PHP PDO fetchOject 的问题

Question

当我将 PHP PDO fetchOject 与下面的查询一起使用时，这似乎是一个错误或问题，

查询：

SELECT 
    p.*,
    t.*

FROM root_pages AS p

LEFT JOIN root_templates AS t
ON p.tmp_id = t.tmp_id

WHERE p.pg_url = ?
AND ? IS NOT NULL

OR p.pg_url = ? 
AND p.pg_hide != ?

从 PHP PDO 数据库类调用，

$page = $this->database->fetch_object($sql,array(
            $pg_url,
            NULL,
            $pg_url,
            1
        ));

结果：

SQLSTATE[HY093]：无效的参数数量：绑定变量的数量 与令牌数量不匹配

PHP PDO FetchOject 方法来自 PDO db 类，

# return the current row of a result set as an object
    public function fetch_object($query, $params = array())
    {
        try
        {
            # prepare the query
            $stmt = $this->connection->prepare($query);

            # if $params is not an array, let's make it array with one value of former $params
            if (!is_array($params)) $params = array($params);

            # execute the query
            $stmt->execute($params);

            # return the result
            return $stmt->fetchObject();
            //return $stmt->fetch(PDO::FETCH_OBJ);
        }
        catch (PDOException $e) 
        {
            # call the get_error function
            $this->get_error($e);
        }
    }

只有我调用这样的方法才可以，

$page = $this->database->fetch_object($sql,array(
            $pg_url,
            1,
            $pg_url,
            1
        ));

但是当我使用 phpMyAdmin 测试下面的查询之一时，我可以获得结果，没有任何错误，

SELECT 
    p.*,
    t.*

FROM root_pages AS p

LEFT JOIN root_templates AS t
ON p.tmp_id = t.tmp_id

WHERE p.pg_url = 'exhibition sample 6' 
AND '1' IS NOT NULL

OR p.pg_url = 'exhibition sample 6' 
AND p.pg_hide != '1'

或者

SELECT 
    p.*,
    t.*

FROM root_pages AS p

LEFT JOIN root_templates AS t
ON p.tmp_id = t.tmp_id

WHERE p.pg_url = 'exhibition sample 6' 
AND NULL IS NOT NULL

OR p.pg_url = 'exhibition sample 6' 
AND p.pg_hide != '1'

任何想法我在使用时错过了fetchOject？

编辑：

$sql ="
SELECT 
    p.*,
    t.*

FROM root_pages AS p

LEFT JOIN root_templates AS t
ON p.tmp_id = t.tmp_id


WHERE p.pg_url = 'exhibition sample 6' 
AND ? IS NOT NULL

OR p.pg_url = 'exhibition sample 6' 
AND p.pg_hide != '1'
";

没有错误，

$item = $connection->fetch_assoc($sql,1);

$item = $connection->fetch_assoc($sql,NULL);

fetch_assoc 方法

    # fetch a single row of result as an array ( =  one dimensional array)
public function fetch_assoc($query, $params = array())
{
    try
    {
        # prepare the query
        $stmt = $this->connection->prepare($query);

        # if $params is not an array, let's make it array with one value of former $params
        if (!is_array($params)) $params = array($params);

        # execute the query
        $stmt->execute($params);

        # return the result
        return $stmt->fetch();
    }
    catch (PDOException $e) 
    {
        # call the get_error function
        $this->get_error($e);
    }


}

Answer 1

您尝试执行的操作（将 null 作为参数传递给 execute）是不可能的。正如文档所述：

输入参数

值的数组，其元素数量与绑定参数的数量相同
在正在执行的 SQL 语句中。 所有值都被视为
PDO::PARAM_STR。

如果要传入 null，则必须将参数与

$stmt->bindValue(1, null, PDO::PARAM_NULL);

命名参数绑定或使用等效语法。