【发布时间】:2019-07-04 12:08:03
【问题描述】:
我正在尝试将图像从 AJAX 传递到后端并上传它们,将它们的名称保存到数据库中。
我的 AJAX 代码:
var form_data = new FormData($('.updateForm')[0]);
$.ajax({
method: "POST",
url: "/updateMyPost",
data: form_data,
cache : false,
contentType: false,
processData: false,
success: function (result) {
console.log(result);
}
});
Laravel 上传文件的函数:
public function updateMyPost(Request $request){
$name = $request->input('text');
$images = $request->file('images');
foreach($images as $image){
$image_name = time() . $image->getClientOriginalName();
$path = public_path('images');
$img->move($path,$image_name);
}
}
这就是我打印出来时 $request->file("images") 给我的。
Array
(
[0] => Illuminate\Http\UploadedFile Object
(
[test:Symfony\Component\HttpFoundation\File\UploadedFile:private] =>
[originalName:Symfony\Component\HttpFoundation\File\UploadedFile:private] => images.jpg
[mimeType:Symfony\Component\HttpFoundation\File\UploadedFile:private] => image/jpeg
[error:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 0
[hashName:protected] =>
[pathName:SplFileInfo:private] => C:\xampp\tmp\phpAB87.tmp
[fileName:SplFileInfo:private] => phpAB87.tmp
)
[1] => Illuminate\Http\UploadedFile Object
(
[test:Symfony\Component\HttpFoundation\File\UploadedFile:private] =>
[originalName:Symfony\Component\HttpFoundation\File\UploadedFile:private] => iphonex-TA.jpg
[mimeType:Symfony\Component\HttpFoundation\File\UploadedFile:private] => image/jpeg
[error:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 0
[hashName:protected] =>
[pathName:SplFileInfo:private] => C:\xampp\tmp\phpAB88.tmp
[fileName:SplFileInfo:private] => phpAB88.tmp
)
)
如何获取每个图像名称并上传它们?
【问题讨论】:
-
您要上传多张图片吗?
-
是的,在这种情况下是两张图片
-
检查我的答案