Question

有人可以看看这个并告诉我它为什么不起作用。这是我正在研究的课程的一部分。插入查询工作正常，如果我调用$ char-＆gt; db-＆gt; insert_id;从它外面工作（$ char是班级）

function createChar($charName, $charRace, $charClass) {
    //put userId, charName and charRace into database
    $sql="insert into characters (userID, name, race, class) values ('$this->userID', '$charName', '$charRace', '$charClass') ";
    $result=$this->db->query($sql);
    if($this->db->affected_rows == '1')
        {
            return '1';
        }
        else
        {
            return '0';
        }

    //get last insert id     
    $this->charID=$this->db->insert_id;

}

Answer 1

尝试在if / else之前移动$this->charID = $this->db->insert_id;，因为两者都会在到达insert_id行之前返回。

function createChar($charName, $charRace, $charClass) {
    //put userId, charName and charRace into database
    $sql = "insert into characters (userID, name, race, class) values ('$this->userID', '$charName', '$charRace', '$charClass') ";
    $result = $this->db->query($sql);

    //get last insert id     
    $this->charID = $this->db->insert_id;

    if($this->db->affected_rows == '1')
    {
        return '1';
    }
    else
    {
        return '0';
    }
}

Answer 2

由于return语句在返回发生之前尝试将其放在部件上，因此无法达到insert_id的赋值。

Answer 3

请尝试以下。

正确引用值：

  $sql="insert into characters (userID, name, race, class) values 
    ('".$this->userID."', '".$charName."', '".$charRace."','".$charClass."') ";
  $result=$this->db->query($sql);
  $this->charID=$this->db->insert_id;

mysqli insert_id无效

3 个答案: