Php Codegniter-访问数据库时出现内部错误
我使用CodeIgniter访问数据库,但每次运行代码时,它都会收到一个内部错误。这是我的jQuery代码:Php Codegniter-访问数据库时出现内部错误,php,ajax,database,codeigniter,Php,Ajax,Database,Codeigniter,我使用CodeIgniter访问数据库,但每次运行代码时,它都会收到一个内部错误。这是我的jQuery代码: function showCadeiras() { $.ajax({ type: "POST", url: base_url + "api/teacher/getCadeiras", data: {id: localStorage.user_id}, success: function(data) { for(var i = 0; i
function showCadeiras() {
$.ajax({
type: "POST",
url: base_url + "api/teacher/getCadeiras",
data: {id: localStorage.user_id},
success: function(data) {
for(var i = 0; i < data.cadeiras_id.length; i++) {
var cadeira_id = data.cadeiras_id[i].cadeira_id;
$.ajax({
type: "POST",
url: base_url + "api/teacher/getCadeiraInfo",
data: {id: cadeira_id},
success: function(data) {
console.log(data);
},
error: function(data) {
console.log("error");
}
})
}
},
error: function(data) {
console.log(data);
}
});
}
最后,示范代码:
public function getCadeiras($id) {
$this->db->select("cadeira_id");
$this->db->where(array('user_id =' => $id));
$query = $this->db->get('professor_cadeira');
return $query->result_array();
}
public function getCadeiraInfo($id) {
$query = $this->db->get_where('cadeira', array('id =' => $id));
return $query->result_array();
}
每次我对第二个函数(getCadeiraInfo)进行注释时,代码都可以正常工作,但是第二个函数不起作用。它们是以类似的方式编写的,我不理解错误。查看chrome工具中的请求和响应。您应该看到其中的错误。上面写着什么?@Juakali92我忘了一个“;”,我们喜欢php。非常感谢。
public function getCadeiras($id) {
$this->db->select("cadeira_id");
$this->db->where(array('user_id =' => $id));
$query = $this->db->get('professor_cadeira');
return $query->result_array();
}
public function getCadeiraInfo($id) {
$query = $this->db->get_where('cadeira', array('id =' => $id));
return $query->result_array();
}