Php 使用cURL将表单数据发布到服务器

我有一张简单的表格。。用户将在其中输入数据，我希望在单击“保存”时将数据发送到服务器

这是我的表格：

<form method="POST" name="admin_form" action=" ">
                    <div class="form-row">
                        <div class="form-group col-md-12">
                            <label for="inputEmail4">Product Name</label>


                            <input type="name" class="form-control" name="name" id="inputproductname"
                                placeholder="Product name">
                        </div>
                        <div class="form-group col-md-12">
                            <label for="inputPassword4">Description</label>
                            <textarea class="form-control" name="description" id="exampleFormControlTextarea1"
                                rows="3"></textarea>
                        </div>
                    </div>

                    <!-- start select shop -->
                    <div class="form-group">
                        <label for="exampleFormControlSelect2">Select shop</label>
                        <?php 


                $q = "SELECT * FROM heroku_50fc799819d.shops ORDER BY  name ASC";        
                $r = mysqli_query($conn,$q);
                    ?>
                        <select class="form-control" name="shops" id="shops" required>
                            <!-- <option value="0">Select category.</option> -->
                            <?php while ($data = mysqli_fetch_array($r)){
                $id = $data['id'];
                $name = $data['name']; 
                ?>
                            <option value="<?php echo $id;?>"> <?php echo $name; ?> </option>
                            <?php } ?>
                        </select>
                    </div>

                    <!-- end shop select -->

                    <!-- image select -->
                    <form>
                        <div class="form-group">
                            <label for="exampleFormControlFile1">Upload image product</label>
                            <input type="file" class="form-control-file" id="exampleFormControlFile1">
                        </div>
                    </form>

                    <!-- end image select -->

                    <!-- start category select -->
                    <div class="form-group">
                        <label for="exampleFormControlSelect1">Select one or multiple categories</label>

                        <?php 
                $q = "SELECT * FROM heroku_50fc799819d.categories ORDER BY  name ASC";        
                $r = mysqli_query($conn,$q);
                    ?>
                        <select multiple class="form-control" name="category" id="product_category" required>
                            <!-- <option value="0">Select category.</option> -->
                            <?php while ($data = mysqli_fetch_array($r)){
                $id = $data['id'];
                $name = $data['name']; 
                ?>
                            <option value="<?php echo $id;?>"> <?php echo $name; ?> </option>
                            <?php } ?>
                        </select>
                    </div>


                    <!-- end category select -->


                    <div class="form-group">
                        <label for="inputPrice">Price</label>
                        <input type="text" class="form-control" name="price" id="inputPrice" placeholder="12000">
                    </div>
                    <div class="form-group">
                        <label for="inputPrice">Discount</label>
                        <input type="text" class="form-control" name="discount" id="inputPrice" placeholder="10">
                    </div>

                    <div class="form-group">
                        <label for="inputPrice">Qty</label>
                        <input type="text" class="form-control" name="qty" id="inputPrice" placeholder="10">
                    </div>
                    <button type="submit" name="submit" class="btn btn-primary">
                        Save
                    </button>
                </form>

我当前正在从服务器返回此错误消息。那么，我如何通过单击save来传递表单中的数据呢？？

---

{“错误”：“未传递任何产品名称”}

这是否回答了您的问题？是的，我找到了如何处理错误的答案…所以我的问题解决了一半。。。。问题是当用户单击save（保存）按钮时，如何从表单中获取变量值在您尝试使用json上传图像时-图像是二进制数据，json不是二进制安全的，如果您处理二进制数据，请不要使用json，使用

多部分/表单数据

或

应用程序/x-www-form-urlencoded

，它们都完全能够处理二进制数据（如图像）（多部分/表单数据的带宽开销低于x-www-form-urlencoded tho，后者在最坏的情况下每字节二进制数据的开销约为66%，多部分/表单数据每字节的开销约为0%，它只有一些头开销，但没有每字节的开销）@hanshenrik非常好的观察结果…实际上我计划将图像发送到firebase存储并从firebase收集图像URL，这就是我将发送的内容-暂时忽略图像位..关于如何发送表单数据的任何建议…我对此非常了解

$post_data = new \stdClass();
$final_post_data = new \stdClass();
$post_data ->name = $_POST['name']?? '';
$post_data ->description = $_POST['description']?? '';
$post_data ->shopId = $_POST['shops']?? '';
$post_data ->categories = [2, 16, 278];
$post_data ->price = $_POST['price']?? '';
$post_data ->quantity = $_POST['qty']?? '';

$post_dataJSON = json_encode($post_data);
$final_post_data ->body = $post_dataJSON;
$final_post_dataJSON = json_encode($final_post_data);
//initialize
$ch = curl_init('https://admin2go.herokuapp.com/api/admin/products/create');
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
//adding the post variable to the request
curl_setopt($ch, CURLOPT_POSTFIELDS, $final_post_dataJSON);
//Return output instead of outputting it
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_HTTPHEADER, array(                                                                          
    'Content-Type: application/json',                                                                                
    'Content-Length: ' . strlen($final_post_dataJSON)
    )                                                                       
);  

$result = curl_exec($ch);

if ($result === FALSE) {
    echo "cURL Error: " . curl_error($ch);
}
//close and free up the curl handle
curl_close($ch);
//Display row output
print_r($result);

?>