PHP - Pictures From A Database

Hi guys,

A beginner at php/mysql here...

For a project I'm doing, I would like to display search results with an image, small description, then a 'read more' link.
After googling this issue, I can't find much information on it, which surprised me.  Someone did mention blobs for images... (whatever they are!)
I'm wondering if anyone knows of any tutorials, or has any code to do this?

Your help is much appreciated.

Thanks.

I want to display pictures in a table using a PHP loop or something, for example 3 pictures in one row, and then when that row has filled, move to the next row, So I can have a grid of pictures, (similar to that on facebook, when your viewing photo albums)... But I've no idea how to do it!

Hi, I cant quite figure out how to  assign an unique id to the picture I upload to my page, all help would be appreciated.

Here's the upload code:
 
 //This is the directory where images will be saved 
 $target = "images/"; 
 $target = $target . basename( $_FILES['photo']['name']); 
 
 //This gets all the other information from the form 
 $name=$_POST['name']; 
 $email=$_POST['email']; 
 $phone=$_POST['phone']; 
 $pic=($_FILES['photo']['name']); 
 
 // Connects to your Database 
 mysql_connect("your.hostaddress.com", "username", "password") or die(mysql_error()) ; 
 mysql_select_db("Database_Name") or die(mysql_error()) ; 
 
 //Writes the information to the database 
 mysql_query("INSERT INTO `employees` VALUES ('$name', '$email', '$phone', '$pic')") ; 
 
 //Writes the photo to the server 
 if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) 
 { 
 
 //Tells you if its all ok 
 echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory"; 
 } 
 else { 
 
 //Gives and error if its not 
 echo "Sorry, there was a problem uploading your file."; 
 } 
 ?> 

Hello there,

I am working on a pagination script and have hit a bit of a wall. Here is the script:
Code: [Select]
<?php include("header.php"); ?>
<?php


$host = "localhost"; //

$user = ""; //username to connect to database

$pass = ""; //password to connect to database

$db = ""; //the name of the database



mysql_connect($host,$user,$pass) or die("ERROR:".mysql_error());

mysql_select_db($db) or die("ERROR DB:".mysql_error()); 





$max = 25; //amount of articles per page.

$p = $_GET['p'];

if(empty($p))

{

$p = 1;

}

$limits = ($p - 1) * $max; 

//view the article!

if(isset($_GET['act']) && $_GET['act'] == "view")

{

$id = $_GET['id'];

$sql = mysql_query("SELECT * FROM 3dpics WHERE id = '$id'");

while($r = mysql_fetch_array($sql))

{

$title = $r['title'];

$url = $r['url'];


echo "<div align='center'><img src='$url' width='500' height='500'><br />$title<br /><br /><a href='test.php'>Back</a></div>";

}



}else{



//view all the articles in rows

$sql = mysql_query("SELECT * FROM 3dpics LIMIT ".$limits.",$max") or die(mysql_error());

//the total rows in the table

$totalres = mysql_result(mysql_query("SELECT COUNT(id) AS tot FROM 3dpics"),0);


$totalpages = ceil($totalres / $max);
    
echo "<table name='myTable' cellpadding='5' cellspacing='5'>";
echo "<tr>";
//the table


while($r = mysql_fetch_array($sql))

{

$id = $r['id'];

$title = $r['title'];

$url = $r['url'];
echo "<div id='piccontentalign' align='center'>";
echo "<td><img src='$url' width='225' height='225'><br /><div align='center'><a href='test.php?act=view&id=$id'>$title</a></div></td>";
}
echo "</tr></table></div>";
//close up the table


for($i = 1; $i <= $totalpages; $i++){ 

//this is the pagination link

echo "<a href='test.php?p=$i'>$i</a>|";

}

}



?>

All this does is pull the url's of the images out of the database. But, it all shows up on 1 line across the screen! I tried echoing the rows instead, but then it just shows up as 1 line going horizontal. So i'm trying to figure out a way to get 5 pictures per row.
This is the code that displays them on the main page:
Code: [Select]
echo "<table name='myTable' cellpadding='5' cellspacing='5'>";
echo "<tr>";
//the table


while($r = mysql_fetch_array($sql))

{

$id = $r['id'];

$title = $r['title'];

$url = $r['url'];
echo "<div id='piccontentalign' align='center'>";
echo "<td><img src='$url' width='225' height='225'><br /><div align='center'><a href='test.php?act=view&id=$id'>$title</a></div></td>";
}
echo "</tr></table></div>";
//close up the table
Thanks for all your help!

