Question

当我将PHP PDO fetchOject与下面的查询一起使用时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 db类调用

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

结果：

SQLSTATE [HY093]：参数号无效：绑定变量数与令牌数量不匹配

来自PDO db类的

PHP PDO FetchOject方法

# 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);
        }
    }

如果我调用这样的方法

，那就没关系了

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

但是，当我使用phpMyAdmin测试以下查询之一时，我可以毫无错误地获得结果，

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

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'

使用fetchOject时我错过了哪些想法？

修改

$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);

或

$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

您要做的事情（将null作为参数传递给execute）是不可能的。正如documentation所述：

input_parameters

具有与绑定参数一样多的元素的值数组   在正在执行的SQL语句中。 所有值都被视为   PDO :: PARAM_STR

如果要传递null，则必须使用
绑定参数
$stmt->bindValue(1, null, PDO::PARAM_NULL);

或使用命名参数的等效语法。

PHP PDO fetchOject的问题

1 个答案: