我是PHP新手,我正在尝试编写一个函数,该函数将从reviews
MySQL表中获取所有结果。我已经发布了我尝试过的内容,但它说我在 101 行有一个问题,这是"$sql = 'SELECT * FROM reviews WHERE id = "'.$review_id.'";'"
,我看不出问题是什么,有人可以帮忙。IM 使用 Apigee 测试 API
/*reviews*/
function reviews_get() {
$this->load->database();
$sql = 'SELECT * FROM reviews;';
$query = $this->db->query($sql);
$data = $query->result();
$this->response($data, 200);
}
function review_get($review_id) {
$this->load->database();
$sql = 'SELECT * FROM reviews WHERE id = "'.$review_id.'";';
$query = $this->db->query($sql);
$data = $query->row();
$this->response($data, 200);
}
请从查询字符串的末尾删除";",您的问题将得到解决
你的 SQL 语句是错误的。它应该是
$sql = "SELECT * FROM reviews WHERE id = ".$review_id;