PHP - Display Image From Blob In Mysql

Hi All,

This is my first post on here. I'd really appreciate any help with this, it's driving me mad.

I'm trying to create a form to save a image into a mySQL database.

I have a website hosted by UK2.net which has a mysql db with a table called gallery(name (varchar 30), size (int), type varchar(30), thePic(mediumBlob)).

I have a form with this:

Code: [Select]
<td><p><label>Pic: </label></td><td><input name="userfile" type="file" id="userfile" /></td></p>
Which when submitted actions addImage.php. The code for this looks like this:

Code: [Select]
<?php


$fileName = $_FILES['userfile']['name'];
$tmpName  = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];

$fp      = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);

if(!get_magic_quotes_gpc())
{
    $fileName = addslashes($fileName);
}

$con = mysql_connect("localhost", "userName", "password") or die(mysql_error()); 

mysql_select_db("dbName", $con) or die(mysql_error()); 

$query = "INSERT INTO gallery (name, size, type, thePic) ".
"VALUES ('$fileName', '$fileSize', '$fileType', '$content')";

if (!mysql_query($query,$con))
  {
  die('Error: ' . mysql_error());
  }

echo "<br>File $fileName uploaded<br>";


?>
I get the following error message:

Quote

Warning: fopen() [function.fopen]: Filename cannot be empty in /home/hiddenje/public_html/addImage.php on line 13


Does anyone know what this is about? I've been on out friend google and a lot of people seem to be pointing to permissions but I can't seem to apply it to my scenario and just can't get it to work. Im a developer by trade, but this is my first step into the...interesting world of PHP and mySQL.

Again, I'd appreciate any help with this. I'd love someone to talk me through exactly what I'm missing or doing wrong.

Thanks in advance.

Okay, so here is the deal.

Have a table which stores image as blob files. Now i want to read the image width and height directly from the blob field.
Is this possible and if yes, how?

Things i tried so far;
list($size[0],$size[1],$type, $attr) = getimagesize('image.php?i=26ddd45b02859e836d13d4b9fde34281');      
print_r($size);

$img = 'image.php?i=26ddd45b02859e836d13d4b9fde34281';
echo imagesy($img);

image.php grabs the image from DB and show's it with header("Content-type: image/jpg");   
It works for just showing the images with the <img> tag.

Any ideas of help would be great!

I have created a image using imagepng() and then right-click'd and saved it. I then added it to the database by using phpMyAdmin to upload it directly into a field called 'avatar' which is a BLOB. The code the view the image back is the following:

header("Content-type: image/png");
echo mysql_result(mysql_query("SELECT `avatar` FROM `rscd_players` WHERE `username` = 'Kryptix'"), 0);

Now when I try and add the imagepng() directly to the database it doesn't work. I've tried multiple things. Here's the code:

ob_start(); imagepng($char_image); $pngimagedata = ob_get_contents(); ob_end_clean();

mysql_query("UPDATE `rscd_players` SET `avatar` = " . $pngimagedata . " WHERE `username` = 'Kryptix'");

Here's the full code: http://pastebin.com/jqsKc9Qd

This is the query that phpMyAdmin produces: http://pastebin.com/nuypNuwJ

This is the query that the above code produces: http://pastebin.com/NPWmi5eb

phpMyAdmin is 2.2kB, my code is 4.4kB

Any idea? I've been trying all night...

Hi guys,

This is my first time to insert PDF into MySQL BLOB.
Below is my form that i used

Code: [Select]
<?php
<form enctype="multipart/form-data" name="frmUploadFile" action="ulf-exec.php" method="post">

<select name="title"  id="title">
                      <option>xxx</option>
                      <option>yyy</option>
                      <option>zzz</option>
                    </select>
                  </label>
<input name="des" type="text" class="dropdownlists1" id="des"></td>
<input name="fileUpload" type="file" class="dropdownlists1" id="fileUpload" size="20" border=""></td>
<input type="submit" name="button" id="button" value="Submit">

