我是PHP的新手，需要为项目做好准备 - 我已经在HTML中创建了一个图像提交表单，将div中的img更改为使用表单选择的图像 . 以下是我如何实现这一目标：
echo '<div class="img-container">';
echo '<img class="userimg" src="../images/backgroundplanet.png" />';
echo '<img class="testimg" src="" />'; //stores image from php file
echo '</div>';
echo '<div class="upload-button">Edit Profile</div>';
echo '<input class="file-upload" enctype="multipart/form-data" type="file" name="submit" accept="image/*"/>';
echo '</div>';
和
$(document).ready(function() {
var readURL = function(input) {
if (input.files && input.files[0]) {
var reader = new FileReader();
reader.onload = function (e) {
$('.userimg').attr('src', e.target.result);
}
reader.readAsDataURL(input.files[0]);
}
}
$(".file-upload").on('change', function(){
readURL(this);
});
$(".upload-button").on('click', function() {
$(".file-upload").click();
});
});
我需要将此图像上传到我为每个用户的MySql数据库中为中型BLOB创建的列 . 我尝试了很多不同的方法（试过 $sql = new mysql 等），但不知道我在做什么 .
我现在正在学习本教程 - http://www.sevenkb.com/php/how-to-insert-upload-image-into-mysql-database-using-php-and-how-to-display-an-image-in-php-from-mysql-database/并编写了以下应该将图像上传到数据库的内容：
if(!isset($_FILES['upload_image']))
{
echo '<p>Please Select Image to Upload</p>';
}
else
{
try {
upload();
echo '<p>Image Uploaded into MySQL Database as LONGBLOB Using PHP </p>';
}
catch(Exception $e)
{
echo '<h4>'.$e->getMessage().'</h4>';
}
}
function upload(){
/*** check if a file was uploaded ***/
echo '<p>you uploaded</p>';
if(is_uploaded_file($_FILES['upload_image']['tmp_name']) && getimagesize($_FILES['upload_image']['tmp_name']) != false)
{
/*** get the image info. ***/
$size = getimagesize($_FILES['upload_image']['tmp_name']);
/*** assign our variables ***/
$type = $size['mime'];
$imgfp = fopen($_FILES['upload_image']['tmp_name'], 'rb');
$size = $size[3];
$name = $_FILES['upload_image']['name'];
$maxsize = 99999999;
/*** check the file is less than the maximum file size ***/
if($_FILES['upload_image']['size'] < $maxsize )
{
/*** connect to db ***/
$dbh = new PDO("mysql:host=localhost;dbname=sqlserver", 'username', 'password');
/*** set the error mode ***/
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
/*** our sql query ***/
$stmt = $dbh->prepare("INSERT INTO img (image_type ,image, image_size, image_name) VALUES (? ,?, ?, ?)");
/*** bind the params ***/
$stmt->bindParam(1, $type);
$stmt->bindParam(2, $imgfp, PDO::PARAM_LOB);
$stmt->bindParam(3, $size);
$stmt->bindParam(4, $name);
/*** execute the query ***/
$stmt->execute();
}
else
{
/*** throw an exception is image is not of type ***/
throw new Exception("File Size Error");
}
}
else
{
// if the file is not less than the maximum allowed, print an error
throw new Exception("Unsupported Image Format!");
}
}
但是图像没有上传，也没有出现在我在数据库中创建的新列中 . 页面上显示的唯一内容是“请选择要上传的图像”
这有什么不对？最后我需要在div中回应这个问题 . 我应该怎么做呢？
尝试运行示例时：
输入我的表名时出错：
什么时候去phpadmin：
1 回答
我使用过这个链接的代码“http://www.formget.com/ajax-image-upload-php/ " to get what you needed. This code apart from saving the file into a folder will save it into a database. I have made a folder images to store image files and two php files one which shows frontend html form with javascript code and the other for upload and display. The HTML code " index.php”就是这个
对于上传脚本和上传后显示（upload.php），您可以使用
您可以从这里下载整个代码http://www.filehosting.org/file/details/569640/upload.zip . 更改数据库详细信息，并且显示的图像是表单中的最后一个插入ID . 我希望你能完成剩下的工作 .