mysqli_query（）期望參數1是mysqli，在/ Applications/MAMP/htdocs/

Question

中給出的對象

它給了我這個警告： 警告：mysqli_query（）期望參數1是mysqli，在/ Applications/MAMP中給出的對象在線路45 /htdocs/dashboard/pages/index.php

警告：mysqli_error（）預計參數1是上線路50

在/Applications/MAMP/htdocs/dashboard/pages/index.php給出的mysqli，對象

我試圖將項目插入表稱爲產品。但不知何故，這個錯誤不會走的方式。

你能告訴我我做錯了什麼嗎？非常感謝。

這是dbconfig.php

<?php 
$servername = "localhost"; 
$username = "root"; 
$password = "root"; 
$dbname = "products"; 
try { 
    $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password); 
    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
    } 
catch(PDOException $e) 
    { 
    echo "Connection failed: " . $e->getMessage(); 
    } 

？>

這是表

這是PHP代碼

<?php 

require_once('dbconfig.php'); 
$upload_dir = 'uploads/'; 

if(isset($_POST['submit'])){ 
    $product_picture = $_FILES['product_picture']['name']; 
    $imgTmp = $_FILES['product_picture']['tmp_name']; 
    $imgSize = $_FILES['product_picture']['size']; 
    $product_name = $_POST['product_name']; 
    $product_number = $_POST['product_number']; 

    $product_quantity = $_POST['product_quantity']; 
    $product_price = $_POST['product_price']; 
    $product_description = $_POST['product_description']; 

    if(empty($product_name)){ 
     $errorMsg = 'Please input product name'; 
    }elseif(empty($product_number)){ 
     $errorMsg = 'Please input product number'; 
    }elseif(empty($product_quantity)){ 
     $errorMsg = 'Please input product quantity'; 
    }elseif(empty($product_price)){ 
     $errorMsg = 'Please input product price'; 
    }elseif(empty($product_description)){ 
     $errorMsg = 'Please input product description'; 
    }else{ 
     $imgExt = strtolower(pathinfo($product_picture, PATHINFO_EXTENSION)); 
     $allowExt = array('jepg', 'jpg', 'png', 'gif'); 
     $productPic = time().'_'.rand(1000,9999).'.'.$imgExt; 
     if(in_array($imgExt, $allowExt)){ 
      if($imgSize < 5000000){ 
       move_uploaded_file($imgTmp,$upload_dir.$productPic); 
      }else{ 
       $errorMsg = 'Image too large'; 
      } 
     }else{ 
      $errorMsg = 'Please select a valid image'; 
     } 
    } 

    if(!isset($errorMsg)){ 
     $sql = "insert into products(product_picture, product_name, product_number, product_quantity, product_price, product_description) 
       values('".$productPic."', '".$product_name."', '".$product_number."', '".$product_quantity."', '".$product_price."', '".$product_description."')"; 
     $result = mysqli_query($conn, $sql); 
     if($result){ 
      $successMsg = 'New record added sucessfully'; 
      header('refresh:5;index.php'); 
     }else{ 
      $errorMsg = 'Error'.mysqli_error($conn); 
     } 
    } 
} 

？>

這是形式部分

<form class="inputarea text-center" method="post" action="" enctype="multipart/form-data"> 
          <p id="addProduct" class="addTitle">Add new product</p> 
          <input id="pName" class="inputs" type="text" placeholder="Name" name="product_name" > 

          <input id="choose" type = "file" name = "product_picture" /> 

          <input id="pNumber" class="inputs" type="number" placeholder="Prodect No." name="product_number" > 
          <input id="pQuantity" class="inputs" type="text" placeholder="Quantity" name="product_quantity" > 
          <input id="pPrice" class="inputs" type="text" placeholder="Price" name="product_price" > 
          <textarea id="pDescription" class="inputs" placeholder="Description" name="product_description"></textarea> 
          <button id="confirmBut" type="submit" name="submit" class="btn btn-primary btn-lg center-block">Confirm</button> 
         </form> 

Answer 1

的代碼混合PDO和mysqli接口。這不會起作用。

$conn = new PDO("..."); 

mysqli_query($conn, ...); 

---

如果我們創建一個PDO連接，然後我們用PDO接口功能。

如果我們使用mysqli_函數，那麼我們創建一個mysqli連接。