Re: Retrieving Image Location in PHP from MySQL

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, -- // Todd