【发布时间】：2013-11-19 02:47:02
【问题描述】：
这是我的代码的一部分
$con = mysqli_connect("localhost","root","","baspdata",3306);
if (mysqli_connect_errno())
{
echo "Error connecting to database: ".mysqli_connect_error();
exit();
}
else
{
$result=mysqli_query($con,"SELECT * FROM member WHERE Username='$username' and Password = '$password'");
$row=$result->fetch_assoc();
$sellerId=$row['MemberId'];
$picturecontent= file_get_contents($productPic);
$query ="INSERT INTO product (ProductName, ProductPicture, ProductDescription, ProductCategory, ProductPrice, UploadedDate, Sold, SellerId) VALUES(?,?,?,?,?,?,?.?)";
$stmt=$con->prepare($query);
$stmt->bind_param("ssssssss", $productName, $picturecontent, $description, $category, $price, $uploadedDate, $sold , $sellerId);
$stmt->execute();
$con->close();
echo "<h1>".$productName." added successfully! =)<br/> </h1>";
}
我收到错误 Fatal error: Call to a member function bind_param() on a non-object on the line $stmt->bind_param("ssssssss", $productName, $picturecontent, $description, $category, $price , $uploadedDate, $sold , $sellerId);但我无法弄清楚。请帮忙。
【问题讨论】：
-
Fatal error: Call to a member function bind_param() on a non-object的意思正是它所说的。$stmt对象不是对象 - 当prepare调用失败并返回false而不是返回对象时会发生这种情况。 -
如果您阅读
prepare的php 文档页面，您会看到它清楚地指出prepare在失败时返回false。这意味着在将其用作对象之前，您始终需要检查返回的值是否为false。
标签： php