Re: Retrieving Image Location in PHP from MySQL

Hi, Yes, the problem was solved, but It did not work fine when I used the same code in my larger file. Now it makes sense. Let me just repeat what you have said just to make sure that I did not misread you. So you say that the solution to this problem is to create another php file with the image fetching header and just write an img tag <img src="myimagescript.php?id=1234" /> in my original php file (with the html tags). This is what I have understood. Regarding the point you have mentioned ( If you set the content type using header() to "image/jpeg", do not use HTML tags to display your image!), I definitely need the HTML tags, because this application works based on the user input. So unless there is not input through a html form, it wont work. Thanks, Sashi

On Mon, Mar 9, 2009 at 1:29 AM, Sashikanth Gurram <sashi34u@vt.edu> wrote:

Hi Nathan, Thanks a lot for the suggestion. It is working fine for an example code I have used to test it. The code I have written after your suggestion is as follows. <?php $location="C:\wamp\bin\apache\apache2.2.8\htdocs\Bldgs_lots\SQUIRES.jpg"; header('Content-Type: image/jpeg'); imagejpeg(imagecreatefromjpeg($location)); ?> The above code is yielding me a picture. Now, When I tried to use it in my original code it is not giving me the image. Instead it is giving me this warning *Warning*: Cannot modify header information - headers already sent by (output started at C:\wamp\www\mysqli.php:65) in *C:\wamp\www\mysqli.php* on line *221* and a lot binary characters again instead of the image. I know something about this warning that we should not output anything to the browser before the header. But I cannot do that, since I need the user input using a html form and I use the input to fetch the data and dispaly it in a table along with the image. I am posting the whole code below. Please do let me know about any changes, corrections or modifications that are needed. The image display code is towards the last. Thanks, Sashi <html> <body> <form action="mysqli.php" method="post"> <br> <div align="center"> Building Name:<select name="name"> <option value=""> Select a Building</option>

    <option value="Williams Hall">Williams Hall</option>
    <option value="Women's Softball Field">Women's Softball Field</option>
    <option value="Wright House">Wright House</option>

</select> </div>

                  <input type="submit" />

</form> <?php // Connects to your Database $host="*********"; $user="*********"; $password="*********"; $dbname="*********"; $cxn=mysqli_connect($host, $user, $password, $dbname) ; if (!$cxn=mysqli_connect($host, $user, $password, $dbname)) {

      $error=mysqli_error($cxn);
      echo "$error";
      die();

} else {

      echo "Connection established successfully";

} //Define the variables for Day and Month $Today=date("l"); $Month=date("F"); $build=$_POST["name"]; $low_speed=2.5; $high_speed=4; $hour=date("G"); $minute=date("i"); if ($minute>=00 && $minute<=14) {

      $minute=00;

} elseif ($minute>=15 && $minute<=29) {

      $minute=15;

} elseif ($minute>=30 && $minute<=44) {

      $minute=30;

} else {

      $minute=45;

} $times="10:$minute"; $sql="SELECT buildingname, parking_lot_name, empty_spaces, distance, round(distance/($low_speed*60),1) AS low_time, round(distance/($high_speed*60),1) AS high_time, Location FROM buildings, buildings_lots, parkinglots, occupancy2, Image where (buildings.buildingcode=occupancy2.building AND buildings.buildingcode=buildings_lots.building_code AND parkinglots.parking_lot_code=buildings_lots.parking_lot_code AND parkinglots.parking_lot_code=occupancy2.parking_lot AND Buildings.BuildingCode=Image.BuildingCode) AND buildingname='$build' AND month='$Month' AND day='Monday' AND Time='$times'"; $data = mysqli_query($cxn,$sql); if (!$data=mysqli_query($cxn,$sql)) {

      $error=mysqli_error($cxn);
      echo "$error";
      die();

} else {

      echo "<br>";
      echo "Query sent successfully";

} echo "<br>"; echo "<h1> PARKING LOT INFORMATION <h1>"; echo "<table border='1' cellspacing='5' cellpadding='2'>"; echo "<tr>\n

      <th>Building</th>\n
      <th>Parking Lot</th>\n
      <th>Estimated Number of Empty Spaces</th>\n
      <th>Distance (Feet)</th>\n
      <th>Estimated walking time to the building</th>\n

</tr>\n"; while ($row=mysqli_fetch_array($data)) {

      extract($row);

$building = $row[0];

    $parking_lot = $row[1];
    $Number_of_Empty_Spaces = $row[2];
    $Distance = $row[3];
    $time_l = $row[4];
    $time_h=$row[5];
    $location=$row[6];   echo "<tr>\n
            <td>$building</td>\n
            <td>$parking_lot</td>\n

<td>$Number_of_Empty_Spaces</td>\n

            <td>$Distance</td>\n
            <td>$time_h - $time_l mins</td>\n

<td>$location</td>\n

            </tr>\n";

} echo "</table>\n"; header('Content-Type: image/jpeg'); imagejpeg(imagecreatefromjpeg("$location")); ?> </body> </html> Nathan Nobbe wrote:

    
On Sat, Mar 7, 2009 at 9:36 PM, Sashikanth Gurram <sashi34u@vt.edu> wrote:

      
The thing is, I do not have a website or a place where I am storing my

images. I am storing the images on my local PC in folder as mentioned in my earlier post. I have tried using only the else { header('Content-Type: image/jpeg'); imagejpeg($img); imagedestroy($img); } Which you have suggested. This is the piece of code which is returning all the binary character stuff. The code is definitely fetching the image. But it is not able to deliver to the browser in the form of an image. I am trying to read the image file as binary, evident from the FILE_BINARY command. So may be, that is causing the image to be displayed as binary. Is there any way we can convert the binary content to jpeg/jpg content again. This is my guess after having gone through the code again.

i spent 2 minutes working on it and read this in the manual, the first param of imagejpeg(), An image resource, returned by one of the image creation functions, such as imagecreatetruecolor()<http://us.php.net/manual/en/function.imagecreatetruecolor.php> . so, then i tossed this together, http://nathan.moxune.com/echoImage.php <?php header('Content-Type: image/jpeg'); imagejpeg(imagecreatefromjpeg('./phpLogo.jpeg'));

I see this one has already been solved, but I felt like I needed to point this out: If you set the content type using header() to "image/jpeg", do not use HTML tags to display your image! The expected output is the actual binary data from the image, not an <img src="asdf" /> tag. The OUTPUT of this script could be used in an IMG tag elsewhere, perhaps, like so: <html> ... <img src="myimagescript.php?id=1234" /> ... </html> Where "myimagescript.php" uses the Content-type header() and outputs the binary JPEG data. If you were using PHP to build a javascript file to be included, you wouldn't wrap it in <script> tags, either. Same principle applies here. HTH, -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Sashikanth Gurram Graduate Research Assistant Department of Civil and Environmental Engineering Virginia Tech Blacksburg, VA 24060, USA