Re: Retrieving Image Location in MySQL

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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@xxxxxxxx] Sent: Monday, March
02, 2009 9:29 AM
To: php-db@xxxxxxxxxxxxx
Subject: Re:  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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