Question

伙计们，我正在尝试为uni项目提供网络服务。一切似乎都没问题，直到我运行代码并获得此错误未捕获错误：在null上调用成员函数fetchAll（）。这是代码可以有人请告诉我它有什么问题吗？

<?php
header("Content-type: application/json");
$conn = new PDO("mysql:host=localhost;dbname=user;", "user", "pass");
$loc = $_GET["location"];
$type = $_GET["type"];
if(isset($_GET["location"]) && isset($_GET["type"]))
{
    $result = $conn->query("Select * from resit_accommodation where location='$loc' and type='$type'")
}
else if (isset($_GET["location"]))
{
    $result = $conn->query("Select * from resit_accommodation where location='$loc'");
}
else if (isset($_GET["type"]))
{
    $result = $conn->query("Select * from resit_accommodation where location='$type'");
}
$rows = $result->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($rows);
?>

Answer 1

如果未设置任何$_GET参数，则永远不会将$result设置为任何内容，因此如果您尝试使用它，则会出现错误。

您还应该使用预准备语句而不是替换变量，以防止SQL注入。

<?php
header("Content-type: application/json");
$conn = new PDO("mysql:host=localhost;dbname=user;", "user", "pass");
$stmt = null;
if(isset($_GET["location"]) && isset($_GET["type"]))
{
    $stmt = $conn->prepare("Select * from resit_accommodation where location= :loc and type= :type");
    $stmt->execute(['loc' => $_GET['location'], 'type' => $_GET['type']]);
}
elseif (isset($_GET["location"]))
{
    $stmt = $conn->prepare("Select * from resit_accommodation where location= :loc");
    $stmt->execute(['loc' => $_GET['location']]);
}
elseif (isset($_GET["type"]))
{
    $stmt = $conn->prepare("Select * from resit_accommodation where location= :type");
    $stmt->execute(['type' => $_GET['type']]);
}
if ($stmt) {
    $rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
} else {
    $rows = [];
}
echo json_encode($rows);
?>

未捕获的错误：调用成员函数fetchAll（）

1 个答案: