Question

我使用以下代码来确定用户是否已登录，我已设置字段＆＃39; first_sign_in＆＃39;在mysql表中为0，但是我仍然收到已经登录的回应，当天它应该返回并且没有在当天开始时登录＆＃39;

有人可以帮我解决我在这里出错的地方。

$time = date('h:i:s', time());

$checkifstaffexists = mysql_query("SELECT user_id from staff WHERE pin =  3012");

if (!$checkifstaffexists) {
die('Failed.');
}

if (mysql_num_rows($checkifstaffexists) > 0) {


$checkfirstsignin = mysql_query("SELECT first_sign_in from staff WHERE pin =  3012");

if ($checkfirstsignin == 0) {
    echo 'not signed in for start of day</br>';

        $checksignintime = mysql_query("SELECT " . date("d") . " " . "_start from staff WHERE pin =  3012");
            if($checksignintime > $time) {
                echo 'user is late';
                $addtolatetable    = mysql_query("INSERT INTO lates (user_id, date_time) SELECT user_id, '2014-05-15 12:00:00' from staff WHERE pin = 3012");

                //$signuserin = mysql_query(" ");

                $changestatustoin = mysql_query("UPDATE staff SET status=1 WHERE pin = 3012");

                //redirect

            } else {
                echo 'user is not late';
                //$signuserin = mysql_query(" ")

                $changestatustoin = mysql_query("UPDATE staff SET status=1 WHERE pin = 3012");

                //redirect
            }

} else {
    echo 'already signed in for start of day</br>';
        $checkifuserisinourout = mysql_query("SELECT status from staff WHERE pin = 3012");

        if ($checkifuserisinourout == 0) {
            echo 'user is not signed in so we will sign you in';

            //$signuserin = mysql_query(" ");
            $changestatustoin = mysql_query("UPDATE staff SET status=1 WHERE pin = 3012");

            //redirect
        } else {
            echo 'user is signed in so we will sign you out';

            //$signuserout = mysql_query(" ");
            $changestatustoout = mysql_query("UPDATE `staff` SET status=0 WHERE pin = '3012'");

            //redirect
        }
    }

} else {
//The user cannot be found
echo 'User doesn\'t exist.';
}

Answer 1

行

$checkfirstsignin = mysql_query("SELECT first_sign_in from staff WHERE pin =  3012");

您将获得必须用于获取数据的资源，例如：

$row = mysql_fetch_assoc($checkfirstsignin);

并使用此数组（$ row），您可以进一步工作。

请查看mysql_query的联机帮助页以获取进一步阅读...

因为这将一直发布：不推荐使用mysql_ *方法，请使用mysqli或pdo。

Answer 2

if ($checkfirstsignin == 0)

如果查询成功，即使没有匹配的结果，也总是等于true。你需要使用mysql_fetch_row或mysql_fetch_array来做到这一点。

while($row = mysql_fetch_assoc($checkfirstsignin)){
    if($row['first_sign_in']==0){
        //do something
    }
}

Answer 3

写下这个var_dump( $checkfirstsignin );

此行$checkfirstsignin = mysql_query("SELECT first_sign_in from staff WHERE pin = 3012");

之后

并且您将看到返回的结果是一个数组，因此如果检查，则下一个结果总是为假。

if语句不适用于MYSQL验证检查

3 个答案: