对PHP函数的AJAX调用不会使它为我的数据库输入值

Question

I've been working on a piece of code for a few days now but I somehow can't get this to pass the value 1 to my database. Basically what I am trying to achieve is that once the user clicks the button "collect coins" it passes the value 1 to my database. Every day at 12 pm the column "dailyfree" is reset to 0 by a MySQL event so the next day the user can click the button again since the value of "dailyfree" in the database is 0 again. Still I don't want the user to be able to click "collect coins" before the value in the database has been reset. The problem is that once the user click the button, it gives him the free coins but the value 1 does not get passed to the database. The column "dailyfree" is still 0 after clicking the button. Does anybody know why that is?

Just for your information: I am very new to coding!

Here is what I am working with right now.

Javascript:

function free(){
var str = "username"; 
var term = "searched_word"; 
var index = str.indexOf(term); 
if (index != -1) { 
var daily = 1;
$.ajax({
url:"/free?daily="+daily,
    success:function(data){     
        try{
            data = JSON.parse(data);
            console.log(data);
            if(data.success){
                bootbox.alert("Success! You've claimed 100 free credits!");                 
            }else{
                bootbox.alert(data.error);
            }
        }catch(err){
            bootbox.alert("Javascript error: "+err);
        }
    },
    error:function(err){
        bootbox.alert("AJAX error: "+err);
    }
});

   }
  else {
    bootbox.alert("Your username does not contain the searched_word!");}
   }


function disDelay(obj){
    obj.setAttribute('disabled','disabled');
    setTimeout(function(){obj.removeAttribute('disabled')},86400)
}

The Button itself

<div class="text-right">
    <button class="btn btn-success btn-block" id="free" onclick="free(); disDelay(this)">Collect free coins</button>
</div>

PHP

$freecoins = 100;
case 'free':
    if(!$user) 
    exit(json_encode(array('success'=>false, 'error'=>'You must login to access the redeem.')));
    $daily = 1;
    if(!preg_match('/^[a-zA-Z0-9]+$/', $daily)) {
        exit(json_encode(array('success'=>false, 'error'=>'Code is not valid')));
    } else {
        $sql = $db->query('SELECT * FROM `users` WHERE `dailyfree` = '.$db->quote($dailyfree));
        if($sql->rowCount() != 0) {
            $row = $sql->fetch();
            if($row['user'] == $user['userid']) 
            exit(json_encode(array('success'=>false, 'error'=>'You have already redeemed your daily reward!')));
            $db->exec('INSERT INTO `users` SET `dailyfree` = '.$dailyfree.' , `balance` = `balance` + '.$freecoins.' WHERE `userid` = '.$db->quote($user['userid']));
            exit(json_encode(array('success'=>true, 'credits'=>$freecoins)));
        } else {
            exit(json_encode(array('success'=>false, 'error'=>'Code not found')));
        }
    }
    break;

I really hope you can help me find out why it won't enter the value into my database! Anyways, I wish you a very nice day :) Thanks for your help and time!

Answer 1

The query is wrong.

Should be a update query.

$db->exec('UPDATE users SET dailyfree = '.$dailyfree.' , balance = balance + '.$freecoins.' WHERE userid = '.$db->quote($user['userid']));

Or if you want to update when the user record present else insert the data then take a look at ON DUPLICATE KEY