Php和Sql，未定义的未定义索引[重复]

Posted 2023-05-09

【中文标题】Php和Sql，未定义的未定义索引[重复]

【英文标题】：Php and Sql, undefined index when it's not undefined [duplicate]

【发布时间】：2019-05-19 11:14:41

【问题描述】：

我试图完成和锻炼，但我得到了这 2 个错误：

Notice: Undefined index: Id_sessione in C:\xampp\htdocs\Challenge\Recensione.php on line 24

Notice: Undefined index: Cod_sessione in C:\xampp\htdocs\Challenge\Recensione.php on line 25

通常我知道如何解决未定义的索引错误，但这次我已经尝试更改语法并检查表/查询，但对我来说一切似乎都是正确的。

$Titolo = $_POST['Titolo'];
$Rate = $_POST['Valutazione'];
$query = "Select Titolo From recensioni where Titolo = '$Titolo' limit 1";
$query2 = "Select MAX(Id_sessione) from sessione";
$query3 = "Select MAX(Cod_sessione) from recensioni";
$result = $conn->query($query);
$result2 = $conn->query($query2);
$result3 = $conn->query($query3);
$row = mysqli_fetch_assoc($result);
$row2 = mysqli_fetch_assoc($result2);
$row3 = mysqli_fetch_assoc($result3);
$controllaTitolo = $row['Titolo'];
$Sessione = $row2['Id_sessione'];
$controllaSessione = $row3['Cod_sessione'];

表格 会议：

+-------------+------+---------------------+
| Id_sessione | Nome | TimeA               |
+-------------+------+---------------------+
|           1 | zhao | 2019-05-17 13:04:19 |
|           2 | zhao | 2019-05-18 18:07:24 |
|           3 | zhao | 2019-05-18 18:47:20 |
|           4 | zhao | 2019-05-19 11:49:54 |
+-------------+------+---------------------+

评论：

+---------------+---------+-------------+--------------+
| Id_recensione | Titolo  | valutazione | Cod_sessione |
+---------------+---------+-------------+--------------+
|             1 | Avenger |           3 |         NULL |
|             2 | Titanic |           4 |            4 |
+---------------+---------+-------------+--------------+

【问题讨论】：

警告：您对SQL Injections 持开放态度，应该真正使用参数化Prepared Statements，而不是像那样手动构建查询。特别是因为您根本没有逃避用户输入！

【参考方案1】：

您需要在查询中使用适当的列别名并将列别名用作索引

例如max_id 和max_cod

Titolo = $_POST['Titolo'];
$Rate = $_POST['Valutazione'];
$query = "Select Titolo From recensioni where Titolo = '$Titolo' limit 1";
$query2 = "Select MAX(Id_sessione) max_id from sessione";
$query3 = "Select MAX(Cod_sessione) max_cod from recensioni";
$result = $conn->query($query);
$result2 = $conn->query($query2);
$result3 = $conn->query($query3);
$row = mysqli_fetch_assoc($result);
$row2 = mysqli_fetch_assoc($result2);
$row3 = mysqli_fetch_assoc($result3);
$controllaTitolo = $row['Titolo'];
$Sessione = $row2['max_id'];
$controllaSessione = $row3['max_cod'];

无论如何，你不应该在 SQL 查询中使用 PHP var .. 你有 sqlinjection 的风险 .. 为避免这种情况，你应该查看你的数据库驱动程序以获取准备好的语句和绑定参数