Hi Sashi:

You're incurring in this error because the headers for the page have
already been issued in the head section of your page.
When you issue the php header() function, it will generate an error.

Anyway, you are storing a path to a file in the database. My question
is: do you want to put your images on a subfolder of your site, where
they will become acessible through regular URLS, or you want to keep
them private and only accessible by scripting?


-----Original Message-----
From: Sashikanth Gurram <sashi...@vt.edu>
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Retrieving Image Location in MySQL
Date: Fri, 06 Mar 2009 12:23:31 -0500

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 
      <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 = $row[4];
echo "<tr>\n
              <td>$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 
  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.
> 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:m...@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

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