2015-03-02 27 views
-1
$sql=mysql_query("SELECT day='$day','$month' day='$day' year='$year' FROM punchin WHERE pass='$pass' '$month' AND day='$day' AND year='$year' "); 


if (mysql_num_rows($sql) === 0) { 
mysql_query("INSERT INTO punchin (count,pass,month,day,year,time,wday) values (null,'$pass','$month','$day','$year','$time','$wday')") or die(mysql_error()); 
t 

}else 
     echo "<a href='loginoption.php'>You can only punch-in once a day!</a>"; 

问题是.. 员工应该每天只打一次,但这个人可以随时打点。如何选择和查看是否有记录?

+0

'SELECT天='$ day''错误的语法。选择的这一部分没有'='。 – developerwjk 2015-03-02 22:32:08

+0

假设你正在使用MySQL,SELECT语句的语法是[记录在这里](http://dev.mysql.com/doc/refman/5.6/en/select.html) – 2015-03-02 22:33:23

+1

但是,如果你只是学习,学习使用[MySQLi](http://www.php.net/manual/en/book.mysqli.php)或[PDO](http://www.php.net/manual/en/book .pdo.php)与[prepared statements and bind variables]接口(http://php.net/manual/en/mysqli.quickstart.prepared-statements.php),因为MySQL接口将[从PHP正式退出] (http://php.net/manual/en/migration55.deprecated.php)在年底之前 – 2015-03-02 22:34:22

回答

1

你在你的SELECT查询的问题,尝试用这样的:

$sql=mysql_query("SELECT day FROM punchin WHERE pass='$pass' AND month = '$month' AND day='$day' AND year='$year'"); 

if (mysql_num_rows($sql) == 0) { 
mysql_query("INSERT INTO punchin (count,pass,month,day,year,time,wday) values (null,'$pass','$month','$day','$year','$time','$wday')") or die(mysql_error()); 

}else 
     echo "<a href='loginoption.php'>You can only punch-in once a day!</a>"; 
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