I am working on this pictures to upload and when I submit it gives the message below for success but I get this error message which is on the line of $place file

Warning: move_uploaded_file(members/5/image01.jpg) [function.move-uploaded-file]: failed to open stream:



        $check_pic = "members/$id/image01.jpg";
$default_pic = "members/0/image01.jpg";


                        $newname = "image01.jpg";
                        $place_file = move_uploaded_file( $_FILES['fileField']['tmp_name'], "members/$id/".$newname);
                        $success_msg = '<font color="#009900">Your image has been updated, it may take a few minutes for the changes to show... please                                                    be patient.</font>';

I need some help working through this. I have an uploader for files, but I'd like it to create a link of the file uploaded and automatically post it to a page. Each user has a profile page and I'd like these pictures to be posted to their profile page. I guess with a session, it would take the logged in username and post the pictures uploaded on the page " member_profile2.php?username=xxxx "

Here is my upload file for reference.
uploads.php
Code: [Select]
        <?php
            $Dir = "temporary";

            
            if (isset($_POST['upload'])) {

             //if file uploaded
                if (isset($_FILES['new_file'])) {

                    //if successful
                    if (move_uploaded_file($_FILES['new_file']['tmp_name'], $Dir . "/" . $_FILES['new_file']['name']) == TRUE){
                        chmod($Dir . "/" . $_FILES['new_file']['name'], 0644);
                        echo "<b>File \"" . htmlentities($_FILES['new_file']['name']) ." was successfully uploaded </b><br />\n";

                        //displays file info
                        echo "Name: ". htmlentities($_FILES['new_file']['name']) . "<br />".
"Size: " . $_FILES['new_file']['size'] . " bytes" . "<br />".
                             "Type: " . $_FILES['new_file']['type'];
                    }
                    //if not successful
                    else {
                        echo "<b>Uploading \"" . htmlentities($_FILES['new_file']['name']) . " was unsuccessful</b><br />\n";
                    } 
                }  
            }  
        ?>

        <form action="uploads.php" method="POST" enctype="multipart/form-data">
            <input type="hidden" name="MAX_FILE_SIZE" value="250000" /><br />
            Select file to upload:<br />
            <input type="file" name="new_file" />
            250kB limit!<br />
            <input type="submit" name="upload" value="Upload File" /><br />         
        </form>

?>

and my member_profile2.php
Code: [Select]
<?php /*?><?php error_reporting(E_ALL ^ E_NOTICE); ?><?php */?>
<?php
# Starting the session
session_start();

# Requiring SQL connection
require_once 'mysql-connect2.php';

# Setting auser as SESSION['user']
$auser = $_SESSION['user'];

# SQL protecting variables
$username = (isset($_GET['username']))?mysql_real_escape_string($_GET['username']):$username;

# Checking through each query
if(isset($auser)) {
  $sql = mysql_query("SELECT * FROM `user` WHERE `username` = '$username'");
  if(mysql_num_rows($sql)) {
    while($row = mysql_fetch_array($sql)) {
      $page = "<h1>User Info</h1>".
              "<b>Username: {$row['username']}<br /><br />".
  "<b>First Name: {$row['firstname']}<br /><br />".
  "<b>Last Name: {$row['lastname']}<br /><br />".
  "<b>Email: {$row['email']}<br /><br />".
              "<form name=\"backlistfrm\" method=\"post\" action=\"members2.php\">".
              "  <input type=\"submit\" value=\"Back to The List\">".
              "</form><br />";
      }
  } else {
    $page = "ERROR: No member found for username: <strong>{$_GET['username']}</strong>.";
  }
} else {
  $page = "ERROR: Not logged in.";
}

# Printing the final output
print $page;

?>
I'd like the pictures to be displayed after the user details.
Any help with this would be great.



I was also thinking, could I just have the uploader save it as a filename with the user's username? For example, "user1-image1.jpg". And then have their profile page pull and display any files with the name user1-image1-10.jpg (10 being max possible uploads for that user).

How would I do this? This sounds maybe easier.

Hello. I'm working on a website with an editor that allows image uploading. Ideally I want to be able to develop a framework for this and use it in later projects. What I have in mind is this:
  Database has an images table Entries contain these fields: id, filename, and a short description Images are referred to by their ids in other parts of the application. That part seems simple to me, but now there are two details I need to determine: the file name, and storing the images.
Big sites like Facebook - as far as I know - parse uploads and store them all in the same format. I can see the huge security benefit there. What are the best ways of doing that? As for the file name, I know PHP has a function to generate a file with a unique name. Is there any benefit to doing that over using the id? (i.e. 1.jpg, 2.png, 3.jpg, etc)

Hi

I am trying to create a dynamic gallery in php with specific order of pictures on the page, but I can't find the function or piece of php code to do so.

