Question

我收到了这个错误。我认为' $_POST放错了$_POST['regno']，但如果我这样做（例如T_ESCAPED），则会出现$sql = "SELECT * FROM students WHERE RegNo='$_POST[regno]' AND password='$_POST[password]' AND Status='Enabled'";错误。我怎么解决这个问题？这是第33行的代码：

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我使用Wamp作为我的本地主机。谢谢！

Answer 1

您需要像这样检查帖子值

 $reg_no = isset($_POST['regno']) ? $_POST['regno'] : '';
 $sql = "SELECT * FROM students WHERE RegNo='".$reg_no."' AND password='".$_POST['password']."' AND Status='Enabled'";

Answer 2

首先确保您的变量在尝试保存之前存在：

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-reverse-engineering PUBLIC "-//Hibernate/Hibernate Reverse Engineering DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-reverse-engineering-3.0.dtd" >

<hibernate-reverse-engineering>
  <table-filter match-catalog="QLSACH" match-schema="dbo" match-name="DanhMuc"/>
  <table-filter match-catalog="QLSACH" match-schema="dbo" match-name="sach"/>
</hibernate-reverse-engineering>

还要记住，像这样使用POST var会暴露于SQL注入。

Answer 3

试试这个：

$sql = "SELECT * FROM students WHERE RegNo='".$_POST['regno']."' AND password='".$_POST['password']."' AND Status='Enabled'";

PHP $ _POST注意：未定义的索引：第33行的regno

3 个答案: