况下返回 false

Question

我有一个准备好的声明:

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);
}

这里是查询:

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，尽管我尝试在 phpMyAdmin 上运行查询并且它运行良好，但准备好的语句返回 false。 如果我在没有错误保护的情况下执行准备好的语句，我会得到错误:bind_param() on boolean。

如果我在查询末尾添加:

GROUP BY q.question_type

然后一切正常。但是，这不是我想要的，因为它为计数返回 null 而不是 0，而且我也不明白没有 GROUP BY 它是如何工作的。

Answer 1

使用不带group by的聚合函数是不规范的，在最新版本的mysql(5.7)中是不允许的 检查你的版本是否正确，无论如何尝试对 group by 中未提及的列使用曾经聚合的函数 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 = ?)"