如果只有两个输入不存在，我如何才能插入数据库？

Question

I want to give "awards" to users who complete certain tasks such as getting to level 50. I also want to log everything in case something goes wrong and an admin has to manually insert it.

The php for awards is linked on every page and checks if the user from 'users' table has reached level 50. Once the user is level 50 it will insert username, award, time, reason, and who gave the award into a table called 'awards'.

But if the user is level 50 it will continue inserting into the 'awards' table every time the user reload or goes to a new page. How can i only update if the user doesn't already have the specific award?

if(empty($_SESSION['user'])){

}else{
    $member_id = htmlentities($_SESSION['user']['id'], ENT_QUOTES, 'UTF-8');

    $db_conn = mysqli_connect("localhost", "root", "password", "a276") or die ("Could not connect to database");
    $set_award_query = mysqli_query ($db_conn, "SELECT * FROM users WHERE id='$member_id'");

    while($row = mysqli_fetch_array($set_award_query, MYSQLI_ASSOC)){
        $member_id = htmlentities($_SESSION['user']['id'], ENT_QUOTES, 'UTF-8');
        $member_username = htmlentities($_SESSION['user']['username'], ENT_QUOTES, 'UTF-8');
        $member_level = $row['level'];
    }

//Award level50
if($member_level == '50'){
    $award_sql = "INSERT INTO `awards` (`awardusername`, `awardby`, `award`, `awardmessage`) VALUES ('$member_username', 'System', 'level50', 'Congratulations on reaching level 50!')";
    $db_conn->query($award_sql);
}else{}

Answer 1

You can create an UNIQUE INDEX over awardusername and award columns:

ALTER TABLE `awards` ADD UNIQUE INDEX awards_key (awardusername, award);

This will prevent second INSERT for the same user and level and will throw 1062 error which you can handle easily I believe.

Answer 2

Create a user_id column in the awards table

Check if the $member_id matches the user_id in the awards table.

Store the user_id($member_id) in the awards table if there is no match.

// Get the user_id from the award table (Example, this code won't work)

$user_id = "SELECT user_id FROM awards WHERE user_id = $member_id";

// Check if they match

if( $member_id !== $user_id ):
    // Insert into awards table
endif;

Answer 3

I found a way myself to fix the issue. It might be a stupid way or something (I don't know) but it works

    $member_id = htmlentities($_SESSION['user']['id'], ENT_QUOTES, 'UTF-8');

    $db_conn = mysqli_connect("localhost", "root", "password", "a276") or die ("Could not connect to database");
    $set_award_query = mysqli_query ($db_conn, "SELECT * FROM users WHERE id='$member_id'");
    while($row = mysqli_fetch_array($set_award_query, MYSQLI_ASSOC)){
        $member_id = htmlentities($_SESSION['user']['id'], ENT_QUOTES, 'UTF-8');
        $member_username = htmlentities($_SESSION['user']['username'], ENT_QUOTES, 'UTF-8');
        $member_level = $row['level'];
    }

    $check_award_exist_query = mysqli_query ($db_conn, "SELECT * FROM awards WHERE awardusername='$member_username'");
    while($row = mysqli_fetch_array($check_award_exist_query, MYSQLI_ASSOC)){
        $award_username = $row['awardusername'];
        $selected_award = $row['award'];
    }

//Award level50
if($member_username){
    if($member_level == '50'){
        if($selected_award == 'level50'){

        }else{
            $award_sql = "INSERT INTO `awards` (`awardusername`, `awardby`, `award`, `awardmessage`) VALUES ('$member_username', 'System', 'level50', 'Congratulations on reaching level 50!')";
            $db_conn->query($award_sql);
        }
    }else{}
}else{}

Thanks to everyone who took their time to try to help me out! :)