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Javascript Wordpress-针对单个帖子,而不是帖子类型。AJAX/PHP_Javascript_Php_Ajax_Wordpress - Fatal编程技术网

Javascript Wordpress-针对单个帖子,而不是帖子类型。AJAX/PHP

Javascript Wordpress-针对单个帖子,而不是帖子类型。AJAX/PHP,javascript,php,ajax,wordpress,Javascript,Php,Ajax,Wordpress,我试图在单击文章标题时将单击时的单个文章信息加载到页面中。我已经设法让我的脚本在帖子数组中循环浏览标题,然后单击将帖子标题再次加载到一个div中。我希望这样做是为了在单击时加载单个帖子信息 PHP Post数组函数- add_action( "wp_ajax_success_stories", "sa_get_paginated_posts" ); add_action( "wp_ajax_nopriv_success_stories", "sa_get_paginated_posts" );

我试图在单击文章标题时将单击时的单个文章信息加载到页面中。我已经设法让我的脚本在帖子数组中循环浏览标题,然后单击将帖子标题再次加载到一个div中。我希望这样做是为了在单击时加载单个帖子信息

PHP Post数组函数-

add_action( "wp_ajax_success_stories", "sa_get_paginated_posts" );
add_action( "wp_ajax_nopriv_success_stories", "sa_get_paginated_posts" );

function sa_get_paginated_posts(){

      $args = array(
          'posts_per_page'  => -1,
          'post_type'       => 'success-stories',
          'orderby'         => 'post_date',
          'order'           => 'DESC' 
      );

      $the_query = new WP_Query( $args );
      $queryitems = $the_query->get_posts();
      $alldata = array();

      if ($queryitems){

        foreach ($queryitems as $query_item){

          $success_story = array();

          $success_story["ID"] = $query_item->ID;
          $success_story["title"] = $query_item->post_title;

          $alldata[] = $success_story;

        }

      }

      $encoded_data = json_encode($alldata);

      echo $encoded_data;

      die();

  }



function sa_get_posts($post_amount, $post_type){

    $args = array(
        'posts_per_page'  => $post_amount,
        'post_type'       => $post_type,
        'offset'          => 0,
        'orderby'         => 'post_date',
        'order'           => 'ASC'

    );
使用此函数循环通过-

      <? if ($success_stories): //success story posts ?>
            <ul class="small-block-grid-3">
                <? foreach ($success_stories as $success_story): 
                    $title = $success_story->post_title; ?>

                    <li><a data-id="<? echo $success_story->ID; ?>" class="success-extra" href="#" title=""><? echo $title; ?></a></li>

                <? endforeach; ?>
            </ul>

          <div id="success-list"></div>
          <a class="success-close" href="#" title="">CLOSE</a>

    <? endif; ?>

Javascript/AJAX调用

  var $json_response_box = $("#success-list");

  function get_ajax_success_posts() {

    //this line gets the id attribute from the link
    var current_story = $(this).attr("data-id");
    console.log(current_story);

    jQuery.ajax({
        type: "GET",
        url: "WORDPRESSDIRECTORY/admin-ajax.php",
        cache: true,
        data: {
            action: "success_stories"
        },
        dataType: "JSON",
        success: function(data) {

            var arrayLength = data.length;

            for (var i = 0; i < arrayLength; i++) {


                var success_id = data[i].ID;
                var success_title = data[i].title;

                console.log(success_id);

                story = '<ul>';
                  story += '<li>';
                    story += success_title;
                  story += '</li>';
                story += '</ul>'; 

                $json_response_box.append(story).fadeIn(1000);

            }
        },
        error: function(errorThrown) {
            console.log("fail-posts");
        }
    });
  }
var$json_response_box=$(“#成功列表”);
函数get\u ajax\u success\u posts(){
//此行从链接中获取id属性
var current_story=$(this.attr(“数据id”);
console.log(当前_故事);
jQuery.ajax({
键入:“获取”,
url:“WORDPRESSDIRECTORY/admin ajax.php”,
是的,
数据:{
行动:“成功故事”
},
数据类型:“JSON”,
成功:功能(数据){
var arrayLength=data.length;
对于(变量i=0;i”;
故事+=“
  • ”; 故事+=成功与头衔; 故事+='
  • '; 故事+=””; $json\u response\u box.append(story).fadeIn(1000); } }, 错误:函数(错误抛出){ 控制台日志(“失败日志”); } }); }
    感谢您的建议


    D

    如果我正确理解了您的问题,您只需添加另一个php函数和AJAX调用,就可以在单击时获得帖子,类似这样的内容(我还添加了nonce以添加一些secutiry,将其添加到AJAX调用中总是一个好主意)

    HTML:

    JS:


    希望这能有所帮助。

    嗯,我不太明白。你的PHP代码是什么?您可以轻松地将数据id添加到标题中,单击“通过AJAX将其提交到您的插件”,使用get_post($id)获取帖子详细信息并将其返回到AJAX成功函数……您好,我刚刚用所使用的PHP函数更新了帖子。我只是想知道如何编辑AJAX函数来针对特定的帖子。是的,关于这个问题,代码多于解释。我不明白这一点:“我希望这是加载个人张贴信息,而不是点击”。阅读代码不会告诉我任何问题。观察结果:1)函数sa_get_posts()与问题有什么关系?2) 我只需要做
    $queryitems=get_posts($args)完美答案,非常感谢
    
    <li><a data-id="<? echo $success_story->ID; ?>" data-nonce="<?php wp_create_nonce('get_post'.$success_story->ID); ?>" class="success-extra" href="#" title=""><? echo $title; ?></a></li>
    
    add_action( "wp_ajax_sa_get_post", "sa_get_post" );
    add_action( "wp_ajax_nopriv_sa_get_post", "sa_get_post" );
    
    function sa_get_post(){
        if ( ! wp_verify_nonce( $nonce, 'get_post_'.$POST['id'] ) ) {
        // This nonce is not valid.
        $response['status'] = 'fail'; 
        } else {
        // The nonce was valid.
        $post = get_post($POST['id']);
        $response['status'] = 'success'; 
        $response['post'] = $post;        
        } 
        echo json_encode($response);
        exit(); // this is needed, as otherwise the AJAX returns an extra 0       
    }
    
     $('.success-extra').on('click', function () {
         jQuery.ajax({
            type: "POST",
            url: "WORDPRESSDIRECTORY/admin-ajax.php",
            cache: false,
            data: {
                action: "sa_get_post",
                id: $(this).attr('data-id'),
                nonce: $(this).attr('data-nonce')
            },
            success: function(data) {
                var result = $.parseJSON(data);
                if (result.status != 'success') {
                   // display error messago or something
                } else {
    
                  // do whatever you want with your result.post
                }
            }
    
     })