mysqli_query() expects parameter 1 to be mysqli, array given in - php

Question

When I run the following query
$sql ="SELECT * FROM user_info JOIN Notifications ON user_info.user_info_id =Notifications.Sender_id AND Notifications.STATE=0 LIMIT 1";
$query=mysqli_query($con,$sql);
$num_rows=mysqli_num_rows($query);
$message=''
if($num_rows > 0){
$con=mysqli_fetch_assoc($query);
switch ($con['Notification_Type']){
case'events':
$var="Notifications".$con['user_info_id'];
$sql_shown="SELECT *FROMevent WHERE event.notification_shown = 0 AND event.user_info_id='$var'LIMIT 1";
$query_vi=mysqli_query($con,$sql_shown);
$num_rows_vi=mysqli_num_rows($query_vi);
$message.=$con['Sender_id']."has created a Event";
echo $message;
break;
}
}
I get the following error:
mysqli_query() expects parameter 1 to be mysqli, array given in
This is my user table:
CREATE TABLE IF NOT EXISTS user_info(
user_info_id INT(11) NOT NULL AUTO_INCREMENT,
u_first_name VARCHAR(255) NOT NULL,
u_last_name VARCHAR(255) NOT NULL,
u_email VARCHAR(255) NOT NULL,
u_mobile VARCHAR(255) NOT NULL,
role ENUM('1yearM','1yearN','1yearR','1yearU','1yearS','2M','2R','2N','2U','2S','3M','3R','3N','3U','3S','4M','4R','4N','4U','4S','professor','librarian','admission_department') NOT NULL,
password VARCHAR(255) NOT NULL,
u_ip VARCHAR(255) NOT NULL,
signup_date DATETIME NOT NULL,
last_login DATETIME NOT NULL,
act_code VARCHAR(255) NOT NULL,
activation enum('1','0') NOT NULL DEFAULT '0',
PRIMARY KEY (user_info_id),
UNIQUE KEY (u_email,u_mobile)
)
This is my event table:
CREATE TABLE IF NOT EXISTS Event(
Event_id INT(11) NOT NULL AUTO_INCREMENT,
Event_Name VARCHAR(255) NOT NULL,
Event_location VARCHAR(255) NOT NULL,
Event_Organizer VARCHAR(255) NOT NULL,
Event_Date_Posted DATETIME NOT NULL,
Event_Starting_timings DATETIME NOT NULL,
Event_Ending_timings DATETIME NOT NULL,
Event_Day VARCHAR(255) NOT NULL,
Event_Description text NOT NULL,
Event_file_name VARCHAR(255) NOT NULL,
Event_file_path VARCHAR(255) NOT NULL,
Event_file_size VARCHAR(255) NOT NULL,
user_info_id INT(11) NOT NULL,
notification_shown ENUM('1','0') NOT NULL DEFAULT '0',
PRIMARY KEY (Event_id),
FOREIGN KEY (user_info_id) REFERENCES user_info(user_info_id)
)
This is my notification table:
CREATE TABLE IF NOT EXISTS Notifications(
Notifications_id INT(11) NOT NULL AUTO_INCREMENT,
Notification_Type VARCHAR(255) NOT NULL,
Notification_Content VARCHAR(255) NOT NULL,
Notification_Created_Date DATETIME NOT NULL,
Notification_State ENUM('read','unread') NOT NULL DEFAULT 'unread',
Is_Notification_Delete ENUM('1','0') NOT NULL DEFAULT '0',
Notification_Deleted_Date DATETIME NOT NULL,
Sender_id INT(11) NOT NULL,
STATE ENUM('1','0') NOT NULL DEFAULT '0',
Recipient_id INT(11) NOT NULL,
PRIMARY KEY (Notifications_id)
)

Answer 1

You overwrite your $con variable with this line:
$con=mysqli_fetch_assoc($query);
Before that line your $con variable was a valid MySQLi connection handle. After that line it is not an array. Therefore your mysqli_query() call complain that the first argument isn't a mysqli connection/object/handle anymore, but instead is now an array (which it shouldn't).
Change your $con=mysqli_fetch_assoc($query); so it doesn't overwrite your $con variable with the MySQLi connection/object/handle, but instead write the result in a different variable (maybe $userinfo?).

Answer 2

Check if your $con it's been created with properly method mysqli_connect('').