如何解决 Uncaught SyntaxError: Unexpected token < in JSON at position 0 控制台错误？

Question

我很困扰。 我的论坛昨天和今天运行良好，然后现在我收到此错误，我的论坛不再显示或工作。 我还有另一个 js 和 php 用于发布新评论。 该评论仍在添加到我的数据库中，只是没有出现。 有谁知道这是为什么？ 这是我得到的完整错误：

VM232:1 Uncaught SyntaxError: Unexpected token < in JSON at position 0
    at JSON.parse (<anonymous>)
    at Object.success (comments.js:13)
    at u (jquery.min.js:2)
    at Object.fireWith [as resolveWith] (jquery.min.js:2)
    at k (jquery.min.js:2)
    at XMLHttpRequest.<anonymous> (jquery.min.js:2)

HTML

          <div id="content" class="content">
                <form id="forumPost" method='POST'>
                   <textarea rows="3" col="60" name="comment" placeholder="Create a Post..." id="comment"></textarea>
                   <button><input type='submit' name='submit' value='Post' class="post"></button>
                </form>
                <p id="error" class="errormessage"></p>
                <p id="allcomments" class="postmessage"></p>

                <div class="comment-container">
                    <div class="username"><!--obj.username--></div>
                    <div class="comment"><!--obj.comment--></div>
                    <div class="date"><!--obj.commDate--></div>
                    <div class="like"><!--obj.sentiment--></div>
                </div>
            </div>

JS

$(document).ready(function() {

     var comments = document.getElementById("allcomments").value; 

     //Get Storage 
                var username = window.localStorage.getItem("username");

        // Call Ajax for existing comments
        $.ajax({
        type: 'GET',
        url: 'comments.php',
        success: function(result) {
            var arr = JSON.parse(result);

            for(var i = 0; i < arr.length; i++) {
                var obj = arr[i];   

                var output = document.getElementById("allcomments");  

                output.innerHTML += '<div class="comment-container"><div class="username">'+obj.username+'</div><div class="comment">'+obj.comment+'</div><div class="date">'+obj.commDate+'</div><div class="like">'+obj.sentiment+'</div></div>';

            }

        }
    });

    return false;
}); 

PHP

<?php

require_once('checklog.php');
require_once("db_connect.php");
require_once("functions.php");

session_start();

$username = $_POST['username'];

// Print out existing comment
$query  = "SELECT comments.commDate, comments.ID, comments.username, comments.comment, users.username, comments.sentiment FROM comments LEFT JOIN users ON comments.username = users.username"; 
$result = mysqli_query($db_server, $query);
if (!$result)
    die("Database access failed: " . mysqli_error($db_server));
while ($row = mysqli_fetch_array($result)) {

    $comments[] = $row; 
//CHECK THAT THE COMMENT USERID MATCHES SESSION USER ID
if ($row['username'] == $username['username']){
$comments .=" <a href='delete_post.php?pID=".$row['ID']."'>Delete</a>";
}
        if(!isset($username["liked_" . $row['ID']])){
                $comments .= "<a href='like.php?likeid=" . $row['ID'] ."'>Like</a>";
        }else{  
                $comments .="Liked"; 
        } 
}

mysqli_free_result($result);

require_once("db_close.php");

echo json_encode($comments);

?>

Answer 1

看看这一行：

$comments[] = $row; 

您将$comments声明为数组，然后：

$comments.=" <a href='delete_post.php?pID=".$row['ID']."'>Delete</a>";

您将字符串附加到数组变量，这是绝对错误的

您从服务器返回的结果是数组和字符串的组合，结果格式无效，因此它不会转换为 js object。