 </form>
?>
I have prepared my database based on the required but decided to test with echo just to confirm there's no issue with the code
The action="ulf-exec.php" :
Code: [Select]
<?php

function clean($str) {
$str = @trim($str);
if(get_magic_quotes_gpc()) {
$str = stripslashes($str);
}
return mysql_real_escape_string($str);
}

//Sanitize the POST values
$title = clean($_POST['title']);
$des = clean($_POST['des']);
$fileUpload = $_POST['fileUpload'];

if(empty($des) || $fileUpload == "none")

die("You must enter both a description and file");


$fileHandle = fopen($fileUpload, "r");
$fileContent = fread($fileHandle, $fileUpload_size);
$fileContent = addslashes($fileContent);


$date = date('d').'-'.date('m').'-'.date('y');
$time = date('h').':'.date('i').':'.date('s');

echo "<h1>File Uploaded</h1>";

echo "The details of the uploaded file are shown below:<br><br>";

echo "<b>File name:</b> $fileUpload_name <br>";

echo "<b>File type:</b> $fileUpload_type <br>";

echo "<b>File size:</b> $fileUpload_size <br>";

echo "<b>Uploaded to:</b> $fileUpload <br><br>";

echo "<a href='uploadfile.php'>Add Another File</a>";
?>

This is the error:
Code: [Select]
Warning: fopen() [function.fopen]: Filename cannot be empty in /ulf-exec.php on line 30

Warning: fread(): supplied argument is not a valid stream resource in ulf-exec.php on line 31

Hello,
I am storing files that my users upload to the website as a blob in mysql database. Here is the code that does uploading:
$fileName = $_FILES['userfile']['name'];
$tmpName  = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];
$fp      = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = mysql_real_escape_string($content);
fclose($fp);
if(!get_magic_quotes_gpc())
{
   $fileName = addslashes($fileName);
}
$query = mysql_query("INSERT INTO documents (idapplicant, fileType, fileSize, fileData, fileName)
         VALUES ('$idapplicant','$fileType','$fileSize', '$content','$fileName')",$db);

Here is the code for download:
echo "File found:".@mysql_result($result, 0, "fileID"); //always correct
$data = @mysql_result($result, 0, "fileData");
$name = @mysql_result($result, 0, "fileName");
$size = @mysql_result($result, 0, "fileSize");
$type = @mysql_result($result, 0, "fileType");

header("Content-length: $size");
header("Content-type: $type");
header("Content-Disposition: attachment; filename=$name");
echo $data;

I can upload/download files no problem. However, I noticed problem with GIF files. All GIF files are just dark screen after downloading. All other files (PDF, JPEG) looked just fine. I compared files (before upload and after download) in HEX editor and found that the first byte on all files is set to A0 (it is added in front of all the data) and the last byte is deleted!
I have tried:
$content = addslashes($content);

instead of:
$content = mysql_real_escape_string($content);
with the same results
I also tried removing this line of code and in that case no file would be uploaded at all!
Any idea?
Thank you

Hi everyone, i am just trying to learn php for a bit of fun really and started making a sort of 'facebook' website. I am having trouble however trying to display different users images, for example when trying to find a correct 'friend' only the image of the last result is being shown for all people with the same name... here is my code below, if anyone can help me out that would be great

file 1

$count=1; 

while ($numids>=$count){
  echo "<form method=\"post\" action=\"friendadded.php\">";
  $frienduserid=$_SESSION["passedid[$ii]"];
  $friendfirstname=$_SESSION["passedfirstname[$ff]"];
  $friendlastname=$_SESSION["passedlastname[$ll]"];
  $_SESSION['friendsuserpicid'] = $frienduserid;
  echo "<table width=\"700\" height=\"50\" border=\"1\" align=\"center\">";
  echo "<tr>";
  echo "<th></th>";
  echo "<th>First Name</th>";
  echo "<th>Last Name</th>";
  echo "</tr>";
  echo "<tr>";
  echo "<td><center>";
  echo "<img border=\'0\' src=\"frienduserpic.php\" width=\"80\" height=\"80\" align=\"middle\"/>";
  echo "</center></td>";
  echo "<td><center>";
  echo $friendfirstname;
  echo "</center></td>";
  echo "<td><center>";
  echo $friendlastname;
  echo "</center></td>";
  echo "</tr>";
  echo "</table>";
  echo "<center><input type=\"submit\" value=\"Add this friend\" name=\"Add Friend\"></center><br/>";  

  $ii=$ii+1;
  $ff=$ff+1;
  $ll=$ll+1;
  $count=$count+1; 
  
  echo "</form>";
}




file 2

session_start();
$passeduserid=$_SESSION['friendsuserpicid'];
$timespost=$_SESSION['postednum'];
$host= 
$username=
$password=
$db_name= 
$tbl_name=
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$query = mysql_query("SELECT * FROM $tbl_name WHERE picid='".$passeduserid."'");
$row = mysql_fetch_array($query);
$content = $row['image'];
header("Content-type: image/jpeg");
echo $content;


Thanks in advance

Hi everyone,

I've read lots of tutorials on this, but something is not clicking in my brain with it.  I think my coding is close, but, as of right now, all I get is a red x for the image when I try to display the image.  Let me share my code as a starting point - I would truly appreciate any helpful comments or blatant errors that are pointed out or shared with me.

Here is the upload image form and code (so they clicked the item name and then go into this):

if($_REQUEST['modifyfeatured'])
{
$id=$_REQUEST['modid'];

$sqlid="SELECT * FROM product WHERE id='$id'";
$resultid=mysql_query($sqlid, $dbh);
$idrow = mysql_fetch_array($resultid);

echo "<p>";
echo "Fill out the form below to add a short text line (i.e. 20% Off!) and/or an image (like the new note).<br>";
echo "You are modifying the following item: <p><b>";
echo $idrow['product_id'];
echo " ";
echo $idrow['title'];
echo "</b><p>";

?>
<table border="0" cellpadding="2" cellspacing="0">
<tr>
<td>
<form method="post" enctype="multipart/form-data">
<input type="hidden" name="id" value="<? echo $id; ?>">
Short Text Message:
</td><td>
<input type="Text" name="message"></td></tr>
<tr><td>Small Image:</td><td>
<input type="hidden" name="MAX_FILE_SIZE" value="2000000">
<input name="userfile" type="file" id="userfile"></td></tr>
<tr><td>&nbsp;</td><td>
<input name="upload" type="submit" class="box" id="upload" value="Submit">
</td></tr></table>
</form>
<?

}
//then when the click the upload link, here is the code for that:

if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0)
{
$fileName = $_FILES['userfile']['name'];
$tmpName  = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];
$message=$_REQUEST['message'];
$id=$_REQUEST['id'];

$fp      = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);

if(!get_magic_quotes_gpc())
{
    $fileName = addslashes($fileName);
}

include 'library/config.php';
include 'library/opendb.php';

//$query = "INSERT INTO featured_prods (name, size, type, image, message ) ".
//"VALUES ('$fileName', '$fileSize', '$fileType', '$content', '$message')";

$query="UPDATE featured_prods SET name='$fileName', size='$fileSize', type='$fileType', image='$content', message='$message' WHERE id='$id'";

//$sqldone="UPDATE product SET product_id='$product_id', title='$title', description='$description', regular_price='$regular_price', sale_price='$sale_price', stat='$stat', weight='$weight', close_out='$close', additional='$additional', additionalpix='$additionalpix_name' WHERE  id='$id'";

//$resultdone=mysql_query($sqldone, $dbh);

mysql_query($query) or die('Error, query failed'); 
include 'library/closedb.php';

echo "<br>File $fileName uploaded<br>";
} //end if upload is hit


Okay, now here is the code trying to display the image:

<? $featuredquery="SELECT * FROM featured_prods";
$featuredresult=mysql_query($featuredquery, $dbh);
while($featuredrow=mysql_fetch_array($featuredresult)){

