简体   繁体   中英

Select Mysql to get Image url based on id

I am just creating a mysql query to get imagename from database, but i can't select on the right way. I want to get imagename based on ID of member at my website. I did try many times, but i failed many times. If anyone can help me.. Thank you very much then.

If i do this mysql query:

$GetBannerImageSql = $database->database_query("SELECT banner FROM banner_images WHERE banner_id='".$owner->user_info['user_id']."'");

$smarty->assign('bannerexists', $bannerexists);
$smarty->assign('GetBannerImage', $GetBannerImage);

I get:

resource(200) of type (mysql result)

And if i try this query:

$GetBannerImageSql = $database->database_query("SELECT banner FROM banner_images WHERE banner_id='".$owner->user_info['user_id']."'");
$GetBannerImage = $database->database_fetch_assoc($GetBannerImageSql);
var_dump($GetBannerImage);

$smarty->assign('bannerexists', $bannerexists);
$smarty->assign('GetBannerImage', $GetBannerImage);

I get this then:

array(1) { ["banner"]=> string(19) "banner-animated.gif" } 

Second is fine, but its only imagename, but i dont see id.

echo this... ->

  $GetBannerImage['banner']

It will return ... "banner-animated.gif"

Then use it in your template as needed.

Thanks!

@leo.

CREATE TABLE IF NOT EXISTS `PicturePath` (
  `ID` int(255) NOT NULL AUTO_INCREMENT,
  `Path` varchar(255) NOT NULL,
  PRIMARY KEY (`ID`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=0;

    --
INSERT INTO `PicturePath` (`ID`, `Path`) VALUES
(1, 'img/picture.png');
//End DB
// Start script


    $PictureID = $_GET['ID']; // example, user is navigating to http://www.mysite.com/Picture.php?ID=1

    $Get_Picture = $Conn->prepare("SELECT PicturePath FROM pictures WHERE ID=?");
    $Get_Picture->bind_param('i', $PictureID);
    $Get_Picture->execute();
    $Get_Picture->bind_result($Path);
    $Get_Picture->close();
    echo "<img src='$Path'></img>";

My example given is in MYSQLI, adapt this to your requirements

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM