php - Strict standards error?
414
The code below was working correctly a time ago, however, when I tried to use it now, it says that there is an error in this line :
$date1 = new DateTime(array_shift(array_values($array_of_dates)));
The error says :
Strict Standards: Only variables should be passed by reference in /home/...
Below is my code :
public function get_user_weeks($user_id,$getdays = NULL,$before_num_days = NULL) {
$weeks_between =0;
$pgql = mysql_query("SELECT at_time FROM users WHERE track_id='$user_id' ORDER BY at_time ASC");
$array_of_dates = array();
while ($row = mysql_fetch_array($pgql, MYSQL_ASSOC)) {
$array_of_dates[] = $row['at_time'];
}
$date1 = new DateTime(array_shift(array_values($array_of_dates)));
if ($before_num_days==NULL) {
$date2 = new DateTime(date("Y-m-d h:m:s")); }
else {
$date2 = date("Y-m-d h:m:s");
$date2 = strtotime('+'.$before_num_days.' day', strtotime($date2));
$date2 = date('Y-m-d h:m:s',$date2);
$date2 = new DateTime($date2);
}
$interval = $date1->diff($date2);
if(NULL == $getdays) {
$weeks_between = (($interval->d) + (30.5 * $interval->m) + (365 * $interval->y))/7; }
else {
$weeks_between = (($interval->d) + (30.5 * $interval->m) + (365 * $interval->y));
}
return $weeks_between;
}
Answer
Solution:
It is triggered by, because it needs a variable, not a value, as an argument.
Also, please note that MySQL extension is officially deprecated one. Use
MySQLiorPDOinstead.Answer
Solution:
array_shift takes a reference, so you need to change your code like this:
Answer
Solution:
That is because a reference is returned... In such case, assign that to a variable and overcome this issue.
You can do like this...