$id=$featuredrow['id'];
$prodquery="SELECT * FROM product WHERE id='$id'";
$prodresult=mysql_query($prodquery, $dbh);
$prodrow=mysql_fetch_array($prodresult);

echo $prodrow['title'];
echo "<br>";
echo $featuredrow['message'];
echo "<br>";
?>
<img src="getimage.php?id=<?echo $id;?>" alt="cover" />
<?
}


And here is the code for getimage.php:

<?
$id=$_GET["id"];
$result = mysql_query("select image from featured_prods where id='$id'"); 
$row = mysql_fetch_row($result); 
         
$data = base64_decode($row[0]); 
         
$im = imagecreatefromstring($row[0]); 
imagejpeg($im);

header('Content-type: ' . image/jpeg); // 'image/jpeg' for JPEG images
echo $data;

?>


Again, any help would be appreciated.  I'm a real rookie here, and I appreciate everyone's time and effort to assist me very much. 

Hi,

I am having an issue with displaying an image that is uploaded in a mysql database.

The table format is

-----------------------------------------------------------
|   id   |  name  |     mime     |   size   |  data  |
-----------------------------------------------------------
|    1   |   test   | image/jpeg| 74857 |  BLOB  |


The data has uploaded fine. I then have the following script on my 'displayimage.php?imgid=1' page...

Code: [Select]
$id = $_GET['imgid'];
$result = mysql_query("SELECT * FROM images WHERE id = '$id'");
while($row = mysql_fetch_array($result))
$size = $row['size'];
$mime = $row['mime'];
$name = $row['name'];
{
header("Content-length: $size");
header("Content-type: $mime");
header("Content-Disposition: attachment; filename=$name");
echo $row['data'];

However everytime i try to open this page, Firefox or Internet Explorer just ask's me to try and download the file as a php file. Any ideas what is going wrong?

Thanks!

Matt

I have a base64 encoded image -> http://pastebin.com/698ES5t0   Im trying to export this as a png file.  

 
  $picture =  {data on paste bin}; 
 
 
 $picture = base64_decode($picture); 
 
file_put_contents('/home/picture.png', $Picture);
 
 

    Now this creates a file picture.png and i can open it in Preview/Gimp etc.   However, if i upload the file to my website, the image does not render in chrome/firefox, however it does render in Safarai.    It seems that there is no height/width/depth data.

Code: [Select]
echo '<img src="data:image/jpg/png/jpeg;base64,' . base64_encode( $row['image'] ) . '" height="150" />';
This is showing up images great in firefox, safari and chrome, but in internet explorer it shows a nice red cross, and I assume it is because of the encoding?

Does anyone know how to get working in internet explorer as well?

Pretty urgent job!

Much appreciated for your help!

Hi,

I have managed to get the code working to store a .jpg file in the database under the longblob type.

Now all i have left to do is to retrieve that image and display it.

So far i have this:

list.php
Code: [Select]
while($r = mysql_fetch_array($sql)) { //for each record
...
echo "     <img src=  getoutside.php?id='".$r[apartmentId].  " '>     ";

getoutside.php
Code: [Select]
<?php
header("Content-type: image/jpg"); // act as a jpg file to browser
    $nId = $_GET['id'];
include 'dbase.php'; //connect to database
    $sqlo = "SELECT outside FROM apartment WHERE apartmentId = $nId";

$oResult = mysql_query($sqlo);

$oRow = mysql_fetch_array($oResult);
$sJpg = $oRow["outside"];
echo $sJpg;
?>
The result from this is a box with a red cross in it.
Can anyone find a problem with this code please?

Thanks

hi to all.Im currently displaying the images from my upload directory but the image does not display.please help thanks

Hello,
Five images will be displayed inside a division.
There will be a previous and next button/link.
If someone click the next button the next image will be added in that div and the first image will be gone from that div.
The previous button/link will do the same thing.

Is it possible with php?
I am confused if it's a javascript or ajax question.

Thanks.

Would like to be able to click on a radio button that represents an image. Once selected and submitted, have that image display on another page. I have an idea, but need some guidance.

BTW, is using php only doable? Is there a simpler or more elegant way to do this?

Thanks all!