Hi JP,
I am just storing the location of the image on the database. I want to
keep them private and retrieve them through scripting. The images are
stored in a folder on my PC. Just to give you an example, this is the
location of a particular image as stored in the database.
C:\Users\Sashikanth\Desktop\Bldgs_lots\Burruss.jpg
Thanks,
Sashi
Joao Gomes Madeira wrote:
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?
Cheers.
JP
-----Original Message-----
From: Sashikanth Gurram <sashi34u@xxxxxx>
To: php-db@xxxxxxxxxxxxx
Subject: Re: 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.
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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