我正在尝试完成获取当前用户会话(如果存在)然后将其与数据库进行比较,以及loggedIn值= 1(表示loggedIn)并计算存在与此where子句匹配的单行。如果存在,则显示已登录,否则,显示登录表单(如下if{}
else{}
所示)
我有一个具有功能的模型:
function check_if_loggedin(){
#get user session id from db.
$sess = $this->session->userdata('session_id');
#set numerical value of successfully logged in equal 1
$loggedInSetSuccess = 1;
#compare session id to match in db (can only have only 1 match)
$sessionsDbCompare = $this->db->get_where('Client',array('session_id'=>$sess,'loggedIn'=>$loggedInSetSuccess));
if($sessionsDbCompare->count_all_results() = 1) {
# User has valid session(valid sessID+loggedIn=0, show welcome
echo 'Logged in baby';
# show account pref
}
else{
# user doesnt have both a valid session and loggedIn is set to 0, show login form
echo 'login form here';
}
}
问题是 -if($sessionsDbCompare->count_all_results() = 1) {
我的 IDe 正在警告我,我应该使用count_all_results() == 1
or count_all_results() === 1
. 抱歉,这有点新手,我想我有点困惑,因为 count_all_results() 是一个函数。(我确实理解比较运算符,例如“1”==“01”)