Conditions:
1. The gallery must be dynamic, pictures will be coming from a folder via php, because when I'll add/upload more pictures in the folder they must be displayed on the page without adding manually the html tags in. ( this part is easy, problem is the condition 2 ).
2. The first row will have 5 pictures, the second - 4 pictures (important for the specific effect).

My code:

$files = glob("layout/gallery/*.jpg");

      rsort($files, SORT_NATURAL);

      for ($i=0; $i < count($files); $i++) {

        for( ; $i<5; $i++){
        $one = $files[$i];
        echo '<img src="'.$one.'">' . '<br><br>';
        }

echo "<br>";

        for( ; $i<9; $i++){
        $two = $files[$i];
        echo '<img src="'.$two.'">' . '<br><br>';
        }

      }

The code works well, but it just displays 9 pictures obviously. I was unable to make it dynamic displaying 5 pictures first, 4 pictures after and stay this way in a loop till displays all pictures from that folder.

Hi there,

Can you guys/girls tell me if this is even possible:
- take pictures with eye-fi card in the camera, which uploads the images to a server
- on the server, automatically add a watermark, and post them on 3 different facebookpages

Hi everyone!

I'm trying to create a form with an upload photo bit in it.

The photo i know, has to be saved on the server somewhere. Iv'e created an images folder both in my host's control pannel and in my dreamwever files and FTP'd it to the server. I've tried both ways of sending the test .jpg to each folder but i'm getting the desgined mySQL error.

What's going worng here?

Code: [Select]
<?php 
 
 //This is the directory where images will be saved 
 $target = "/images/COMPANIES"; 
 $target = $target . basename( $_FILES['photo']['name']); 
 
 //This gets all the other information from the form 
 $company_name=$_POST['company_name']; 
 $basicpackage_description=$_POST['basicpackage_description']; 
 $location=$_POST['location']; 
 $postcode=$_POST['postcode'];
 $upload=($_FILES['photo']['name']); 
 
 // Connects to your Database 
 mysql_connect("******host", "****username", "****password") or die(mysql_error()) ; 
 mysql_select_db("****DB") or die(mysql_error()) ; 
 
 //Writes the information to the database 
 mysql_query("INSERT INTO `Companies` VALUES ('$company_name', '$basicpackage_description', '$location', '$postcode', '$upload')") ; 
 
 //Writes the photo to the server 
 if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) 
 { 
 
 //Tells you if its all ok 
 echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory"; 
 } 
 else { 
 
 //Gives and error if its not 
 echo "Sorry, there was a problem uploading your file."; 
 } 
 ?>


Thanks

How would I write PHP code to echo the number of pictures I have in a folder on my hosting account?   Thanks in advance.

Hi,

You may noticed that i've posted before and closed them. It was too hard. I've been searching around for something else and i found it. But it's not quite perfect.
What i want to do is make thumbnails smaller (i can do that on my own) but displaying the bigger picture above the thumbnails. And aligning the thumbnails next to each other. So how do i do that?

The current code:
<?php

$result = mysql_query("SELECT reference FROM user_photos WHERE`profile_id`='".$row['id']."'");

while ($row2 = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "<a href=\"".$_GET['username']."/pics/".$row2['reference']."\">
<img src=\"".$_GET['username']."/pics/thumbs/".$row2['reference']."\"></a><br/><br/>";

}
}
?>

The files are refering to the mysql in reference but are actually in a subfolder with the username on the server.

I hope somebody wants to help me,

Best regards,

Martijn (the netherlands)

Hello everyone, I'm new here   My website was working fine until a few days ago, Up until then my str_replace('s130x130', 's800x800') and str_replace('_s', '_o') solved the picture thing. Now the images are broken (or very small) and I can find no documentation on how to enlarge the pictures. Thus my question: how do I retrieve larger pictures from the Facebook PHP SDK? Code

foreach($pagefeed['data'] as $post) {
if ($post['type'] == 'status' || $post['type'] == 'link' || $post['type'] == 'photo') {
if ($post['type'] == 'photo') {
echo "<div class='col-md-6 col-sm-6'>";
echo "<div class='thumbnail'>";
echo "<img src=" . str_replace('s130x130', 's800x800',
str_replace('_s', '_o', $post['picture'])) . " />";
echo "<div class='caption'>";
echo "<sup>Geplaatst op: " . date("d-m-Y, (strtotime($post['created_time']))) . "</sup>";

if (empty($post['message']) === false) {
echo "<p><span class='glyphicon glyphicon-comment'></span> " . $post['message'] . "</p>";}

echo "</div>";
echo "</div>";
echo "</div>";
$i++;
}
}
}

Thanks in advance!

