全部。我正在尝试学习 Laravel,并且正在努力上传图片。我收到以下错误:
“在非对象上调用成员函数 getClientOriginalName()”
我正在使用这些包:
"anahkiasen/former": "dev-master",
"intervention/image": "dev-master",
"intervention/imagecache": "2.*"
使用 SO 后,我已验证以下内容不会导致上述错误:
我的表格是:
<form enctype="multipart/form-data" accept-charset="utf-8" class="form-horizontal" id="create_form" method="POST" action="/elements">
<div class="control-group"><label for="img[]" class="control-label">Upload Image</label><div class="controls"><input multiple="true" class="myclass" accept="image/gif|image/jpeg|image/png" id="img[]" type="file" name="img[]"></div></div>
<div class="form-actions"><input class="btn-large btn-primary btn" type="submit" value="Submit"> <input class="btn-large btn-inverse btn" type="reset" value="Reset"></div>
<input type="hidden" name="_token" value="B0AJ0Y5LMrMng6CsePeZfNSvRQ0KexowOGTK99Gm">
</form>
产生错误的代码是:
$image = Input::file('img');
$filename = $image->getClientOriginalName();
print_($filename);
如果我使用以下方法打印对象:
打印_r($图像);
...然后我得到:
Array
(
[0] => Symfony\Component\HttpFoundation\File\UploadedFile Object
(
[test:Symfony\Component\HttpFoundation\File\UploadedFile:private] =>
[originalName:Symfony\Component\HttpFoundation\File\UploadedFile:private] => storageunit.jpg
[mimeType:Symfony\Component\HttpFoundation\File\UploadedFile:private] => image/jpeg
[size:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 8734
[error:Symfony\Component\HttpFoundation\File\UploadedFile:private] => 0
[pathName:SplFileInfo:private] => /tmp/php9AU1OE
[fileName:SplFileInfo:private] => php9AU1OE
)
)
所有这些在我看来都是正确的,所以我很困惑。
如果有人对接下来要尝试什么有任何想法,我将不胜感激。
最佳答案
看看打印出来的是什么。 $image 是一个对象数组,而不是一个对象。尝试:
$filename = $image[0]->getClientOriginalName();
关于php - 拉维尔 "File Upload Error Call to a member function getClientOriginalName() on a non-object",我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27452716/