[英]Web checking in game, help with the logic of it
我正在建立一個包含游戲的網絡。
該游戲旨在獲取在網絡上查找某些信息的積分。
當用戶找到令牌時,一條信息會進行檢查,他/她會獲得該檢查的積分。
這是到目前為止我建立的數據庫:
CREATE TABLE IF NOT EXISTS `user` (
`userid` int(11) NOT NULL AUTO_INCREMENT,
`points` int(11) NOT NULL,
PRIMARY KEY (`userid`)
) ENGINE=MyISAM DEFAULT AUTO_INCREMENT=1 ;
CREATE TABLE IF NOT EXISTS `checkings` (
`checkingsid` int(11) NOT NULL AUTO_INCREMENT,
`userid` int(11) NOT NULL,
`tokensid` int(11) NOT NULL,
`checked` int(11) NOT NULL,
PRIMARY KEY (`checkingsid`)
) ENGINE=MyISAM DEFAULT AUTO_INCREMENT=1 ;
CREATE TABLE IF NOT EXISTS `tokens` (
`tid` int(11) NOT NULL AUTO_INCREMENT,
`tokensid` int(11) NOT NULL,
`name` varchar(255) NOT NULL,
`points` int(11) NOT NULL,
PRIMARY KEY (`tid`)
) ENGINE=MyISAM DEFAULT AUTO_INCREMENT=1 ;
這是我的代碼。
它的問題是:我怎么知道是否已經為該令牌給出了積分? 請提供邏輯幫助,我已經失去了范圍,我看不到做到這一點的方法。 我可能已經開始朝錯誤的方向編碼。 我解釋了我要在代碼中嘗試做的事情。
非常感謝
<?php
//connected to database and obtained $userId and $tokenID already
$query0 = "select * from checkings where userid=".$userId." and tokensid=".$tokenId."";
$result0 = mysql_query($query0);
$row0 = mysql_fetch_array($result0);
if (($row0['checked']!=' ')|| ($row0['checked']!='0')){ // if checked is empty or 0 the user didn't do the checking in this token yet
if (($row0['checked']==1)&&($row0['userid']==$userId)&&($row0['tokensid']==$tokenId) ){//the checking has already been inserted
//how many points are given for checking in this token?
$query2 = "select points from tokens where tokensid='".$tokenId."'";
$result2 = mysql_query($query2);
$row2 = mysql_fetch_array($result2);
$pointstoken = $row2['points'];
//How many points have the user?
$query3 = "select points from user where userid='".$userId."'";
$result3 = mysql_query($query3);
$row3 = mysql_fetch_array($result3);
$pointsUser = $row3['points'];
//user points are updated after checking in this token
$pointsTotal = $pointsUser+$pointstoken;
$query4 = "update user set points=".$pointsTotal." where userid=".$userId." ";
$result4 = mysql_query($query4);
}else{//here I do the checking in the token inseting the ids plus a 1 as for checked
$query1 = "insert into checkings (userid, tokensid,checked ) values (".$userId.",".$tokenId.",1)";
$result1 = mysql_query($query1);
}
}else{ echo "Hi, user ".$userId." you've already done the checking in token id ".$tokenId." ";}
?>
檢查表是否存儲用戶找到的檢查?
在這種情況下,請在授予用戶任何積分之前先搜索該表。
如果不是,請為用戶+檢查/令牌創建一個表,以跟蹤用戶找到並被授予的每個項目,並在用戶每次提交發現時搜索該表。
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