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在我的應用程序用戶創建文章並向其添加圖像,如果用戶不會添加圖像,該應用程序必須在谷歌圖像中搜索並將該圖像添加到用戶文章。但是,當我試圖從谷歌獲取圖像。林得到這個錯誤:Laravel iIntervention圖像。圖像源不可讀
AbstractDecoder.php line 302:
Image source not readable
控制器的方法:
public function store(ArticleRequest $request)
{
if ($request->hasFile('file')) {
$file = Input::file('file');
$imgTitle = $request->title;
$imagePath = 'uploads/' . $imgTitle . '.jpg';
$request->image_path = $imagePath;
Article::create(array('title' => $request->title,
'body' => $request->body,
'image_path' => $imagePath));
Image::make($file)->resize(300, 200)->save($imagePath);
} else {
// $file = Input::file('file');
$imgTitle = $request->title;
$query = $imgTitle;
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, "https://ajax.googleapis.com/ajax/services/search/images?v=1.0&q=" . urlencode($query));
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$output = json_decode(curl_exec($ch));
// $file = file_get_contents($output);
curl_close($ch);
$imagePath = 'uploads/' . $imgTitle . '.jpg';
$request->image_path = $imagePath;
Article::create(array('title' => $request->title,
'body' => $request->body,
'image_path' => $imagePath));
Image::make($output)->resize(300, 200)->save($imagePath);
}
}
現在,我得到了一個未定義的變量:輸出 – qr11
添加新行'$輸出= json_decode下( curl_exec($ ch));' – mimo
如果你正確地得到了'$ output = json_decode(curl_exec($ ch)); $ output = $ output-> responseData-> results [0] - > url;'Bur now Im get:試圖獲取非對象的屬性 – qr11