Hi Guy's,

I've recoded allot with 3 scripts i had to my disposal. I've fixed 95% and i'm proud allready, but stuck at the following code. It does'nt execute any errors and refreshes the page without uploading the picture to the desired folder. Also folders are'nt created. I've turned on error reporting at level -1, but those errors i fixed.

Can you please help me out?

note: this is a partial code of a tiny usermanagement 'cms'.

Code: [Select]
<?php
//action: add picture for user -----------------------------------------------------------------------------
if (isset($_GET['editpic']) && isset($_GET['id'])) {
$id = (int) $_GET['id'];
if ($id == 0) {
die("Invalid ID provided.");
}
$sql = "SELECT username FROM `users` WHERE `id`='".$_GET['id']."'";
$res = mysql_query($sql) or die(mysql_error());
//execution when completed the add picture form and pressed submit button ---------------------
if (isset($_POST['addPic'])) {

$row = mysql_fetch_assoc($res);

$title = protect($_POST['title']);

if(!$title)
{
echo "<script language=\"Javascript\" type=\"text/javascript\">
alert(\"You must choose a title for your picture!\")
document.location.href='profilecp.php'</script>";

}

$target = $row['username'];
if(!is_dir($target)) @mkdir($target);
$target = $target . '/pics';

if(!is_dir($target)) @mkdir($target);

$target = $target."/".basename($_FILES['pics']['name']) ;
$size = $_FILES['pics']['size'];
$pic = $_FILES['pics']['name'];
$type = $_FILES['pics']['type'];

$sql2= "INSERT INTO `user_photos` (`profile_id`,`title`,`size`,`type`,`reference`)
VALUES ('".$_GET['id']."','$title','$size','$type','$pic'); ";

$res2 = mysql_query($sql2) or die(mysql_error());

if(move_uploaded_file($_FILES['pics']['tmp_name'], $target))
{
echo "<script language=\"Javascript\" type=\"text/javascript\">
alert(\"Your picture has been uploaded\")
document.location.href='profilecp.php'</script>";
}
else
{
echo "<script language=\"Javascript\" type=\"text/javascript\">
alert(\"There was an error, try again\")
document.location.href='profilecp.php'</script>";
}


$target2 = $row['username'];
$target2 = $target2 . '/pics';
$target2 = $target2 . '/thumbs';
if(!is_dir($target2)) @mkdir($target2);
$target2 = $target2."/".basename($_FILES['pics']['name']) ;
createthumb($target,$target2,150,150);

}


<!-----------------------ADD PICTURES-----------------------//-->

<div class="dividerp">
<form enctype="multipart/form-data" method="POST" action="administrator.php?editpic&amp;id=<?php echo $id;?>"><br/>
<strong>Upload Pictures</strong><br/><br/>

<div class="formElm">
<label for="title">Title</label>
<input id="title" type="text" name="title" maxlength="32"><br/>
</div>

<div class="formElm">
<label for="file">File</label>
<input id="file" type="file" name="pics" maxlength="255"><br/>
</div>

<input type="submit" name="picAdd" value="Add">
</form>
</div>

</div>
</body>
</html>
}?>

and the mysql tables:

Code: [Select]
CREATE TABLE `users` (
`id` INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY ,
`first` VARCHAR( 32 ) NOT NULL ,
`last` VARCHAR( 32 ) NOT NULL ,
`username` VARCHAR(32) NOT NULL,
`password` VARCHAR(255) NOT NULL,
`email` VARCHAR(255) NOT NULL,
`about` TEXT NOT NULL,
`level` int(4) default '1'
) ENGINE=InnoDB AUTO_INCREMENT=27 DEFAULT CHARSET=latin1;
Code: [Select]
CREATE TABLE `user_photos` (
`id` INT NOT NULL AUTO_INCREMENT PRIMARY KEY ,
`profile_id` INT NOT NULL ,
`title` VARCHAR( 128 ) NOT NULL ,
`size` INT NOT NULL ,
`type` VARCHAR( 128 ) NOT NULL ,
`reference` VARCHAR( 255 ) NOT NULL
) ENGINE=InnoDB AUTO_INCREMENT=27 DEFAULT CHARSET=latin1;
Best regards,

Martijn