Question

我想弄清楚这段代码中有什么不对，但无法使其正常工作

你能帮我吗？

$email = $_SESSION['email'];
$email = mysql_real_escape_string($email); 
$depst = "SELECT dept FROM  stud_reg WHERE email='$email'";     
$colls = "SELECT coll FROM  stud_reg WHERE email='$email'";
$query="SELECT * FROM  stud_reg WHERE coll='$coll' AND dept='$depst'";
$evesel="SELECT id FROM  events WHERE `group`='($depst)' AND coll_id='($colls)'";   
$studsel="SELECT drs_id FROM event_reg WHERE eve_id='$evesel'";
$query="select * from students WHERE nsite_id='$studsel'";
$result=mysql_query($query) or die(mysql_error());
$num=mysql_num_rows($result);

这是我得到的错误

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'k@***.com')' AND coll_id='(SELECT coll FROM stud_reg WHERE email='k@****' at line 1

PS-此代码中的所有表和行都退出

Answer 1

尝试他的

$query="select * from students WHERE nsite_id='$studsel';";

并且第一个查询从未实际执行，并且您似乎覆盖了第一个$查询。要么“加入”，要么使用子查询，比如

SELECT * FROM t1 WHERE column1 = (SELECT column1 FROM t2);

Answer 2

"SELECT id FROM  events WHERE `group`='".$depst."' AND coll_id='".$colls."'";

Answer 3

根据您收到的错误消息，我认为这只是单引号和双引号问题这是更正的查询：

$email = $_SESSION['email'];
$email = mysql_real_escape_string($email); 
$depst = 'SELECT dept FROM  stud_reg WHERE email="$email"';     
$colls = 'SELECT coll FROM  stud_reg WHERE email="$email"';
$query="SELECT * FROM  stud_reg WHERE coll='$coll' AND dept='$depst'";
$evesel="SELECT id FROM  events WHERE `group`='($depst)' AND coll_id='($colls)'";   
$studsel="SELECT drs_id FROM event_reg WHERE eve_id='$evesel'";
$query="select * from students WHERE nsite_id='$studsel'";
$result=mysql_query($query) or die(mysql_error());
$num=mysql_num_rows($result);

在查询中放置quotes时一定要小心。

Answer 4

您尝试运行的查询是

select * from students WHERE nsite_id='SELECT drs_id FROM event_reg WHERE eve_id='SELECT id FROM  events WHERE `group`='(SELECT dept FROM  stud_reg WHERE email='$email')' AND coll_id='(SELECT coll FROM  stud_reg WHERE email='$email')'''

但即使您确实修复了带引号字符串中引用字符串的问题，这也无法满足您的期望。

我怀疑相反想要将所有这些结合起来，所以也许有点像：

SELECT      s.* 

FROM        students AS s

INNER JOIN  event_reg AS er
    ON      er.drs_id = s.nsite_id

INNER JOIN  events AS e
    ON      er.eve_id = e.id

INNER JOIN  stud_reg AS grp
    ON      grp.dept = e.group
    AND     grp.coll = e.coll_id

WHERE       grp.email='$email'

与SQL的任何语法错误一样，当从PHP运行时，最好先在MySQL中使用查询，然后再尝试将其插入应用程序。

Answer 5

为什么不尝试单一查询。问题在于您的情况。

$query = "SELECT * FROM students WHERE nsite_id =
    (   SELECT drs_id FROM event_reg WHERE eve_id = (
        SELECT id FROM  events WHERE
        `group` = (SELECT dept FROM  stud_reg WHERE email = '".$email."')
        AND
        coll_id = (SELECT coll FROM  stud_reg WHERE email = '".$email."')
        )
     )";
$result=mysql_query($query) or die(mysql_error());
$num=mysql_num_rows($result);

得到我的Sql错误

5 个答案: