Yii2 Display image stored in BLOB database field

Question

Greetings

I need to display images stored in BLOB database fields in a LISTVIEW or similar widget from Yii2

I have the actionCreate that saves the image in table named honra as a BLOB

public function actionCreate()
{
    $model = new Honra();

    if ($model->load(Yii::$app->request->post()) && $model->save()) {
        $file = UploadedFile::getInstance($model,'binaryfile');
        $model->binaryfile=$file;
        return $this->redirect(['view', 'id' => $model->id]);
    } else {
        return $this->render('create', [
            'model' => $model,
        ]);
    }
}

I also have the model named Honra

class Honra extends \yii\db\ActiveRecord{

public static function tableName()
{
    return 'honra';
}

/**
 * @inheritdoc
 */
public function rules()
{
    return [
        [['nome', 'texto'], 'required'],
        [['texto', 'binaryfile'], 'string'],
        [['ativo'], 'integer'],
        [['nome'], 'string', 'max' => 255],
        [['fileName'], 'string', 'max' => 100],
        [['fileType'], 'string', 'max' => 50]
    ];
}

/**
 * @inheritdoc
 */
public function attributeLabels()
{
    return [
        'id' => Yii::t('app', 'ID'),
        'nome' => Yii::t('app', 'Nome'),
        'texto' => Yii::t('app', 'Texto'),
        'fileName' => Yii::t('app', 'File Name'),
        'fileType' => Yii::t('app', 'File Type'),
        'binaryfile' => Yii::t('app', 'Binaryfile'),
        'ativo' => Yii::t('app', 'Ativo'),
    ];
}

}

The images are successfully stored in the table field named binaryfile

And i need to display each blob image stored in database in a view named view2 inside a LISTVIEW or similar widget

Anyone knows what block of code i need to put in view2 and the HonraController to achieve this ??

EDIT

SOLVED

HERE IS THE CODE THAT SOLVED THE ISSUE, NOW EVERYTHING IS WORKING WELL

public function actionCreate()
{
    $model = new Honra();

     if (Yii::$app->request->isPost) {

            $model->file = UploadedFile::getInstance($model, 'file');
            $filePath = 'uploadsimg/'  . $model->file->baseName . '.' . $model->file->extension;
            $model->fileName = $filePath;
            $model->save();

            $model->file->saveAs('uploadsimg/' . $model->file->baseName . '.' . $model->file->extension);
            if ($model->load(Yii::$app->request->post()) && $model->save()) {
                return $this->redirect(Url::toRoute(['create']));
            }
        }
        return $this->render('create', ['model' => $model]);
}

MANY THANKS FOR THE HELP MIHAI

Well you have to make another call to an action in a controller that will give you the image. That is for each image in the list. You cannot directly show it in the list I believe as html does not work that way. Why did you choose to put the images in the DB? It will make your db huge very fast, you are much better to create a unique link (maybe with a hash so people do not guess the path of other images) and just store the path in the db. — Mihai P.
– Mihai P., Commented Dec 23, 2014 at 23:42
More opinions on keeping the images in the db stackoverflow.com/questions/3748/… — Mihai P.
– Mihai P., Commented Dec 23, 2014 at 23:43
I put the images in the db because i want a listview to store and show the name of a person and also show its photo. The photo as to appear visible from the first moment, not a link. — André Castro
– André Castro, Commented Dec 23, 2014 at 23:48
So, you are telling me that it is impossible to display in a listview for each person in the database the matching photo stored as a BLOB ? — André Castro
– André Castro, Commented Dec 23, 2014 at 23:53
Well, you can embed the images directly into html. See here techerator.com/2011/12/… — Mihai P.
– Mihai P., Commented Dec 23, 2014 at 23:56

Mihai P. · Accepted Answer · 2014-12-24 21:37:00Z

1

Here is the official documentation on Yii2 https://github.com/yiisoft/yii2/blob/master/docs/guide/input-file-upload.md It is quite simple.

some comments:
1) the pics might be private and it might be a big problem if somebody guesses the path of others. In this case you can put the file someplace NOT in the web folder. If you do this then you have to use the assets to show them assetManager->getPublishedUrl('@app/folder')?>/fav.ico
2) if the pics are not so sensitive then you can just save them someplace in the web folder.

As you can see you might have problems with uploading the same file name 2 times, as 1 will override the other. You can always rename the file, I would actually change their code to something like this

        if ($model->file && $model->validate()) {
$model->file = UploadedFile::getInstance($model, 'file');
            $filePath = 'uploads/' . hash('ripemd160', microtime()) . $model->file->baseName . '.' . $model->file->extension;
            $model->fileName = $filePath;
            $model->save();

            $model->file->saveAs('uploadsimg/' . $model->file->baseName . '.' . $model->file->extension);
            if ($model->load(Yii::$app->request->post()) && $model->save()) {
                return $this->redirect(Url::toRoute(['create']));
            }
        }

edited Dec 24, 2014 at 21:37

answered Dec 24, 2014 at 0:23

Mihai P.

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5 Comments

André Castro Over a year ago

Many thanks for the answer, Mihai. I will give it a try in a short time:)

André Castro Over a year ago

Hi Mihai, i edited your code so you can see my actual state and problem.

André Castro Over a year ago

SOLVED. Please see my complete solution in the above EDIT. Many thanks to Mihai that pointed the right path to follow.

Mihai P. Over a year ago

I would still do the hash part mate, or add the id of the model before the name. Sooner or later you will overwrite an image because the name will be the same.

André Castro Over a year ago

Yes, you are right. I will implement the hash to save in local folder. I intended to do it later. Many thanks for the reminder.

Rob Mies · Accepted Answer · 2016-09-06 20:21:58Z

1

The shortest way I found to display a image from a BLOB field (in a MySQL database) was using this code (in my view):

$sql = "SELECT FieldName FROM TableName"; //optional where statements etc.
$connection = Yii::$app->getDb();
$command = $connection->createCommand($sql);
$imgLogo = $command->queryScalar();
echo '<img src="data:image/jpeg;base64,'.base64_encode($imgLogo).'"/>';

Note that in this example I have only one row in the table 'TableName' as this example was used to retrieve the logo from the company. And there is only one company stored in that table.

answered Sep 6, 2016 at 20:21

Rob Mies

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