Question

我修复了所有语法错误，但是我试图在表上插入的项目没有传递到数据库中的表。这是代码：

   <?php
   session_start();
   if(!isset($_SESSION["manager"])){
   header("location:admin_login.php");
   exit();
    }

   $id=preg_replace('#[^0-9]#i','',$_SESSION["id"]);
   $manager=preg_replace('#[^A-Za-z0-9]#i','',$_SESSION["manager"]);
   $password=preg_replace('#[^A-Za-z0-9]#i','',$_SESSION["password"]);
   include "connect_to_db.php";
   $sqlcommand="SELECT COUNT(id) FROM admin WHERE id='$id' AND username='$manager'    AND      password='$password' LIMIT 1";
   $counting=mysqli_query($connect_dude,$sqlcommand);
   $numrow=mysqli_num_rows($counting);
   if($numrow==0){
   echo "you better get out of here";
   exit();
   }
   ?>
   <?php 
   // Script Error Reporting
  error_reporting(E_ALL);
  ini_set('display_errors', '1');
    ?>
  <?php

     if(isset($_POST["product_name"])){

    $productname=preg_replace('#[^A-Za-z]#i','',$_POST["product_name"]);
    $productprice=preg_replace('#[^0-9]#i','',$_POST["product_price"]);
    $productdetails=preg_replace('#[^A-Za-z0-9]#i','',$_POST["product_detail"]);
    $productdivision=$_POST["product_division"];
    $productsubdivision=$_POST["product_subdivision"];
    $sql="SELECT COUNT(id) FROM products WHERE product_name='$productname' LIMIT 1";
    $counting=mysqli_query($connect_dude,$sql);
    $numrow=mysqli_num_rows($counting);
    if($numrow>0){
    echo "you have added the product before";
    }
    else{
    $sql="INSERT INTO products (product_name,price,details,category,subcategory,date_added) VALUES ($productname,$productprice,$productdetails,$productdivision,$productsubdivision,now())";
    mysqli_query($connect_dude,$sql);
    $pid=mysqli_insert_id();
    $newfile=$pid.jpg;
    move_uploaded_file($_FILES['image']['tmp_name'],"$productsubdivision/$newfile");
    header("location: inventory_list.php"); 
    exit();
       }
    }

    ?>
    <html>
    <head>
    </head>
     <body>

    <fieldset>
      <legend value="add something">Add product,dude</legend>
   <form id="formo" action="inventory_list.php" method="post" enctype="multipart/form-data">
   Product Name:<input type="text" id="product_name" name="product_name" placeholder="Type Product Name" /></br>
   Product price:<input type="text" id="product_price" name="product_price" placeholder="Type Product price" /></br>
   Product division:
  <select id="product_division" name="product_division">
    <option>select</option>
    <option>food</option>
 <option>apparel</option>
<option>beauty</option>
<option>toys</option>
   </select>
  </br> 
       Product type:
        <select id="product_subdivision" name="product_subdivision">
       <option>select</option>
          <option>Rice</option>
        <option>milk</option>
       <option>Meat</option>
       <option>sweet</option>
     <option>fruits</option>
      <option>cap</option>
      <option>shirt</option>
     <option>pant</option>
     </select>
  </br>
     Product Detail:<textarea id="product_detail" name="product_detail" placeholder="Type Product detail,bro" ></textarea> 
    </br>
    product Image:<input type="file" id="image" name="image" />
     </br>
    <input type="submit" name="button" id="button" value="Add This Item Now" />

     </form>
  </fieldset>
  </body>
   </html>

向我显示消息“您之前已添加产品”，但未插入表格。我是初学PHP程序员。请帮助。

Answer 1

您缺少字段上的name属性。 $_POST['product_name']未设置。

<input type="text" id="product_name" placeholder="Type Product Name" />

应该是

<input type="text" name="product_name" id="product_name" placeholder="Type Product Name" />

您的条件是检查isset($_POST['product_name'])以运行插入，因为缺少name属性，该条件永远不会评估为true。

旁注，您也在尝试接受文件。您需要在表单属性中使用enctype="miltipart/form-data"：

<form id="formo" action="inventory_list.php" method="post" enctype="miltipart/form-data">

另外，请确保您拥有所有字段的名称属性。

您还要检查计数中的布尔值：

$numrow=mysqli_num_rows($counting);
if($counting==1){

应该是

$numrow=mysqli_num_rows($counting);
if($numrow > 0){

mysql数据库不会在表上插入任何项目，浏览器也不会显示任何错误

1 个答案: