[英]In Yii how to display image as output whose path is stored in database
I am creating a project in Yii. 我正在Yii创建一个项目。 I am having table as Dnymedia
with fields as, 我有Dnymedia
表,字段为
Dnymedia Dnymedia
-mediaId -mediaId
-mediaPath -mediaPath
-contentTitleId -contentTitleId
Suppose I have media path stored as=" F:\\Documents and Settings\\All Users\\Documents\\My Pictures\\Sample Pictures
" 假设我的媒体路径存储为=“ F:\\Documents and Settings\\All Users\\Documents\\My Pictures\\Sample Pictures
”
How do I display this image as output? 如何显示此图像作为输出? I don't need to display it in view. 我不需要在视图中显示它。 I just want to display the image as output of the controller's action in Yii. 我只想在Yii中将图像显示为控制器动作的输出。
instead of read file use file_get_contents 代替读取文件,使用file_get_contents
header('Content-Type: image/jpeg');
echo file_get_contents('F:\Documents and Settings\All Users\Documents\My Pictures\Sample.jpg');
die;
This worked for me.Please check it.
public function actionTest()
{
header('Content-Type: image/jpeg');
readfile('http://localhost/myAppetite/images/bodyBg.jpg');
die;
}
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