Question

我正在尝试将更多评论功能放在一起。所以我的原始帖子id为comment_streamitem是原始帖子ID，而comment_id是评论ID。我在帖子上显示两条评论，然后查看所有链接显示。

我目前能够通过$ _GET JSON数组从数据库中获取所有注释，并在Firebug中显示。 comment_list_ID删除两个原始注释并删除视图链接，但不会从数据库中插入4个检索到的注释，它们应插入到div中的区域保持空白。

有人可以查看我的代码是否存在任何可能的疏忽或错误？

谢谢

<script type="text/javascript">
$(function()
{
$(".view_comments").click(function()
{
var ID = $(this).attr("id");

$.ajax({
type: "GET",
url: "viewmorecommentslink.php?comment_streamitem="+ ID,
success: function(data){
$("#comment_list_"+ID).append('<div id="comment_list_'+data['comment_streamitem']+'"><div class="stream_comment" id="comment_'+data['comment_id']+' style="margin-top:0px; background-color:#000000;"><table width=100%><tr><td valign=top width=30px><img class="stream_profileimage" style="border:none;padding:0px;display:inline;" border=\"0\" src=\"userimages/cropped'+data['comment_poster']+'.jpg\" onerror=this.src=\"userimages/no_profile_img.jpeg\" width=\"40\" height=\"40\" ></a><td valign=top align=left>\
<a href="/profile.php?username='+data['username']+'">'+data['first']+' '+ data['middle']+' '+data['last']+'</a> - <abbr class="timeago" title="'+data['comment_datetime']+'">'+data['comment_datetime']+'</abbr><div class="commentholder">'+data['comment_content']+'</div><br/>\<div id="commentactivitycontainer"><a style="cursor:pointer;" onClick=\"deletecomment('+data['comment_id']+',comment_'+data['comment_id']+');\">Delete</a><a id="likecontext_'+data['comment_id']+'" style="cursor:pointer;" onClick=\"likestatuscomment('+data['comment_id']+',this.id);\"><div style="width:80px; position:relative; float:left; left:40px" id="likescommentprint'+data['comment_id']+'">Like</div></a><div style="width:80px; position:relative; float:left; left:40px" id="likescommentprint'+data['comment_id']+'"></div></form>\
<a id="dislikecontext_'+data['comment_id']+'" style="cursor:pointer;" onClick=\"dislikestatuscomment('+data['comment_id']+',this.id);\"><div style="width:90px; position:relative;top:-0px; float:left; left:200px" id="dislikescommentprint'+data['comment_id']+'">Dislike</div></a><div style="width:90px; position:relative; top:-0px; float:left; left:200px" id="dislikescommentprint'+data['comment_id']+'"></div></form></div></div></table></div></div></div></div></table></div></div>');
$("#view"+ID).remove();
$("#comment_list_"+ID).remove();
}
});
return false;
});
});
</script>

Viewcommentslink.php

<?php
require"include/rawfeeds_load.php";

if(isset($_GET['comment_streamitem'])){
$id=$_GET['comment_streamitem'];

$check = "select comment_id,comment_poster,comment_streamitem,comment_datetime,comment_content FROM streamdata_comments WHERE comment_streamitem='$id' order by comment_id";
$check1 = mysqli_query($mysqli,$check) or die(mysqli_error($mysqli));

while($resultArr = mysqli_fetch_array($check1))
{
$json['comment_id'] = $resultArr['comment_id'];
$json['comment_poster'] = $resultArr['comment_poster'];
$json['comment_streamitem'] = $resultArr['comment_streamitem'];
$json['comment_datetime'] = $resultArr['comment_datetime'];
$json['comment_content'] = $resultArr['comment_content'];

$user=$resultArr['comment_poster'];
$check2= "SELECT * FROM user WHERE id='$user'";
$check22 = mysqli_query($mysqli,$check2);
$resultArr = mysqli_fetch_array($check22);
$json['username'] = $resultArr['username'];
$json['id'] = $resultArr['id'];
$json['first'] = $resultArr['first'];
$json['middle'] = $resultArr['middle'];
$json['last'] = $resultArr['last'];

echo json_encode($json);
}}

Answer 1

你在循环中回显json，创建了许多json字符串。你应该只回显一个json字符串 - 为了做到这一点，在循环中添加一个数组，然后在循环后输出该数组：

$temp = array();

while($resultArr = mysqli_fetch_array($check1))
{

    $json['comment_id'] = $resultArr['comment_id'];
    $json['comment_poster'] = $resultArr['comment_poster'];
    $json['comment_streamitem'] = $resultArr['comment_streamitem'];
    $json['comment_datetime'] = $resultArr['comment_datetime'];
    $json['comment_content'] = $resultArr['comment_content'];

    $user=$resultArr['comment_poster'];
    $check2= "SELECT * FROM user WHERE id='$user'";
    $check22 = mysqli_query($mysqli,$check2);
    $resultArr = mysqli_fetch_array($check22);
    $json['username'] = $resultArr['username'];
    $json['id'] = $resultArr['id'];
    $json['first'] = $resultArr['first'];
    $json['middle'] = $resultArr['middle'];
    $json['last'] = $resultArr['last'];


    //echo json_encode($json);

    $temp[]=$json;

}

echo json_encode($temp);
die();

使用Ajax查看所有评论

1 个答案: