MySQL PHP jQuery-行不更新嗎？

Question

我正在編寫CMS，但仍無法更新products表中的現有產品。

問題在於，每個值除“文本”和“庫存”行外還將接收更新...它們變為空白。

這是我的代碼

形成：

<form class="form-horizontal form-inline">
            <div class="form-group">
                <label class="control-label col-sm-3" for="title">Product:</label>
                <input type="text" class="form-control" id="title" name="title" placeholder="Enter new title">
            </div>
            <div class="form-group">
                <label class="control-label col-sm-3" for="price">Price:</label>
                <input type="text" class="form-control" id="price" name="price" placeholder="Enter new price">
            </div>
            <div class="form-group">
                <label class="control-label col-sm-3" for="stock">Stock:</label>
                <input type="text" class="form-control" id="stock" name="stock" placeholder="Enter new stock">
            </div>
            <div class='form-group'>
                    <label class="control-label col-sm-3" for="text">Text:</label>
                    <textarea class='form-control' rows='5' id='newPostText' name='text' placeholder='Enter new text'></textarea>
            </div>
            <div class="form-group">
                <label class="control-label col-sm-3" for="description">Description:</label>
                <input type="text" class="form-control" id="description" name="description" placeholder="Enter new description">
            </div>
            <div class="form-group">
                <label class="control-label col-sm-3" for="brand">Brand:</label>
                <input type="text" class="form-control" id="brand" name="brand" placeholder="Enter new brand">
            </div>
            <div class="form-group">
                <label class="control-label col-sm-3"  for="keywords">Keywords:</label>
                <input type="text" class="form-control" id="keywords" name="keywords" placeholder="Enter new keywords">
            </div>
            <a href="#" id="updateProductRecord" pid="<?php echo $pro_id; ?>"><button type="submit" id="submitUpdateProduct" class="btn btn-default">Submit</button></a>
            </form>

jQuery的：

$("body").delegate("#updateProductRecord","click",function(event){
event.preventDefault();

var product_id = $(this).attr('pid');
updateProductRecord(product_id);

animatedLoading();
});

function updateProductRecord(product){

var title = $("#title").val();
var text = $("#text").val();
var summary = $("#description").val();
var keywords = $("#keywords").val();
var price = $("#price").val();
var stock = $("#stock").val();
var brand = $("#brand").val();

var product = product;

$.ajax({
    url :   "action.php",
    method: "POST",
    data    :   {updateProductRecord:1,pid:product,title:title,text:text,summary:summary,keywords:keywords,price:price,stock:stock,brand:brand},
    success :   function(data){
        $(".row").append(data);
    },
});

}

PHP：

if(isset($_POST['updateProductRecord'])) {

$product_id = $_POST['pid'];
$title = $_POST["title"];
$category = $_POST["brand"];
$text = $_POST['text'];
$summary = $_POST['summary'];
$keywords = $_POST['keywords'];
$price = $_POST['price'];
$stock = $_POST['stock'];
$date=date('d.m.y h:i:s');


$sql = "UPDATE products SET `title`='$title', `desc`='$summary', `price`='$price', `brand`='$category', `keywords`='$keywords', `stock`='$stock', `text`='$text' WHERE id='$product_id'";

$run_query = mysqli_query($connect,$sql);
if($run_query){

    echo "
        <div class='alert alert-success'>
            <a href='#' class='close' data-dismiss='alert' aria-label='close'>&times;</a>
            <b>Your post has been submitted.!</b>
        </div>
        ";
}
else {

    echo "
        <div class='alert alert-warning'>
            <a href='#' class='close' data-dismiss='alert' aria-label='close'>&times;</a>
            <b>Oops, something went wrong..</b>
        </div>
        ";
}

}

Answer 1

更改jQuery文本字段值

var text = $("#newPostText").val();