严格标准:资源ID#73作为偏移量，转换为整数 [英] Strict Standards: Resource ID#73 used as offset, casting to integer

问题描述

我使用此PHP函数获取MySql结果:

i fetch MySql result using this PHP function:

function fetcharray ($query_id)

    {
        if(!$query_id)
        {
            $query_id = $this->query_res;
        }

        if($query_id)

        {
            $this->q_array[$query_id] = @mysql_fetch_array($query_id,MYSQL_ASSOC); // LINE 124
            return $this->q_array[$query_id]; //LINE 125
        }

        else

        {
            return false;
        }
    }

现在，我转到PHP 5.5，然后看到此错误:

Now, I move to PHP 5.5 and see This error:

Strict Standards: Resource ID#73 used as offset, casting to integer (73) in domain.com/includes/functions/db.php on line 124

Strict Standards: Resource ID#73 used as offset, casting to integer (73) in domain.com/includes/functions/db.php on line 125

如何解决此错误?

推荐答案

您正在使用资源，它是 not 整数.即使在检查时看到一个数字，它实际上也不是真正的整数.这是与您的MySQL连接关联的资源ID.

You're using a resource, which is not an integer, as an integer. Even though you see a number when you inspect it, it actually is not a true integer. It's a resource ID that is associated with your MySQL connection.

如果您想将其用作整数，我认为，则可以在使用它之前将其转换为整数(我从未尝试过).

If you want to use it as an integer I think you can cast it to an integer before using it (I never actually tried it).

    if($query_id)
    {
        $id = (int) $query_id;

        $this->q_array[$id] = @mysql_fetch_array($query_id,MYSQL_ASSOC); // LINE 124
        return $this->q_array[$id]; //LINE 125
    }

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