Question

当我编写我的在线商店时，我收到了这个奇怪的错误消息，我之前从未见过它！

致命错误：在第5行的写入上下文中不能使用函数返回值

这是我的代码

    <?php 
                        if (isset($_GET['pro_id'])){
                            $product_id = $_GET['pro_id'];
                            $query = $con->query("SELECT * FROM viewcounter WHERE productcode='$product_id'"); 
                            if(mysqli_num_rows($query) = 0){
                                $insert_id = "insert into viewcounter (productcode) values ('$product_id')";
                                $insert_id = mysqli_query($con, $insert_id);
                                mysql_query("UPDATE viewcounter SET `views`=`views`+1 WHERE productcode='$product_id'")
                            }
                            $get_pro = "select * from products where product_id='$product_id'";
                            $run_pro = mysqli_query($con,$get_pro);
                            while($row_pro = mysqli_fetch_array($run_pro)){

                                $_SESSION['pro_id'] = $row_pro['product_id'];
                                $_SESSION['pro_cat'] = $row_pro['product_cat'];
                                $_SESSION['pro_brand'] = $row_pro['product_brand'];
                                $_SESSION['pro_title'] = $row_pro['product_title'];
                                $_SESSION['pro_price'] = $row_pro['product_price'];

                                echo '....';
";}}?>

这是第5行，我收到错误消息：

if（mysqli_num_rows（$ query）= 0）{

注意：我跳过编写连接数据库的代码，因为它已经没问题了！

Answer 1

input.h

即将函数的值赋值为0

为了比较，您需要双倍或三倍：

if(mysqli_num_rows($query) = 0){

More info on PHP comparison operators

在写上下文错误消息中不能使用函数返回值

1 个答案: