Tutorial: Storing Images in MySQL with PHP / Part II / Display your images

**Astha** · 08-03-2010, 06:03 AM

Hello Jaan,

I got a similar script, it displays the image, but I have to reference the individual images (by its id, i.e, testgetpicture.php?image_id=1) separately. Is there any way where I can display all the images (from database table) in a row or column without specifying a specific id?

Here's my script:
testgetpicture.php

Code:


<?php

    // just so we know it is broken

    error_reporting(E_ALL);

    // some basic sanity checks

    if(isset($_GET['image_id']) && is_numeric($_GET['image_id'])) {

        //connect to the db

        $link = mysql_connect("localhost", "root", "") or die("Could not connect: " . mysql_error());

 

        // select our database

        mysql_select_db("templetree") or die(mysql_error());

 

        // get the image from the db



        $sql = "SELECT content FROM sample_design WHERE image_id=".$_GET['image_id'];

 

        // the result of the query

        $result = mysql_query("$sql") or die("Invalid query: " . mysql_error());

 

        // set the header for the image

        header("Content-type: image/jpeg");

        echo mysql_result($result, 0);

 

        // close the db link

        mysql_close($link);

    }

    else {

        echo 'Please use a real id number';

    }

?>

**pravin3009** · 08-03-2010, 06:25 AM

Hi Astha,

where r u showing the image, is it html image control or anything else? can you share those few lines too....

Thanks
-pravin

**Astha** · 08-03-2010, 09:06 PM

Hi,

Yes, you have to display the image in another file (html/php).

Here's the code:
display_image.html

HTML Code:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html>
<head>
<title>Untitled Document</title>
</head>
<body>
<img src="testgetpicture1.php?image_id=1"  />
</body>
</html>

But make sure that this HTML page (display_image.html) should be in the same folder where your testgetpicture.php file is.

Hope this helps you and let me know.

**pravin3009** · 08-04-2010, 06:40 AM

Thank you Astha,

unfortunatly that did not work, here is my code, what I am getting a is box with cross mark-

Code:


//php code

//getImage

<?php 



mysql_connect("localhost:3306","root","mysql2010") or die("Unable to connect to SQL server"); 

@mysql_select_db("testdb") or die("Unable to select database"); 

$result=mysql_query("select * from tblimagedetails where rowid=1") or die("Can't perform Query"); 



$data=@MYSQL_RESULT($result,0,"data"); 

header('Content-type: image/jpg'); 

echo($data);  



?>

Code:


//html code - the place where I am calling getimage is at the bottom, as you suggested so I tried 

//to keep the image call in the body element, rather than a <td> element as I wanted to...



<html>

 <head>

  <style type="text/css">

   .background { background-image:url(back.png);background-repeat:repeat-x; }

</style>

  <title>PHP Test</title>

 </head>

 <body>

 <div class="background">

  <table style="width:100%; height:100px;">

     <tr>

       <td width="100%" align="center" height="100px">

        <!-- Banner -->

         My Page

      </td>

    </tr>

   </table>

    <table style="width:100%; height:100px;">

     <tr>

       <td>

         <!-- Image1   --> 

                

         

       </td >

       <td width="30%" style="height:100px;">

         <!-- Image2 -->        

       </td>

       <td width="30%" style="height:100px;">

         <!-- Image3 -->

         image 3

       </td>

     </tr>

     </table>

  </div>

   <img src="getImage.php" >   

 </body>

</html>

If you think, there are some environment settings required related to php or IIS let me know.

**Prog4rammer** · 08-06-2010, 03:38 PM

Thanks a lot very Cool

**sundayose** · 09-04-2010, 04:32 AM

Thanks for your thread.
Help me with my code below on how to display an image from Mysql (longblob) in the picture field.
This code is the results page for my search page with search parameter. The problem I have is that when I search, the results page has funny staff in the picture field.
Thank you.

Code:

<?php require_once('Connections/conn_hostels.php'); ?>
<?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;

  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? "'" . doubleval($theValue) . "'" : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}

