Question

好的我正在运行查询而CodeIgniter正在给我这个错误：

错误号码：1064

您的SQL语法有错误;检查手册   对应于您的MySQL服务器版本，以便使用正确的语法   靠近''第4行

SELECT username FROM friend_request JOIN user ON user_id = friend   WHERE（朋友='8'和发件人=

我确实知道，它并没有抓住发件人，但是这个＆gt; session-＆gt; userdata（'user_id'）回声是4（这是我的user_id）... 那怎么会出现什么呢？

$this->db->query('SELECT username
    FROM friend_request
    JOIN user ON user_id = friend
    WHERE (friend = ? AND sender = ?)
    OR (friend = ? AND sender = ?)', 
        $user->row()->user_id, 
        $this->session->userdata('user_id'), 
        $this->session->userdata('user_id'), 
        $user->row()->user_id);

if($this->db->num_rows() > 0) {
    $this->errors->set_error('You either have a pending request from '.ucfirst($this->db->row()->username).' 
        or you have already requested their friendship!');
    return false;
}

我尝试将所有内容放在一行，因为我不知道输入是否会破坏任何内容，但它仍然无法解决任何问题。

Answer 1

您收到错误，因为您应该在查询结果上使用num_rows，而不是在数据库上使用$query = $this->db->query( 'SELECT username FROM friend_request JOIN user ON user_id = friend WHERE (friend = ? AND sender = ?) OR (friend = ? AND sender = ?)', array( $user->row()->user_id, $this->session->userdata('user_id'), $this->session->userdata('user_id'), $user->row()->user_id ) ); if ($query->num_rows() > 0) { $this->errors->set_error('You either have a pending request from '.ucfirst($this->db->row()->username).' or you have already requested their friendship!'); return false; }。此外，您查询绑定是错误的。

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MySQL - CodeIgniter无法插入查询参数

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