Displaying images retrieved from a mysql db using the image path

colcar · June 8, 2008

Hi,

Can anyone help me here, I want to display an image using the path of the image and the image title from a mysql db?

I can upload the info to the db, but I have trouble displaying the image.

Here is the code:

CREATE TABLE `trailer` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`title` varchar(64) NOT NULL,
`model` text NOT NULL,
`year` text NOT NULL,
`price` text NOT NULL,
`location` text NOT NULL,
`path` text NOT NULL,
PRIMARY KEY (`id`)
);

<?php
$db_host = 'localhost'; // don't forget to change 
$db_user = 'root'; 
$db_pwd = 'colum';
$database = 'colum';
$table = 'trailer';// use the same name as SQL table
$password = 'kinefad';// simple upload restriction,// to disallow uploading to everyone
if (!mysql_connect($db_host, $db_user, $db_pwd)) 
die("Can't connect to database");
if (!mysql_select_db($database)) 
die("Can't select database");
// This function makes usage of// $_GET, $_POST, etc... variables
// completly safe in SQL queries
function sql_safe($s){ 
if (get_magic_quotes_gpc()) 
$s = stripslashes($s); 
return mysql_real_escape_string($s);
}
// If user pressed submit in one of the forms
if ($_SERVER['REQUEST_METHOD'] == 'POST'){ 
// cleaning title field 
$title = trim(sql_safe($_POST['title'])); 
$model = trim(sql_safe($_POST['model']));
$year = trim(sql_safe($_POST['year']));
$price = trim(sql_safe($_POST['price']));
$location = trim(sql_safe($_POST['location']));
$path = './images/';
if ($title == '') // if title is not set 
$title = '(empty title)';// use (empty title) string 
if ($_POST['password'] != $password) // cheking passwors 
$msg = 'Error: wrong upload password'; 
else 
{ 
if (isset($_FILES['photo'])) 
{ 
if (!isset($msg)) // If there was no error 
{ 
// Preparing data to be used in MySQL query 
mysql_query("INSERT INTO {$table} SET 

title='$title',model='$model',year='$year',price='$price',location='$location',path='$path'"); 
$msg = 'Success: image uploaded'; 
} 
} 
elseif (isset($_GET['title'])) // isset(..title) needed 
$msg = 'Error: file not loaded';
// to make sure we've using 
// upload form, not form 
// for deletion 
if (isset($_POST['del'])) // If used selected some photo to delete 
{ // in 'uploaded images form'; 
$id = intval($_POST['del']); 
mysql_query("DELETE FROM {$table} WHERE id=$id"); 
$msg = 'Photo deleted'; 
} 
}
}
?>
<html><head>
<title>Administration Page</title>
</head>
<body>
<?php
if (isset($msg)) // this is special section for 
// outputing message
{
?>
<p style="font-weight: bold;"><?=$msg?>
<br>
<a href="<?=$PHP_SELF?>">reload page</a>
<!-- I've added reloading link, because 
refreshing POST queries is not good idea -->
</p>
<?php
}
?>
<h1>Administration Page
</h1>
<h2>Uploaded images:</h2>
<form action="<?=$PHP_SELF?>" method="post">
<!-- This form is used for image deletion -->

</form>
<h2>Upload new image:</h2>
<form action="<?=$PHP_SELF?>" method="POST" enctype="multipart/form-data">
<label for="title">Title:</label><br>
<input type="text" name="title" id="title" size="64"><br><br>
<label for="model">Model:</label><br>
<input type="text" name="model" id="model" size="64"><br><br>
<label for="year">Year:</label><br>
<input type="text" name="year" id="year" size="64"><br><br>
<label for="price">Price:</label><br>
<input type="text" name="price" id="price" size="64"><br><br>
<label for="location">Location:</label><br>
<input type="text" name="location" id="location" size="64"><br><br>
<label for="photo">Photo:</label><br>
<input type="file" name="photo" id="photo"><br><br>
<label for="password">Password:</label><br>
<input type="password" name="password" id="password"><br><br>
<input type="submit" value="upload">
</form>
</body>
</html>

Here is my bad attempt of the display page:

<?
$dbcnx = @mysql_connect("localhost", "root", "colum"); 

if (!$dbcnx) 
{
echo( "connection to database server failed!" ); 
exit(); 
} 

if (! @mysql_select_db("colum") ) 
{ 
echo( "Image Database Not Available!" ); 
exit(); 
} 

$result = @mysql_query("SELECT * FROM trailer WHERE id=1"); 

if (!$result) 
{ 
echo("Error performing query: " . mysql_error() . ""); 
exit(); 
} 

while ( $row = mysql_fetch_array($result) ) 
{ 
$title = $row["title"];
$path = $row["path"];
} 

?> 

<html> 

<body> 

<img src="images/$title.JPG"> 

</body> 

</html>

Can anyone help me as to where or what I am doing wrong here?

Thanks

webent · June 8, 2008

Right off the bat, this looks like it wouldn't work...

<img src="images/$title.JPG">

try ...

