Question

运行我的查询，我收到错误..我的查询写在下面

$qry = "UPDATE Offer  SET offer_year='$offeryear', " .
"course_code='$coursecode', offer_list='$offerlist', " .
"WHERE offer_id ='$offerid'";

我得到的错误是

ERROR: Record could not be added
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 'WHERE offer_id ='1'' at line 1

对我的更新查询进行了一些更改后仍然出现错误..

Answer 1

如果您尝试逐个检查，则在comma子句之前有额外的WHERE

$qry = "UPDATE Offer  SET offer_year='$offeryear', " .
       "course_code='$coursecode', offer_list='$offerlist' ". // remove comma here
       "WHERE offer_id ='$offerid'";

作为旁注，如果变量的值（ s ）来自外部，则查询易受SQL Injection攻击。请查看下面的文章，了解如何防止它。通过使用PreparedStatements，您可以摆脱在值周围使用单引号。

How to prevent SQL injection in PHP?

Answer 2

像这样更新您的更新查询

$qry = "UPDATE Offer  
        SET offer_year='$offeryear',
            course_code='$coursecode', 
            offer_list='$offerlist' 
        WHERE offer_id ='$offerid'";

您在WHERE条件

之前添加了额外的逗号

Answer 3

试试这个（offerlist之后有逗号）：

$qry = "UPDATE Offer  SET offer_year='$offeryear', " .
"course_code='$coursecode', offer_list='$offerlist' " .
"WHERE offer_id ='$offerid'";

在mysql中更新查询

3 个答案: