單擊登錄后,還是提示驗證碼錯誤,但驗證碼是正確,是否是數據庫沒連上
問題描述
<?phpnamespace appindexController;use appindexControllerBase;use thinkRequest;use appindexmodelUser as UserModel;class User extends Base{ public function login() { return $this-> view ->fetch(); }
public function checklogin(Request $request) {//初始返回參數 //$status = 0; // $result = ''; // $data = $request -> param(); $status = 0; //驗證失敗標志$result = '驗證失敗'; //失敗提示信息$data = $request -> param(); // 創建驗證規則 $rule=[ 'name|用戶名'=>'require',//用戶名必填 'password|密碼'=>'require',//密碼必填 'verify|驗證碼'=>'require|captcha',//驗證碼必填 ];
$msg=[ 'name'=>['require'=>'用戶名不能為空請檢查'], 'passwrod'=>['require'=>'密碼不能為空請檢查'], 'verify'=>[ 'require'=>'驗證碼不能為空請檢查', 'captcha'=>'驗證碼錯誤',
], ];
//$result = $this->validate($data,$rule,$msg);
$result=$this->validate($data, $rule, $msg); if ($result===true) { //查詢條件 $map = [ 'name' => $data['name'], 'password' => md5($data['password']) ];
//數據表查詢,返回模型對象 $user = UserModel::get($map); if (null === $user) { $result = '沒有該用戶,請檢查'; } else { $status = 1; $result = '驗證通過,點擊[確定]后進入后臺'; } } return['status' => $status,'message' => $result,'data' => $data]; //return['status'=>$status,'message'=>$result, 'data'=>$data];
}
public function logout() {
}}?
問題解答
回答1:請幫指出,可以嗎,謝謝回答2:jrhhy 你確定打對了????
