Laravel 调用非对象上的成员函数
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Laravel Call to a member function on a non-object
提问by Rafael
I have searched on stack and found many questions like this...
我在堆栈上搜索过，发现了很多这样的问题......
However, I am sure that I am using an object and have verified that in the dump.
但是，我确信我正在使用一个对象并在转储中验证了它。
Error
错误
>16: Call to a member function getRealPath() on a non-object
>16: Call to a member function getRealPath() on a non-object
Line 16
16号线
$image->getRealPath();
$image->getRealPath();
Var Dump $image
变量转储 $image
object(Symfony\Component\HttpFoundation\File\UploadedFile)#9 (7) { ["test":"Symfony\Component\HttpFoundation\File\UploadedFile":private]=> bool(false) ["originalName":"Symfony\Component\HttpFoundation\File\UploadedFile":private]=> string(41) "5a862f92da957da3e0208357ce006afd_970x.jpg" ["mimeType":"Symfony\Component\HttpFoundation\File\UploadedFile":private]=> string(10) "image/jpeg" ["size":"Symfony\Component\HttpFoundation\File\UploadedFile":private]=> int(346047) ["error":"Symfony\Component\HttpFoundation\File\UploadedFile":private]=> int(0) ["pathName":"SplFileInfo":private]=> string(24) "C:\xampp\tmp\phpA5C7.tmp" ["fileName":"SplFileInfo":private]=> string(11) "phpA5C7.tmp" }
If I dump $image->getRealPath();I get a string(24) "C:\xampp\tmp\phpE193.tmp"which is what I want. I have no idea why it's saying $image is not an object when clearly it is.
如果我转储，$image->getRealPath();我会得到一个string(24) "C:\xampp\tmp\phpE193.tmp"我想要的。我不知道为什么它说 $image 显然不是一个对象。
Controller
控制器
# Process Images - Queue?
if ($filesUploaded) {
$images = Input::file('images');
foreach ($images as $image) {
# Record Creation
$record = new Image;
$record->user_id = Auth::id();
$record->ad_id = $adId;
$record->name = Str::random();
$record->save();
# Create TN
$record->createTN($image);
# Create Smalls
# Create Larges
}
}
Image class
图片类
public function createTN($image) {
$PUBLIC_PATH = public_path();
# Load Zebra Image Library
require_once $PUBLIC_PATH.'/uploads/Zebra_Image.php';
$destinationPath = $PUBLIC_PATH.'/uploads/thumbnails/';
$tn = new Zebra_Image();
$tn->source_path = $image->getRealPath();
$tn->target_path = $destinationPath.$this->name.'_sm.jpg';
$tn->jpeg_quality = 60;
$tn->preserve_aspect_ratio = true;
$tn->enlarge_smaller_images = true;
$tn->resize(100, 100, ZEBRA_IMAGE_CROP_CENTER);
}
回答by Alan Storm
Looking at the original loop
查看原始循环
$images = Input::file('images');
foreach ($images as $image) {
# Record Creation
$record = new Image;
$record->user_id = Auth::id();
$record->ad_id = $adId;
$record->name = Str::random();
$record->save();
# Create TN
$record->createTN($image);
# Create Smalls
# Create Larges
}
It looks like you're processing a request with more than a single file upload. The form, javascript code, or flash uploader code you're using to post to this Laravel route looks like it's sending at least one file (or perhaps a blank file entry) that Laravel's request object can't process into an object. A loop like this might reveal more about what's going on
看起来您正在处理具有多个文件上传的请求。您用来发布到此 Laravel 路由的表单、javascript 代码或 Flash 上传器代码看起来至少发送了一个 Laravel 的请求对象无法处理的文件（或者可能是一个空白文件条目）。像这样的循环可能会揭示更多关于正在发生的事情
foreach($images as $image)
{
var_dump($image);
}
Most of those will be Symfony\Component\HttpFoundation\File\UploadedFileobjects, but I bet at least one of them is a null. You could also try var_dump($_FILES)and look for any obvious errors or empty entries -- maybe something that's too large for the server to process?
其中大部分将是Symfony\Component\HttpFoundation\File\UploadedFile对象，但我敢打赌其中至少一个是null. 您还可以尝试var_dump($_FILES)查找任何明显的错误或空条目——也许是服务器无法处理的内容？
The reason you couldn't use lukasgeiter'ssolution is the Symfony\Component\HttpFoundation\File\UploadedFileclass has PHP's internal SplFileInfoclass as an ancsetor, and until very recentlyPHP had a bug/feature where a programmer couldn't cast SplFileInfoas a boolean (using the using !converts the variable to a boolean type for the comparision)
您不能使用lukasgeiter 的解决方案的原因是Symfony\Component\HttpFoundation\File\UploadedFile该类具有 PHP 的内部SplFileInfo类作为祖先，直到最近PHP 有一个错误/功能，程序员无法将其转换SplFileInfo为布尔值（使用 using!将变量转换为布尔值输入比较）
Finally, Laravel relies on Symfony components for its file upload functionality. I believe you can find the Symfony\Component\HttpFoundation\File\UploadedFileinstantiations here
最后，Laravel 依赖 Symfony 组件来实现其文件上传功能。我相信你可以在Symfony\Component\HttpFoundation\File\UploadedFile这里找到实例
#File: vendor/symfony/http-foundation/Symfony/Component/HttpFoundation/FileBag.php
protected function convertFileInformation($file)
{
if ($file instanceof UploadedFile) {
return $file;
}
$file = $this->fixPhpFilesArray($file);
if (is_array($file)) {
$keys = array_keys($file);
sort($keys);
if ($keys == self::$fileKeys) {
if (UPLOAD_ERR_NO_FILE == $file['error']) {
$file = null;
} else {
$file = new UploadedFile($file['tmp_name'], $file['name'], $file['type'], $file['size'], $file['error']);
}
} else {
$file = array_map(array($this, 'convertFileInformation'), $file);
}
}
return $file;
}
So if you're a debug at the source sort of person, start here.
因此，如果您是源头调试人员，请从这里开始。
回答by lukasgeiter
One of your images probably is not set. So when you check (the first) with a var_dumpeverything seems alright. But when the full loop runs it throws an exception on the second, third, ... iteration.
您的其中一张图片可能未设置。因此，当您使用 a 检查（第一个）时，var_dump一切似乎都很好。但是当完整循环运行时，它会在第二次、第三次、...迭代中抛出异常。
This should fix it:
这应该解决它：
foreach ($images as $image) {
if(!is_object($image)) continue;
# Record Creation
$record = new Image;
$record->user_id = Auth::id();
$record->ad_id = $adId;
$record->name = Str::random();
$record->save();
# Create TN
$record->createTN($image);
# Create Smalls
# Create Larges
}