<img src="images/<? echo $title; ?>.JPG">

Didn't really look at too much of the other.

colcar · June 8, 2008

Hi,

Thanks that works...It's not working the way I want though!! It is only showing the last image in the table. I want it show the image for each row in the DB in a table:

See my very poor attempt of code:

<?
$dbcnx = @mysql_connect("localhost", "root", "colum"); 

if (!$dbcnx) 
{
echo( "connection to database server failed!" ); 
exit(); 
} 

if (! @mysql_select_db("colum") ) 
{ 
echo( "Image Database Not Available!" ); 
exit(); 
} 

$result = @mysql_query("SELECT * FROM trailer"); 

if (!$result) 
{ 
echo("Error performing query: " . mysql_error() . ""); 
exit(); 
} 

while ( $row = mysql_fetch_array($result) ) 
{ 
$title = $row["title"];
$path = $row["path"];
$model = $row["model"];
$location = $row["location"];
$price = $row["price"];
$year = $row["year"];
} 

?> 


<html> 

<body> 

<div align="center">
  <center>
  <table border="0" width="31%" height="336" cellspacing="0" cellpadding="0">
    <tr>
      <td width="50%" height="243" align="center">
        <p align="left"><img border="0" src="images/<? echo $title; ?>.JPG" width="305" height="243" align="center"></p>
        <table border="1" width="100%" bordercolor="#0000FF">
          <tr>
            <td width="50%"><b>Name</b></td>
            <td width="50%"><? echo $title; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Model</b></td>
            <td width="50%"><? echo $model; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Year</b></td>
            <td width="50%"><? echo $year; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Price</b></td>
            <td width="50%"><span lang="EN-IE" style="mso-ansi-language:EN-IE">€<o:p>
              </o:p>
              </span><? echo $price; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Location</b></td>
            <td width="50%"><? echo $location; ?></td>
          </tr>
        </table>
      </td>
      <td width="100%" height="243" align="center">
        <p align="left"><img border="0" src="images/<? echo $title; ?>.JPG" width="305" height="243">
        </p>
        <table border="1" width="100%" bordercolor="#0000FF">
          <tr>
            <td width="50%"><b>Name</b></td>
            <td width="50%"><? echo $title; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Model</b></td>
            <td width="50%"><? echo $model; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Year</b></td>
            <td width="50%"><? echo $year; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Price</b></td>
            <td width="50%"> <span lang="EN-IE" style="mso-ansi-language:EN-IE">€<o:p>
              </o:p>
              </span><? echo $price; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Location</b></td>
            <td width="50%"><? echo $location; ?></td>
          </tr>
        </table>
      </td>
    </tr>
  </table>
  </center>
</div>


</body> 

</html>

Can anyone help me? Or point me in the right direction?

Thanks

webent · June 9, 2008

If you want it to print all of the results, then you need to put that section of your html structure in your while loop...

while ( $row = mysql_fetch_array($result) ) 
{
}

colcar · June 9, 2008

Hi,

Thanks webent...that wokred wonders...

I'm wrecking my brain all day in order to find the solution to my next problem. I have 2 tables within 1 big tables, in the 2 tables, the image and other information appear. Now I have the same image showing in both of the 2 smaller tables (i.e. in table 1 and 2, image 1 is being displayed. How can I change my code to show image 1 in table 1 and image 2 in table 2?

I have tried several different attempts, but yet to have some joy. Here is the latest code can anyone point me in the right direction?

<?
$dbcnx = @mysql_connect("localhost", "root", "colum"); 

if (!$dbcnx) 
{
echo( "connection to database server failed!" ); 
exit(); 
} 

if (! @mysql_select_db("colum") ) 
{ 
echo( "Image Database Not Available!" ); 
exit(); 
} 

$result = @mysql_query("SELECT * FROM trailer"); 

if (!$result) 
{ 
echo("Error performing query: " . mysql_error() . ""); 
exit(); 
} 

while ( $row = mysql_fetch_array($result) ) 
{ 
$title = $row["title"];
$path = $row["path"];
$model = $row["model"];
$location = $row["location"];
$price = $row["price"];
$year = $row["year"];
?>

<html> 

<body> 

<div align="center">
  <center>
  <table border="0" width="31%" height="336" cellspacing="0" cellpadding="0">
    <tr>
      <td width="50%" height="243" align="center">
        <p align="left"><img border="0" src="images/<? echo $title; ?>.JPG" width="305" height="243" align="center"></p>
        <table border="1" width="100%" bordercolor="#0000FF">
          <tr>
            <td width="50%"><b>Name</b></td>
            <td width="50%"><? echo $title; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Model</b></td>
            <td width="50%"><? echo $model; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Year</b></td>
            <td width="50%"><? echo $year; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Price</b></td>
            <td width="50%"><span lang="EN-IE" style="mso-ansi-language:EN-IE">€<o:p>
              </o:p>
              </span><? echo $price; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Location</b></td>
            <td width="50%"><? echo $location; ?></td>
          </tr>
        </table>
      </td>
<td width="100%" height="243" align="center">
        <p align="left"><img border="0" src="images/<? echo $title; ?>.JPG" width="305" height="243">
        </p>
        <table border="1" width="100%" bordercolor="#0000FF">
          <tr>
            <td width="50%"><b>Name</b></td>
            <td width="50%"><? echo $title; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Model</b></td>
            <td width="50%"><? echo $model; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Year</b></td>
            <td width="50%"><? echo $year; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Price</b></td>
            <td width="50%"> <span lang="EN-IE" style="mso-ansi-language:EN-IE">€<o:p>
              </o:p>
              </span><? echo $price; ?></td>
          </tr>
          <tr>
            <td width="50%"><b>Location</b></td>
            <td width="50%"><? echo $location; ?></td>
          </tr>
        </table>
      </td>
    </tr>
  </table>
  </center>
</div>


</body> 

</html>

<?
}

Thanks

colcar · June 10, 2008

Hi,

Can anyone help me here? I still can't figure this out.

Thanks

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Displaying images retrieved from a mysql db using the image path

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