How to display images from a database using PHP?

Question

My program needs to upload images, so an image (varchar) location is saved in a MySQL database. It's working so far.

Now I want to display images and this does not work. Here is the code:

include ('connect.php');
if(isset($_POST['submit'])){
    $filetemp= $_FILES['image'] ['tmp_name'];
    $filename= $_FILES['image'] ['name'];
    $filepath= "images/".$filename;

    move_uploaded_file($filetemp,$filepath);
    $sql=mysqli_query($con,"insert into  images (image) value ('$filepath')");
    if($sql){
        echo "uploaded";
    }
    else{
        echo " not uploaded";
    }
}

$sql=mysqli_query($con,"select * from images");
while($row=mysqli_fetch_array($sql)){
    echo "<img src=' images/".$row['image']."'>"; // the problem is here, its just displaying img icon, not actual image 
}   

?>

what is the value stored for image in the db? also, the path in the img tag has a leading space before ` images/` — Professor Abronsius
– Professor Abronsius, Commented May 6, 2017 at 12:16
this is the value that is saved in image column "images/download.jpg" — Maliha Khan
– Maliha Khan, Commented May 6, 2017 at 12:27
What do you mean by "does not work"? What happens instead? Also, be warned that your INSERT query is highly vulnerable for SQL injection - have a look at prepared statements to avoid getting hacked — Nico Haase
– Nico Haase, Commented Mar 1, 2024 at 15:58

Niklesh Raut · Accepted Answer · 2017-05-06 12:17:16Z

2

remove images from image path, you already store this in image column in images table

echo "<img src='/".$row['image']."'>"; // the problem is here, its just displaying img icon, not actual image

answered May 6, 2017 at 12:17

Niklesh Raut

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Maliha Khan Over a year ago

echo "<img src='/".$row['image']."'>"; // your code had '/' below line is correct echo "<img src='".$row['image']."'>"; Thanks

Ahmed Ginani · Accepted Answer · 2017-05-06 12:16:23Z

0

Correction :

$sql=mysqli_query($con,"select * from images");
while($row=mysqli_fetch_array($sql)){
    echo "<img src='".$row['image']."'>"; // the problem is here, its just displaying img icon, not actual image 
}

Also you need to get the actual path via __FILE__ in case if needed.

answered May 6, 2017 at 12:16

Ahmed Ginani

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Comments

supajason · Accepted Answer · 2017-05-06 12:19:13Z

0

You are adding images/ into the database then when you call the row to display the image you're adding images/ adding making it look in images/images/filename

answered May 6, 2017 at 12:19

supajason

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