$colname_rsEresults = "-1";
if (isset($_POST['hostel_name'])) {
  $colname_rsEresults = $_POST['hostel_name'];
}
$colname2_rsEresults = "-1";
if (isset($_POST['institution_nearest'])) {
  $colname2_rsEresults = $_POST['institution_nearest'];
}
$colname3_rsEresults = "-1";
if (isset($_POST['category'])) {
  $colname3_rsEresults = $_POST['category'];
}
$colname4_rsEresults = "-1";
if (isset($_POST['district'])) {
  $colname4_rsEresults = $_POST['district'];
}
mysql_select_db($database_conn_hostels, $conn_hostels);
$query_rsEresults = sprintf("SELECT * FROM hostel_details WHERE hostel_name = %s OR institution_nearest = %s OR category = %s OR district = %s", GetSQLValueString($colname_rsEresults, "text"),GetSQLValueString($colname2_rsEresults, "text"),GetSQLValueString($colname3_rsEresults, "text"),GetSQLValueString($colname4_rsEresults, "text"));
$rsEresults = mysql_query($query_rsEresults, $conn_hostels) or die(mysql_error());
$row_rsEresults = mysql_fetch_assoc($rsEresults);
$totalRows_rsEresults = mysql_num_rows($rsEresults);
?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<table border="1">
  <tr>
    <td>hostel_id</td>
    <td>hostel_name</td>
    <td>institution_nearest</td>
    <td>direction_from_institution</td>
    <td>distance_from_institution</td>
    <td>category</td>
    <td>location</td>
    <td>district</td>
    <td>price_per_sem</td>
    <td>contact_person</td>
    <td>contact_numbers</td>
    <td>address</td>
    <td>remarks</td>
    <td>picture</td>
  </tr>
  <?php do { ?>
    <tr>
      <td><?php echo $row_rsEresults['hostel_id']; ?></td>
      <td><?php echo $row_rsEresults['hostel_name']; ?></td>
      <td><?php echo $row_rsEresults['institution_nearest']; ?></td>
      <td><?php echo $row_rsEresults['direction_from_institution']; ?></td>
      <td><?php echo $row_rsEresults['distance_from_institution']; ?></td>
      <td><?php echo $row_rsEresults['category']; ?></td>
      <td><?php echo $row_rsEresults['location']; ?></td>
      <td><?php echo $row_rsEresults['district']; ?></td>
      <td><?php echo $row_rsEresults['price_per_sem']; ?></td>
      <td><?php echo $row_rsEresults['contact_person']; ?></td>
      <td><?php echo $row_rsEresults['contact_numbers']; ?></td>
      <td><?php echo $row_rsEresults['address']; ?></td>
      <td><?php echo $row_rsEresults['remarks']; ?></td>
      <td><?php echo $row_rsEresults['picture']; ?></td>
    </tr>
    <?php } while ($row_rsEresults = mysql_fetch_assoc($rsEresults)); ?>
</table>
</body>
</html>
<?php
mysql_free_result($rsEresults);
?>

**renu** · 12-21-2010, 06:27 AM

how to display friends thumbnails as in orkut from database?
with my database images are shown in a row and all people in database. How to select friends from database?

**Tech Coder** · 12-28-2010, 08:22 AM

First, THANK YOU for this code - right stuff, right time! I was able to get my database/picture display up and running in no time by following the instructions EXACTLY (those that have trouble, RTFM! - it won't work if you don't follow the instructions! Make changes after the basics work!)

OK, on to what I trust will help several;

STEP-BY-STEP on

HOW TO SHOW SELECTED PICTURES IN A 4-WIDE TABLE

After getting the basics running (showing an image), I wanted what several on this thread have asked for - A WAY TO SHOW SEVERAL IMAGES IN A TABLE. As I found no post on that, I just 'hacked' at it until it worked and I wanted to share that with you here ('cleaner' ways are welcome - I'm pretty new to coding, though get it done!).

My goal was to make a 4-picture wide table so I could see all the pictures from my database. As we do a lot of 'online auction' stuff, mostly this is for a way to separate all the pictures into 'Item' numbers that we assign, though this should work well for those that are looking to do 'friends' or any other partial selection.

1. Add a column to your database with the label 'Item' (or whatever works for you). I used a default of '9999999999' to make it simple to find all pictures that don't have any other 'Item' yet assigned, then as we find them and assign Item numbers, they will be grouped together under that Item number. Be sure to put your 'default' number into the 'Item' database column for ALL previously loaded pictures before trying the code!

2. Create a .php file (I called mine 'tablepic') with EXACTLY the code below.

Code:


 <?php 

 $con = mysql_connect("host","user","pass");

    if (!$con)

    {

    die('Could not connect: ' . mysql_error());

    }

    $db_selected = mysql_select_db("db", $con);

$item = $_GET['Item'];

echo "Your selected item number is $item <br />";

if(!isset($item) || empty($item))

{

    die("Please select your item!");

}

else

{

    $query = mysql_query("SELECT * FROM db WHERE Item = '".$item."' ");

    ?>

    <table border="1" width = "700">

    <?php

    $i=0;

    echo "<tr>";

    while($row = mysql_fetch_array($query))

    {

        $i++;

        $id = $row['id'];

        echo '<td><img src="seepic.php?id=' . $id .'" width="194" height="165" /></td>';

        if ($i / 4 == ceil($i / 4))

        {

            echo "</tr><tr>";

        }

    }

    echo $id."</tr>";

}

mysql_close($con);

?>

3. Make sure your 'seepic.php' file looks like this (this is code taken directly from this thread and renamed)

Code:


<?php 

 $con = mysql_connect("host","user","pass");

    if (!$con)

    {

    die('Could not connect: ' . mysql_error());

    }

    $db_selected = mysql_select_db("db", $con);

$id = $_GET['id'];

if(!isset($id) || empty($id)){

die("Please select your image!");

}else{

$query = mysql_query("SELECT * FROM db WHERE id = '".$id."'");

while($row = mysql_fetch_array($query))

{

    $content = $row['picture'];

    header('Content-type: image/jpg');

    echo $content;

}

}

mysql_close($con);

?>

4. Be sure to change 'db', 'host', 'user' and 'pass' to your local settings!

5. Run the 'tablepics.php' file by going to http://YOURSITE.com/tablepic.php?Item=9999999999

You should see FOUR WIDE pictures all lined up in a nice table! ENJOY!

**John** · 12-28-2010, 05:32 PM

Tech Coder, thanks for sharing your changes.

**batowiise** · 01-10-2011, 09:43 AM

Thanks, this is is very simple and fine tutorial.

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