php - 包含存储为 blob 的图像的 JSON 数组

Question

我使用以下代码从 mySQL 表创建了一个 json 数组:

$sql = "SELECT * FROM new";

if ($result = mysqli_query($con, $sql))
{
$resultArray = array();
$tempArray = array();

while($row = $result->fetch_object())
{
$tempArray = $row;
array_push($resultArray, $tempArray);
}

echo json_encode($resultArray);
}
function_exists('json_encode');

这仅适用于文本输入，当我尝试添加图像(存储为 BLOB)时，JSON 输出完全消失。我很清楚将图像存储在数据库中并不是一个很好的做法，但我需要以这种方式进行。谁能帮我解决这个问题？如何将图像作为 JSON 数组的一部分？

Answer 1

根据评论，这将是一个由两部分组成的答案。

1) 文件上传时，要存放到指定路径。然后应将路径和文件名存储在数据库中。以下示例未经测试，应根据您的环境进行调整:

<?php
// PHP File Upload Handler
// Store File to specific path and Insert Path and details to DB

$uploaddir = '/var/www/uploads/';
$uploadfilepath = $uploaddir . basename($_FILES['userfile']['name']);
$uploadfilename = basename($_FILES['userfile']['name'])
$uploadfiletype = $_FILES['userfile']['type'];
$uploadfilesize = $_FILES['userfile']['size'];

if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfilepath)) {
    $mysqli = new mysqli("localhost", "my_user", "my_password", "world");
    $stmt = $mysqli->prepare(""INSERT INTO mediaTable (name, path, type, size) VALUES (?, ?, ?, ?)");
    $stmt->bind_param("sssi", $uploadfilename, $uploadfilepath, $uploadfiletype, $uploadfilesize);
    $stmt->execute();
    $stmt->close();
    $mysqli->close();
    $results = array(
        "status" => "success",
        "name" => $uploadfilename,
        "type" => $uploadfiletype,
        "size" => $uploadfilesize,
        "path" => $uploadfilepath
    );
} else {
    $results = array(
        "status" => "error",
        "Could not save uploaded file."
    );
}
?>

2) 当我们想要检索文件并通过 JSON 返回它时，它看起来像这样(同样，未经测试):

<?php
// Get Media file (POST input) from database and return data via JSON

if(isset($_POST['fn'])){
    $filename = $_POST['fn'];
    $results = array();
    $mysqli = new mysqli("localhost", "my_user", "my_password", "world");
    $stmt = $mysqli->prepare("SELECT name, path, type, size FROM mediaTable WHERE name=?");
    $stmt->bind_param("s", $filename);
    $stmt->execute();
    $stmt->bind_result($fname, $fpath, $ftype, fsize);
    while ($stmt->fetch()) {
         $results[] = array("name" => $fname, "path" => $fpath, "type" => $ftype, "size" => $fsize, "base64" => '');
    }
    $stmt->close();
    $mysqli->close();

    foreach($results as $file){
        $file['base64'] = file_get_contents($file['path']);
    }
} else {
    $results["error"] = "No file submitted";
}

if(isset($results)){
    echo json_encode($results);
}
?>