how to display data an image using php from a database

[rinsa]

hi..i have a database named abc and a table named xyz.....in the table i have fields such as id no:,name,addres s,profession... i need one more field that displays the image of the person along with the information...

it should display in a table format or any other good format in a page so that the image and the person's information is well displayed...

how can i do this in php...i am using php and mysql...please help asap...

[chathura86]

from php connect to mysql and select the database

http://php.net/manual/en/function.mysql-connect.php

http://php.net/manual/en/function.mysql-select-db.php

select the data from database

http://www.php.net/manual/en/function.mysql-query.php

display the data in a format you like

http://www.php.net/manual/en/function.mysql-fetch-array.php

for more details

http://www.php.net/manual/en/ref.mysql.php

Regrds

[rinsa]

hi below is the code that displays my data from the database...is there a code that i can add to this which will help me display the image of the members along with their information..

<?php

  $con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("speka", $con);
  
  $result = mysql_query("SELECT * FROM members");
  
  echo "<table border='1'>
<tr>
<th>IDNo:</th>
<th>Name</th>
<th>Course Studied </th>
<th>Academic Year</th>
<th>Address</th>
<th>Home Address</th>
<th>Employment/Profession</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['ID No:'] . "</td>";
  echo "<td>" . $row['Name'] . "</td>";
  echo "<td>" . $row['Course Studied'] . "</td>";
  echo "<td>" . $row['Academic Year'] . "</td>";
  echo "<td>" . $row['Address'] . "</td>";
  echo "<td>" . $row['Home Address'] . "</td>";
  echo "<td>" . $row['Employment/Proffession'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>

[chathura86]

i guess you store the image in the database

http://onlamp.com/pub/a/onlamp/2002/05/09/webdb2.html

check the "Delivering an Image From a Database" section

Regards

[rinsa]

hi..thanks chathura for the reply... i tried adding the image url...but still its not displaying the image... :(:(:( ...its showing that image icon with a cut in between for all the members....
please some1 help....

<?php

  $con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("speka", $con);
  
  $result = mysql_query("SELECT * FROM members");
  
  echo "<table border='1'>
<tr>
<th>IDNo:</th>
<th>Image</th>
<th>Name</th>
<th>Course Studied </th>
<th>Academic Year</th>
<th>Address</th>
<th>Home Address</th>
<th>Employment/Profession</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['ID No:'] . "</td>";
  echo "<td>" .'<img src ="'.$imgurl.'">'; "</td>";
  echo "<td>" . $row['Name'] . "</td>";
  echo "<td>" . $row['Course Studied'] . "</td>";
  echo "<td>" . $row['Academic Year'] . "</td>";
  echo "<td>" . $row['Address'] . "</td>";
  echo "<td>" . $row['Home Address'] . "</td>";
  echo "<td>" . $row['Employment/Proffession'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>

[Atli]

Hey.

rinsa, where exactly is the image stored? Is the image inside the database, or does the database contain a location on the file-system where the image is stored?

If it is inside the database, you need to create a separate PHP file to fetch it. I exapain that process in this article. (See phase #4, specifically.)

If it is on the file-system, you need only fetch the location from the database and print a <img>, like you try to do on line #28 of your code. However, it is not enough to just print a $imgurl variable. You need to actually define the variable before using it, or it will be empty and no image will be found.

[rinsa]

hi atli...i have created a column in the table which has the name image and the datatype varchar225...i have given the location of the file-system where the image is stored in the image column....eg: C:\xampp\htdocs \HTML\images\im age1.jpg....and now in the coding how should i declare the imgurl variable?

is it like $imgurl=C:\xamp p\htdocs\HTML\i mages;

sorry i am new to php and mysql...thankyo u for helping me out...

[Atli]

Ok, so you have an absolute path to the image. That's not ideal, but we can work around that. The problem is that when you put a <img> tag in your HTML, the src attribute has to point to a URL location where the image is found. An absolute, physical location doesn't work.

When the browser gets a <img> tag, it reads the src attribute and sends another request to the server, requesting this image. Therefore, the src attribute has to point to a place where the image can be read via the HTTP server.

If you use this tag in a page at http://localhost/index.php:

<img src="C:\xampp\htdocs\HTML\images\image1.jpg">

The browser will try to fetch an image at:
- http://localhost/C:\xampp\htdocs\HTML\images\image1.jpg
Which will obviously not work.

What you need to do is remove the C:\xampp\htdocs part of the page create a tag that reads:

<img src="HTML/images/image1.jpg">

This will make the browser request:
- http://localhost/HTML/images/image1.jpg
Which, if I am understanding you setup correctly, will return the image.


To make this happen in your code, you need to do the following:

1. Inside the while loop on line #24, where you echo the row, you need to fetch the image location from the $row into a variable.
2. Use the str_replace function to remove the C:\xampp\htdocs part of the location.
3. Echo the altered image location into the <img> tag on line #28.

To help you along, here is an example of how this can be done. You will need to adapt this to your own code.
[code=php]<?php
// Connect to a MySQL database.
$dbLink = mysql_connect( 'host', 'user', 'pwd' );
mysql_select_db ( 'myDb', $dbLink );

// Fetch images from the database.
$sql = "SELECT name, imagelocation FROM myImages";
$result = mysql_query( $sql, $dbLink ) or die(mysql_error ());

while($row = mysql_fetch_ass oc( $result )) {
// Fetch the data for the current image.
$name = htmlentities( $row['name'] );
$imageLocation = $row['imageLocation'];

// Remove the junk from the image location
$junk = 'C:\\xampp\\htd ocs';
$imageLocation = str_replace( $junk, '', $imageLocation );

// Replace \ slashes with / slashes.
$imageLocation = str_replace('\\ ', '/', $imageLocation) ;

// Print the <img> tag.
echo "<img src=\"{$imageLo cation}\" title=\"{$name} \">";
}
?>[/code]