Question

在我的应用程序中，如果用户不添加图像，用户创建文章并向其添加图像，该应用程序必须在谷歌图像中搜索它并将该图像添加到用户文章。但是当我试图从谷歌获取图像。我得到这个错误：

     AbstractDecoder.php line 302:
Image source not readable

控制器方法：

 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);

        }

    }

Answer 1

$output = json_decode(curl_exec($ch));

返回带有结果的json对象。如果你想获得第一个图片网址添加

$output = json_decode(curl_exec($ch));
if(!empty($responseData->results))
    $output = $output->responseData->results[0]->url;
else {
    die('no image found');
}

Answer 2

我找到了解决方案

else{
        $url = 'https://ajax.googleapis.com/ajax/services/search/images?v=1.0&rsz=1&q=' . $request->title;
        $url = file_get_contents($url);
        $file = json_decode($url);
        $file = $file->responseData->results[0]->url;
        $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);
    }

Laravel iIntervention Image。图像源不可读

2 个答案: