使用PHP更新mysqli数据库中的产品

尝试使用php一次更新一行

我想让用户能够更新他们已经添加到数据库中的产品，我有一个带有相关字段的简单表单：

 <fieldset><legend><span> Update a product in the database! </span></legend>

        <form  id ="productsform" method="post" onsubmit="return false;" enctype="multipart/form-data">


        <label> Product name:               <input  type="text"     id="name"           name="name"/>           </label>

        <label> Product quantity:           <input  type="number"   id="quantity"       name="quantity"/>       </label>

        <label> Product description:        <input  type="text"     id="description"    name="description"/>    </label>

        <label> Product price:              <input  type="text"     id="price"          name="price"/>          </label>

        </br>

        <input id="submit" name="submit" type="button" class="reg" value="Update Product">

        <div id="update"></div>

            </form>

更新数据库中的产品！
产品名称：
产品数量：
产品说明：
产品价格：

我使用的是ajax，它在控制台上正常工作，但我在php方面难以更新行：

<?php

include("dbase/config_database.php");

$id = $_POST["id"];
$name = $_POST['name'];
$quantity = $_POST['quantity'];
$description = $_POST['description'];
$price = $_POST['price'];

$query = "UPDATE products SET name = '{$name}', quantity = '{$quantity}', description = '{$description}', price = '{$price}' 
WHERE id = {$id}";

mysqli_query($mysqli, $query);

?>

---

您没有从表单中获得
$\u POST[“id”]
，因为没有名为
id
的输入元素

从表中获取所有数据后，将
id
作为隐藏字段放入表单中

像

您是否收到任何错误？这是从哪里来的<代码>$\u POST[“id”]我在updateSQL.php的第5行获得了未定义的index:id
<?php
include("dbase/config_database.php");

//Stores all information passed through AJAX into the query
$name = $_POST['name'];
$quantity = $_POST['quantity'];
$description = $_POST['description'];
$price = $_POST['price'];


//Adds information to database
$query = "INSERT INTO products (name, quantity, description, price) VALUES ('$name','$quantity','$description','$price')";
//Runs the query
$result = $mysqli->query($query) OR die("Failed query $query");
echo $mysqli->error."<p>";

echo "Product Added";

$querynew = ("SELECT id as 'collectid' from products WHERE name = '$name'and quantity = '$quantity'and description ='$description'and price = '$price'");
$resultnew = $mysqli->query($querynew) OR die("Failed query $querynew");

while($info = mysqli_fetch_array( $resultnew)){
    $productid = $info['collectid'];
}

$image = $_FILES['file1']['name'];
$type = $_FILES['file1']['type'];
$size = $_FILES['file1']['size'];
$tmp_name = $_FILES['file1']['tmp_name'];

$imgpath = "images/".$productid.".jpg";

// Run the move_uploaded_file() function here
$moveResult = move_uploaded_file($tmp_name, $imgpath);
// Evaluate the value returned from the function if needed

$querytwo = ("SELECT * FROM products WHERE name = '$name' and quantity = '$quantity' and description = '$description' and price = '$price'");
$resulttwo = $mysqli ->query($querytwo) OR die ("Failed query $querynew");

$info = array();
while($row = mysqli_fetch_assoc($resulttwo)){
    $product = array("id" => $row ['id'],
        "name" => $row ['name'],
        "quantity" => $row ['quantity'],
        "description" => $row ['description'],
        "price" => $row ['price'],

);

    array_push($info,$product);
}

$json_output = json_encode($info);
echo $json_output;
?>

<input type="hidden" name="id" value="<?=$row['id']?>">