Database Image Path

[Salchester]

Hello Everyone,

I currently have a "Path" column in a database, containing paths of image files.
How do I display each image on a page, using the file path contained in the database?

Many Thanks,

[sde]

how would you display the image if it weren't coming from a database? show me. (you'll probably answer your own question in the process)

[Salchester]

<img src="c:/image.jpg">

[sde]

your image value in the db needs to be just the image, or the relative path from your web root.. not your c drive.

<img src="/images/myimage.jpg">

now type code that uses php to print that image tag.

[Salchester]

Sand Man,

<?php echo <img src="/images/myimage.jpg"> ?>

Also, I currently have the following code to upload a file to a specific folder on the server, only how do I save the file path and filename into a database containing a path and filename fields.

<?php
if (($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/pjpeg")
&& ($_FILES["file"]["size"] < 20000))
  {
  if ($_FILES["file"]["error"] > 0)
    {
    echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
    }
  else
    {
    echo "Upload: " . $_FILES["file"]["name"] . "<br />";
    echo "Type: " . $_FILES["file"]["type"] . "<br />";
    echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />";
    echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />";

    if (file_exists("upload/" . $_FILES["file"]["name"]))
      {
      echo $_FILES["file"]["name"] . " already exists. ";
      }
    else
      {
      move_uploaded_file($_FILES["file"]["tmp_name"],
      "upload/" . $_FILES["file"]["name"]);
      echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
      }
    }
  }
else
  {
  echo "Invalid file";
  }
?>

Many Thanks,

[sde]

Originally Posted by Salchester

Sand Man,

<?php echo '<img src="/images/myimage.jpg">'; ?>[/code]

i'm pointing out that you know how to do this stuff already.. you just gotta put it all together. the code above needed single quotes around the string you are echoing.

so now write your query, and print the results as you want them to appear. the image string will just need to be replaced with a database results variable.