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.


<form action="mysqli.php" method="post">
<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>
                   <input type="submit" />
<img src="mysqli.php?">

// Connects to your Database
$cxn=mysqli_connect($host, $user, $password, $dbname) ;
if (!$cxn=mysqli_connect($host, $user, $password, $dbname))
       echo "$error";
       echo "Connection established successfully";
//Define the variables for Day and Month
if ($minute>=00 && $minute<=14)
elseif ($minute>=15 && $minute<=29)
elseif ($minute>=30 && $minute<=44)
$sql="SELECT buildingname, parking_lot_name, empty_spaces, distance, round(distance/($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))
       echo "$error";
       echo "<br>";
       echo "Query sent successfully";
echo "<br>";
echo "<table border='1' cellspacing='5' cellpadding='2'>";
echo "<tr>\n
       <th>Parking Lot</th>\n
       <th>Estimated Number of Empty Spaces</th>\n
       <th>Distance (Feet)</th>\n
       <th>Estimated walking time to the building</th>\n
while ($row=mysqli_fetch_array($data))
$building = $row[0];
     $parking_lot = $row[1];
     $Number_of_Empty_Spaces = $row[2];
     $Distance = $row[3];
     $time_l = $row[4];
$location=$row[6]; echo "<tr>\n
<td>$parking_lot</td>\n <td>$Number_of_Empty_Spaces</td>\n
             <td>$time_h - $time_l mins</td>\n
   echo "</table>\n";
if ($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>';
 header('Content-Type: image/jpeg');

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.


Fortuno, Adam wrote:

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,

-----Original Message-----
From: Mattyasovszky Janos [] Sent: Monday, March 02, 2009 9:29 AM
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";

Sorry, but this won't work, since you don't map the value of the escaped $value to the %s, lets say with sprintf()...


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

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