如何在数据库中显示 BLOB 图像？

Question

我有一个 JSON 编码的数组，我想显示数据。有一个标题'varchar'，描述'varchar'和learnimage'blob'。我想显示这是我的 html，但出现错误。如何使用 PHP、JS 和 HTML 显示图像？我目前收到此错误：

Uncaught SyntaxError: Unexpected end of JSON input
    at JSON.parse (<anonymous>)
    at Object.success (learn.js:7)
    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 class="resources">

                    <div class="tite"><!--obj.title--></div>
                    <div class="description"><!--obj.description--></div>
                    <div class="learnimage"><!--obj.learnimage--></div>

                <p id="error" class="errormessage"></p>
                <p id="resources" class="postmessage"></p>
         </div>

PHP

<?php

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

session_start();

// Print out existing tips and articles
$query = "SELECT title, description, learnimage FROM learn ORDER BY title";
$result = mysqli_query($db_server, $query);
if (!$result)
    die("Database access failed: " . mysqli_error($db_server));
while ($row = mysqli_fetch_array($result)) {

    $learn[] = $row; 

}

mysqli_free_result($result);

require_once("db_close.php");

echo json_encode($learn);

?>

JS

$(document).ready(function() {

        $.ajax({
        type: 'GET',
        url: 'learn.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("resources");  

                output.innerHTML += '<div class="comment-container"><div class="title">'+obj.title+'</div><div class="description">'+obj.description+'</div><div class="learnimage">'+obj.learnimage+'</div></div>';

            }
        }
    });
});

Answer 1

无论出于何种原因，您的 PHP 都没有返回有效的 JSON。

只需记录您的结果并查看它。它可能会回答你的问题。

$.ajax({
    type: 'GET',
    url: 'learn.php',
    success: function(result) {
        console.log(result);
    }
});