我为用户创建了一个表单,可以上传多张图片,并将上传的图片移到“上传”状态。文件夹并将其名称存储在数据库中。这是我的代码
class ImageController extends Controller
{
public function imageAction()
{
// $mgr_com = new Commons();
if ($this->request->hasFiles() == true) {
$uploads = $this->request->getUploadedFiles();
//print_r($uploads);
$isUploaded = false;
foreach ($uploads as $upload) {
$mgr_com = new Commons();
$path = strtolower($upload->getname());
//$path = 'temp / ' . md5(uniqid(rand(), true)) . ' - ' . strtolower($upload->getname());
//print_r($path);
$sql= "call img('$path')";
$results = $mgr_com->getReadConnection()->query($sql);
($upload->moveTo($path)) ? $isUploaded = true : $isUploaded = false;
}
($isUploaded) ? die("Files successfully uploaded.") : die("Some error ocurred.");
} /*else {
die("You must choose at least one file to send. Please try again.");
}*/
}
}
上传后,所有图片都成功移至“上传”状态。但是,在数据库中它只存储一个图像名称。那么如何将所有图像名称存储在数据库中?请帮助我,谢谢你的帮助。
这是我的jquery
$(document).ready(function () {
$("#imageupload").change(function () {
if (typeof (FileReader) != "undefined") {
var dvPreview = $("#preview-image");
dvPreview.html("");
var regex = /^([a-zA-Z0-9\s_\\.\-:])+(.jpg|.jpeg|.gif|.png|.bmp)$/;
$($(this)[0].files).each(function () {
var file = $(this);
if (regex.test(file[0].name.toLowerCase())) {
var file_data = $('#imageupload').prop('files')[0];
var form_data = new FormData();
form_data.append('file', file_data);
for (var i = 0; i < file_data.length; i++) {
form_data.append("UploadedImage" + i, file_data[i]);
}
$.ajax({
url:'/image/image',
method: "POST",
data: form_data,
processData: false,
contentType: false,
multiple: true,
success: function(response){
var result= $.parseJSON(response);
if(result.success){
$("#image1").html("<img src=http://localhost/template/fileupload/"+result.userObj.file+">");
}else{
$("#error").show().html(result.message);
}
}
});
var reader = new FileReader();
reader.onload = function (e) {
var img = $("<img />");
img.attr("style", "height:100px;width: 100px");
img.attr("src", e.target.result);
dvPreview.append(img);
}
reader.readAsDataURL(file[0]);
} else {
alert(file[0].name + " is not a valid image file.");
dvPreview.html("");
return false;
}
});
} else {
alert("This browser does not support HTML5 FileReader.");
}
});
});
答案 0 :(得分:0)
尝试添加以下内容:
$results = new Resultset(null,
$mgr_com,
$mgr_com->getReadConnection()->query($sql));