Question

我已经快速制作了一个显示数据库中当前项目的页面。我已经创建了调整库存水平的功能，但我现在正在尝试添加删除项目的功能。由于某些原因，它所做的只是将库存水平降低到0，但不会从数据库中删除该项目。

还可以/更好地将生成表的代码放入另一个文件并在此处加载吗？因为我需要它来运行库存更新或删除功能时自动更新。

AJAX / HTML CODE

<!DOCTYPE html>
<html>
<head>
<script     src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js">    </script>
<script>

$(document).ready(function(){
$('#dataSubmit').on('submit',function(e){
    $.ajax({
        url:'files/update.php',
        data:$(dataSubmit).serialize(),
        type:'POST',
        success:function(data){
            console.log(data);
            if(data != "Error") {
               $("#data").html(data).show().fadeOut(9000);
            }
            else {
               $("#data").html(data).show().fadeOut(9000);
            }
        },
        error:function(data){
            $("#error").show().fadeOut(5000);
        }
    });
    e.preventDefault();
    document.getElementById("dataSubmit").reset();
});
});
$(document).ready(function(){
$('#deleteItem').on('dsubmit',function(e){
    $.ajax({
        url:'files/delete.php',
        data:$(deleteItem).serialize(),
        type:'POST',
        success:function(data){
            console.log(data);
            if(data != "Error") {
                   $("#datad").html(data).show().fadeOut(9000);
                }
                else {
                   $("#datad").html(data).show().fadeOut(9000);
                }
        },
        error:function(data){
            $("#error").show().fadeOut(5000);
        }
    });
    e.preventDefault();
    document.getElementById("deleteItem").reset();
});
});
</script>

</head>
<body>
<h2> Update Stock Levels </h2>
<form name="dataSubmit" id="dataSubmit" action="">
Product ID: <input type="number" name="Product_ID" value=""><br>
New Stock Amount: <input type="text" name="Product_Stock" value=""><br>
<input type="submit" name="submit" > 
<div id="data"></div>

<h2> Delete Items </h2>
<form name="deleteItem" id="deleteItem" action="">
Product ID: <input type="number" name="Product_ID" value=""><br>
<input type="submit" name="dsubmit"  >
<div id="datad"></div>





<?php
echo "<table style='border: solid 1px black;'>";
echo "<tr><th>Id</th><th>Product Name</th><th>Product Description</th>    <th>Product Price</th><th>Product Stock Amount</th></tr>";

class TableRows extends RecursiveIteratorIterator { 
function __construct($it) { 
    parent::__construct($it, self::LEAVES_ONLY); 
    }

function current() {
    return "<td style='width:150px;border:1px solid black;'>" . parent::current(). "</td>";
}

function beginChildren() { 
    echo "<tr>"; 
} 

function endChildren() { 
    echo "</tr>" . "\n";
} 
} 

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "cms";

try {
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username,         $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$stmt = $conn->prepare("SELECT Product_ID, Product_Name, Product_Desc, Product_Price, Product_Stock FROM Products"); 
$stmt->execute();

// set the resulting array to associative
$result = $stmt->setFetchMode(PDO::FETCH_ASSOC); 
foreach(new TableRows(new RecursiveArrayIterator($stmt->fetchAll())) as     $k=>$v) { 
    echo $v;
}
}
catch(PDOException $e) {
echo "Error: " . $e->getMessage();
}
$conn = null;
echo "</table>";
?>
</body>
</html>

PHP代码

 <?php

$dsn = 'mysql:host=localhost;dbname=cms';
$user = 'root';
$password = '';
try {
$pdo = new PDO($dsn, $user, $password);
$pdo ->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch (PDOException $e) {
echo 'Connection failed: ' . $e->getMessage();
}

$sql = "DELETE FROM Products WHERE Product_ID = :Product_ID";
$stmt = $pdo->prepare($sql);
$stmt->bindParam(':Product_ID', $_POST['Product_ID'], PDO::PARAM_INT);   
$stmt->execute();



?>

Answer 1

您没有绑定到删除操作的提交事件，您有一个拼写错误

$('#deleteItem').on('dsubmit',function(e){
                     ^

表单只是提交。

为什么php脚本不删除表？

1 个答案: