为什么我看不到任何结果 PHP mysqli prepare

Question

这是PHP代码：

<?php  
    include "connection.php";

    if(isset($_REQUEST['q'])){
         
    $INTHECART = array();
    include "connection.php";
  
     $result = $connection->prepare("SELECT * FROM `laptop` WHERE id_no=?");
     $result->bind_param("i",$_REQUEST['q']);
     $result->execute();

    $dirToImg = "img/";
    if ($result->num_rows > 0) {
        while($row = $result->fetch_assoc()) {
            $product_id = $row['id_no'];
            $laptopName = $row["laptopName"] ;
            $laptopImage = $row["laptopImages"];
            $laptopFrontImage = $row["laptopFrontImages"]; 
            $laptopFeatures = $row["laptopFeatures"]; 
            $laptopOrginalPrice = $row["laptopPriceOrginal"] ;
            $laptopDiscountPrice  = $row["laptopPriceDiscounted"]; 
            $n =  $row["laptopRating"];
            if($_REQUEST['q']==$product_id){

            ?>
        <div class="container">
        <input type="text" hidden style="display: none;" data-id="<?php echo $product_id;   ?>">
         <img src="<?php echo "{$dirToImg}{$laptopFrontImage}" ;?>">
        <div class="btm">
            <h2 class="heading"><?php echo $laptopName ; ?></h2>
         <small>
            <?php
            while ($n>0){
                echo("⭐");
                $n = $n-1;
            }
    echo "inside if";

            ?>
         </small>
         <small class="orginal_price">
             <small class="high_price">
               RS.  <?php echo $laptopOrginalPrice ; ?>
            </small>
            RS. <?php echo  $laptopDiscountPrice ; ?>
            </small>
            <br>
                
    <?php
            // making a query to check fetch the data from the userinfo tab.
            if(isset($_SESSION["is_login"])){
                $checkQ = "SELECT * FROM `userinfo` WHERE emailAddress='{$_SESSION['email']}'";
            //    checking the output of the query
                $output =  $connection->query($checkQ);
                if($output->num_rows > 0){
                    // echo "it has some value";
                  
                    while($cart = $output->fetch_assoc()){
                        array_push($INTHECART , $cart['cart']);
                    }
                    $unique_array_IN  = array_unique($INTHECART);
                    
                         if(in_array($product_id, $unique_array_IN) ){
                          
                                ?>
                                
                                <a class="btn_add_cart disabled " href="#cart">
                                    In the Cart <i class="fas fa-cart-plus"></i>
                                </a>
                                <?php
                        }
                        else {
                            ?>
                            <a class="btn_add_cart " href="#cart">
                                        Add to cart <i class="fas fa-cart-plus"></i>
                            </a>
                            <?php
                    }}
                    else {
                        ?>
                        <a class="btn_add_cart " href="#cart">
                                    Add to cart <i class="fas fa-cart-plus"></i>
                        </a>
                        <?php
                    }}
                    else{
                        ?>
                        <a class="btn_add_cart not_logged_in" href="#cart">
                                    Add to cart <i class="fas fa-cart-plus"></i>
                        </a>
                        <?php
                    }
            ?>
        </div>
    </div>
<?php 
            }  }
    }else{
        ?>
        <h3 style="margin-top:20px">
            Try adding items in the favurite list.
        </h3>
        <?php
        echo $_REQUEST['q'] . " are in the fav list ";
    } 
}else{
        echo "this is not working";
    }         
    ?>

这是我在点击事件中调用的 AJAX 的一部分：

document.querySelector(".fa-shopping-cart").addEventListener("click", function(){
window.location.href = "show_cart.php";
});

document.querySelector(".fa-user-circle").addEventListener("click",function(){
    window.location.href = "user_profile.php";
});

let below_nav = document.querySelectorAll(".childs_of_the_user_detail_nav");

below_nav.forEach(function(elm , i ){
  elm.addEventListener("click",function(){

    if(i == 1 ){
      for (let [key, value] of Object.entries(localStorage)) {
        let minfied = key.replace("?keyword=", "");
        
        if(isNaN(minfied)){
        }else{
          var xmlhttp = new XMLHttpRequest();
          // console.log(`${key}: ${value}`);
        xmlhttp.open("GET", "load_fav.php?q=" + minfied, true);
        xmlhttp.send();
        xmlhttp.onreadystatechange = function() {
        if(this.readyState === 4 && this.status === 200) {
        document.querySelector(".fav_container_user_page").innerHTML = this.responseText;
        }}
      }
  }
    }
    below_nav.forEach(elme =>{
      elme.style.border ="none";
    });
    elm.style.borderBottom = "5px solid rgba(51, 51, 51, 0.802)";
  })
})

我无法找到错误。 为什么我的代码不起作用。

我正在尝试制作一个网络应用程序，这个错误并没有让我工作，但如果我在没有准备语句的情况下这样做，那么它就可以工作。

我认为这对安全性不利。

Answer 1

您为变量指定了错误的名称。 prepare()的结果是一个准备好的语句，即mysqli_stmt类型的mysqli_stmt 。 这不是数据的结果。 您必须在执行后得到结果。

$stmt = $connection->prepare("SELECT * FROM `laptop` WHERE id_no=?");
$stmt->bind_param("i",$_REQUEST['q']);
$stmt->execute();

$result = $stmt->get_result();