Re: Retrieving Image Location in MySQL

Hi, Thanks a lot for the suggestion. It is working fine but there is some problem. When I introduced the piece of code you have suggested into my code, I am getting a space for the picture but nothing is there in that space. It is just blank. I do not know the reason for this. Further, this is the piece of code I have written to display the pic. echo "<img src='$location' border='1' height='150' width='200' alt='$build' >"; Any help would be greatly appreciated. Thanks, Sashi mrfroasty wrote:

Hello, Storing only location of the image in dB wouldnt be any different than other string/data retrieved from the dB.That means if you have something like this in html documents whereby $image_location comes from the dB It should work without ....header stuffs Example <img src="$image_location" border="1" height="150" width="200" alt="Image" ISMAP> P:S Replace $image_location with image location from your dB. GR Muhsin Sashikanth Gurram wrote:

Dear all, I have been trying to retrieve the location of a image from database and display the image in the browser using PHP. I have written a sort of code for the purpose. All I am getting in my browser after using the code is a long set of ASCII characters with the following 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 *237 * I am herewith attaching my code. If anyone can point out the error or give some kind of advice that would be really great. The location of the image is stored in a table by the name IMAGE under the column name Location and here in the code, the location is retrieved under the varibale name $Location. 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="Agnew Hall">Agnew Hall</option>

     <option value="Rector Field House">Rector Field House</option>
     <option value="Richard B. Talbot Educational Resources

Center">Richard B. Talbot Educational Resources Center</option>

     <option value="Robeson Hall">Robeson Hall</option>
     <option value="Sandy Hall">Sandy Hall</option>
     <option value="Saunders hall">Saunders hall</option>
     <option value="Seitz Hall">Seitz Hall</option>
     <option value="Shanks Hall">Shanks Hall</option>
     <option value="Shultz Hall">Shultz Hall</option>
     <option value="Skelton Conference Center">Skelton Conference

Center</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> <img src="mysqli.php?"> <?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;

} $times="$hour:$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='$Month' AND day='$Today' 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"; $err=1; if ($img = file_get_contents($location, FILE_BINARY)) { if ($img = imagecreatefromstring($img)) $err = 0; } if ($err) { header('Content-Type: text/html'); echo '<html><body><p style="font-size:9px">Error getting image...</p></body></html>'; } else { header('Content-Type: image/jpeg'); imagejpeg($img); imagedestroy($img); } ?> </body> </html>/ Sashikanth Gurram wrote:

    
Hello guys,

Thanks to you all for your kind replies. I will try the steps and get back to you if I encounter more problems. Thanks, Sashi Fortuno, Adam wrote:

      
Matya,

Ha, ha, ha! Thank you good friend. I did say I didn't try the code :-) AF> Please note, I haven't tried this. It just seems plausible. I apologize if I confused anyone. I just meant to show how the parameter could help retrieve a picture. I wasn't too concerned with the particulars. Hopefully the concept is sound. If not, please do flame my note so no one attempts it. Be Well, A- -----Original Message----- From: Mattyasovszky Janos [mailto:mail@matya.hu] Sent: Monday, March 02, 2009 9:29 AM To: php-db@lists.php.net Subject: Re: [PHP-DB] Retrieving Image Location in MySQL Fortuno, Adam írta:

        
        //Write a query to pull out the picture's path
        $sql = "SELECT path FROM Image WHERE ID = %s";
        mysql_real_escape_string($value);
              
Sorry, but this won't work, since you don't map the value of the

escaped $value to the %s, lets say with sprintf()... Regards, Matya

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Sashikanth Gurram Graduate Research Assistant Department of Civil and Environmental Engineering Virginia Tech Blacksburg, VA 24060, USA