php - 尝试将图像上传到数据库，然后从数据库中显示。遇到 fatal error

Question

错误:

Fatal error: Uncaught Error: Call to a member function bind_param() on boolean in C:\xampp\htdocs\FinalProject\upload-profilePic.php:22 Stack trace: #0 {main} thrown in C:...

我的代码:

上传图片

<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
if (file_exists($_FILES['upload_file']['tmp_name'])) {
    // get temp name and file name
    $tmpName = $_FILES['upload_file']['tmp_name'];
    $fileName = $_FILES['upload_file']['name'];
    $type = $_FILES['upload_file']['type'];

    $handler = fopen($tmpName, 'r'); // opens image file stream
    $data = fread($handler, filesize($tmpName)); // read byte data
    fclose($handler); // close image file stream

    $stmt = $con->prepare('insert into images (image_type,
        filename, image_data) values(?, ?, ?)');
    $stmt->bind_param('sss', $type, $fileName, $data);
    $stmt->execute();
    $result = $stmt->get_result();
    if (!$result) {
        echo $con->error;
    }
  }
}

$result = $con->query('select * from images');
if ($result) {
    foreach ($result as $img) {
        $name = $img['filename'];
        $id = $img['image_id'];
        $url = "get-profilePic.php?id=$id";
        echo "<h3>$name</h3>";
        echo "<img src=\"$url\" alt=\"$name\" />\n\n";
    }
}
?>

获取图像

<?php

$imageId = $_GET['id'];
$result = $con->query("select * from images
where image_id = $imageId");

if ($result) { 
    $image = $result->fetch_object();
    $type = $image->image_type;
    $data = $image->image_data;
    header('Content-Type: ' . $type);
    echo $data;
}
?>

Answer 1

结合你的场景，问题出在:

$stmt = $con->prepare('insert into images (image_type, filename, image_data) values(?, ?, ?)');
$stmt->bind_param('sss', $type, $fileName, $data);

prepare()方法可以返回 false 并且需要检查它为什么返回 false，也许表名或列名(在 SELECT 或 WHERE 子句中)不正确？主要是您查询中的一些错误。

另外，尝试使用类似 $stmt ->error_list 的东西检查解析 SQL 时发生的错误。