Question

我无法弄清楚为什么我会收到错误＆＃34; JSON.parse：意外字符＆＃34;在Firefox控制台..看起来问题是围绕jQuery.parseJSON（数据））;在comment.js文件..如果我删除我的PHP文件中的两个if语句一切正常，但这不是我需要的..

谢谢！

comment.php

<?php

require_once ("../modules/comments.php");
require_once ("../includes/db_connect.php");

 if(isset($_POST['task']) && $_POST['task'] == 'comment_insert' ) {

    $userId = (int)$_POST['userId'];
    //$comment = addslashes(str_replace("\n", "<br>", $_POST['comment']));

    $comment = strip_tags($_POST['comment']); // remove html tags
    $comment = ucfirst(strtolower($comment)); 
    //$comment = addslashes(str_replace("\n", "<br>"));

    $std = new stdClass();
    $std->commentId = 24;
    $std->userId = $userId;
    $std->comment = $comment;
    $std->userName = "John Stu";
    $std->profile_img = "https://s.ytimg.com/yts/img/avatar_48-vfllY0UTT.png";

     if(class_exists('Comments') && class_exists('Subscribers')) {

        $commentInfo = Comments::insertComments($comment, $userId);

        if($commentInfo != null) {


        }

    }   
        echo json_encode($std); 



 } else {

   //header("Location: /");

 }

?>

comment.js

$(document).ready(function(){

    $('#post-comment-btn').click(function(){
        comment_post_btn_click();

    });

});


function comment_post_btn_click(){

var _comment = $('#comment-post-text').val();
        var _userId = $('#userId').val();
        var _userName = $('#userName').val();

        if(_comment.length > 0 && _userId != null) {

        $.post("ajax/comment_insert.php", {
            task    : "comment_insert",
            userId  : _userId,
            comment : _comment,

        }).done(function(data) {

         comment_insert(jQuery.parseJSON(data));    
         console.log("ResponseText:" + data);

            });

            console.log(_comment + " Username: " + _userName + " User Id: " + _userId);

        } else {

            console.log("The text area is empty..");

        }

        $('#comment-post-text').val("");

}

function comment_insert(data) {


    var t = '';

    t += '<li class="comment-holder" id="_'+data.comment_id+'">';                       
    t += '<div class="user-img">';
    t += '<img src="'+data.profile_img+'">';
    t += '</div>';
    t += '<div class="comment-body">';
    t += '<h3 class="username-field">'+data.userName+'</h3>';
    t += '<div class="comment-text">'+data.commen+'</div>';
    t += '</div>';
    t += '<div class="comment-buttons-holder">';
    t += '<ul>';
    t += '<li class="delete-btn">X</li>';
    t += '</ul>';
    t += '</div>';
    t += '</li>';

    $('.comments-holder-ul').prepend(t);
}

Answer 1

您的PHP文件中看起来有一些错误/警告。试试这个， Inspect > Network >Select XHR并进行AJAX调用。现在，您将在Response标签中看到PHP响应。（更好地使用谷歌浏览器）

因为PHP错误/警告包含了从"<"开始的HTML标记。（示例<div>Undefined Index</div>）

Answer 2

首先清理输出缓冲区，然后终止当前脚本。尝试以下代码：

<强> comment.php

require_once ("../modules/comments.php");
require_once ("../includes/db_connect.php");

 if(isset($_POST['task']) && $_POST['task'] == 'comment_insert' ) {

    $userId = (int)$_POST['userId'];
    //$comment = addslashes(str_replace("\n", "<br>", $_POST['comment']));

    $comment = strip_tags($_POST['comment']); // remove html tags
    $comment = ucfirst(strtolower($comment)); 
    //$comment = addslashes(str_replace("\n", "<br>"));

    $std = new stdClass();
    $std->commentId = 24;
    $std->userId = $userId;
    $std->comment = $comment;
    $std->userName = "John Stu";
    $std->profile_img = "https://s.ytimg.com/yts/img/avatar_48-vfllY0UTT.png";

     if(class_exists('Comments') && class_exists('Subscribers')) {

        $commentInfo = Comments::insertComments($comment, $userId);

        if($commentInfo != null) {


        }

    }   

    // Clean the output buffer
    ob_clean();

    echo json_encode($std); 

    // Terminate the current script
    die();

 } else {

   //header("Location: /");

 }

?>

Answer 3

如果要返回json数据，则必须在添加最后一个参数时将其指示给$ .post函数（否则，data将不是json对象）...

$.post("ajax/comment_insert.php", {
    task : "comment_insert",
    userId  : _userId,
    comment : _comment,
},,"json").done(...

更改并检查错误是否仍然存在......

JSON.parse：意外的字符问题

3 个答案: