Why this if else condition is not not working properly - php

Question

$taskid=$this->privacy();
if($taskid){I returning something in one function and then by reference to this return,I am fetching data from another table in second function using if condition in such a way that if data(return) found in first table then execute else leave empty
But I data found in first table then every thing is fine,hower data not found in first table then it shows as Notice: Undefined variable: task in....
I think error is in if else condition.Please help or suggest any alternative approach.
Plz also note that that I am working on pricacy table
first function returning rows
function privacy()
{
$session=new session();
$sql2=mysqli_query($this->db->connection,"SELECT * from privacy where viewerid='$session->userid'");
while($row2=mysqli_fetch_array($sql2)){
$task[]=$row2['task'];
}
return $task;
}
function 2.. if found something then do something else do nothing(but error appears as undefined variable)
function showactivityo4apply()
{
$session=new session();
$taskid=$this->privacy();
if($taskid){
foreach($taskid as $fvid){
$sql=mysqli_query($this->db->connection,"SELECT * FROM activity where id='$fvid'");
while($row=mysqli_fetch_array($sql)){
$name_id=$row['sub_id'];
echo $name;
}
}
}
else{}
}

Answer 1

if($taskid=$this->privacy()) is assigning a value of $this->privacy() to $taskid
I think you mean
if($taskid == $this->privacy())

Answer 2

If condition have == sign not = sign
if($taskid==$this->privacy()){
}
else
{
}
= means assign value to another
== check the value to other variable

Answer 3

it is showing notice because php is not getting value in array on key "task" use isset() function to check whether data is available or not
use like this
$task[] = (isset($row2['task'])?$row2['task']:'';
if value is not set in $row2['task'] then it will assign blank value in $task[] array.
and one more change you have to do
instead of if($taskid) condition use
if( count($taskid) > 0){
}
so that if record not found in array it will not inter in to the loop

Answer 4

please change this line of your code like:
$sql2=mysqli_query($this->db->connection,"SELECT * from privacy where
viewerid='$session->userid'");
=========================================
to
=============================
$sql2=mysqli_query($this->db->connection,"SELECT * from privacy where
viewerid='" . $session->userid . "'");
=======================
conclution: Here $session->userid is not found thats why you not get result.