Wordpress 如何使用 Ajax 显示成功 insert_post() 返回的新数据？

Question

成功 Ajax 插入条目后，我想查看同一条目的 ID 和 url 是什么，并以模式 window 显示它而不刷新页面

有什么方法可以从 success: function (response) {} 中获取此数据？这是我必须使用 ajax 创建一个新条目的代码，它工作得很好：

  <script>
        
        $("#enquiry_email_form").on("submit", function (event) {
            event.preventDefault();
 
            var form= $(this);
            var ajaxurl = form.data("url");
            var detail_info = {
                post_title: form.find("#post_title").val(),
                post_description: form.find("#post_description").val()
            }
 
            if(detail_info.post_title === "" || detail_info.post_description === "") {
                alert("Fields cannot be blank");
                return;
            }
 
            $.ajax({
 
                url: ajaxurl,
                type: 'POST',
                data: {
                    post_details : detail_info,
                    action: 'save_post_details_form' // this is going to be used inside wordpress functions.php// *esto se utilizará dentro de las functions.php*
                },
                error: function(error) {
                    alert("Insert Failed" + error);
                },
      success: function(response) {
                modal.style.display = "block";  * abre la ventana modal*

                body.style.position = "static";
                body.style.height = "100%";
                body.style.overflow = "hidden";
                     
                     
                }
            });
        })
    </script>
<button id="btnModal">Abrir modal</button> 
<div id="tvesModal" class="modalContainer">
 <div class="modal-content">
 <span class="close">×</span> <h2>Modal</h2> * Ventana modal mostrar le url y ID generado *
 <p><?php ***echo $title_post, $url, $ID*** ?></p> 
 
 </div>
 </div>

存档funtions.php

 function save_enquiry_form_action() {
     
        $post_title = $_POST['post_details']['post_title'];
        $post_description = $_POST['post_details']['post_description'];
        $args = [
            'post_title'=> $post_title,
            'post_content'=>$post_description,
            'post_status'=> 'publish',
            'post_type'=> 'post',
            'show_in_rest' => true,
            'post_date'=> get_the_date()
        ];
     
        $is_post_inserted = wp_insert_post($args);
     
        if($is_post_inserted) {
            return "success";
        } else {
            return "failed";
        }
    }

Answer 1

当您使用 wp_insert_post^Docs 功能时，它将 return post id。

It returns the post ID on success. The value 0 or WP_Error on failure.

首先，您可以启动一个空数组并调用它，比方说，$response 并根据 wp_insert_post 函数的 returned 值填充它。
然后，我们也可以使用 id 来获取固定链接，使用 get_permalink^Docs.
最后，我们可以使用 wp_send_json_success^Docs 函数将该数组发送回客户端。

所以你在 php 端的代码应该是这样的：

function save_enquiry_form_action() {

    $response = array(
    'error'             => '',
    'success'           => '',
    'post_id'           => '',
    'post_url'          => '',
  );
     
    $post_title       = sanitize_text_field($_POST['post_details']['post_title']);
    // Note we could have used 'sanitize_title()' function too!

    $post_description = sanitize_textarea_field($_POST['post_details']['post_description']);

    $args = array(
            'post_title'   => $post_title,
            'post_content' => $post_description,
            'post_status'  => 'publish',
            'post_type'    => 'post',
            'show_in_rest' => true,
            'post_date'    => get_the_date()
    );

    $post_id = wp_insert_post($args);

    if($post_id){
        $response['success']       = true;
        $response['error']         = false;
        $response['id']            = $post_id;
        $response['post_url']      = get_permalink($post_id);
    }else{
        $response['success']       = false;
        $response['error']         = true;
    }

    wp_send_json_success($response);

    exit;

}

注：

我已经使用 sanitize_text_field^Docs function to sanitize the $_POST['post_details']['post_title'] value and sanitize_textarea_field^Docs 函数来清理 $_POST['post_details']['post_description'] 值。
当您在客户端收到响应时，您可以检查 $response['success'] 和 $response['error'] 值。

在javascript那边

如您在以下屏幕截图中所见，数据 return 为 data object。要访问数据，您可以使用 response.data.success、response.data.error、response.data.id 和 response.data.post_url.

Wordpress 如何使用 Ajax 显示成功 insert_post() 返回的新数据？

Wordpress how to use Ajax to show new data returned from successful insert_post()?

javascript

php

ajax

wordpress

custom-wordpress-pages