PHP - View Display Image Blob With Php/oracle!

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 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 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.

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!

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.

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!

Hi.
idea:
View image from /skins/ folder. The id for the image is taken from mysql DB field called PlayerDefaultSkin

So, lets say a user has 240 in the field in database and in /skins/ folder I have image called 240.png
What I want is to php read from PlayerDefaultSkin field and view the image from /skins/

Right now I have this:
query to update:
Code: [Select]
`PlayerDefaultSkin` = $data[PlayerDefaultSkin]'
and the output code

Code: [Select]
<img src="skins/<?php echo '$PlayerDefaultSkin' ?>.png">
which simply doesnt work and is probably very wrong.  So can anyone help me with this?

Hi Guys,

I am trying to use a script that is available on the net to upload an image into a mysql using BLOB.

However it wont upload it keep getting

Warning: fread() [function.fread]: Length parameter must be greater than 0 in /home/theacidf/public_html/attendance/upload.php on line 17

Here is the code...

<?php

// Connect to database

$errmsg = "";
if (! @mysql_connect("localhost","theacidf_admin","password")) {
        $errmsg = "Cannot connect to database";
        }
@mysql_select_db("theacidf_attendanc");

// Insert any new image into database

if ($_REQUEST[completed] == 1) {

        move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img");
        $instr = fopen("latest.img","rb");
        $image = addslashes(fread($instr,filesize("latest.img")));
        if (strlen($image) < 149000) {
                mysql_query ("insert into pix (title, imgdata) values (\"".
                $_REQUEST[whatsit].
                "\", \"".
                $image.
                "\")");
        } else {
                $errmsg = "Too large!";
        }
}

// Find out about latest image

$gotten = @mysql_query("select * from pix order by pid desc limit 1");
if ($row = @mysql_fetch_assoc($gotten)) {
        $title = htmlspecialchars($row[title]);
        $bytes = $row[imgdata];
} else {
        $errmsg = "There is no image in the database yet";
        $title = "no database image available";
        // Put up a picture of our training centre
        $instr = fopen("../images/logo_png.png","rb");
        $bytes = fread($instr,filesize("../images/logo_png.png"));
}

// If this is the image request, send out the image

if ($_REQUEST[gim] == 1) {
        header("Content-type: image/jpeg");
        print $bytes;
        exit ();
        }
?>

<html><head>
<title>Upload an image to a database</title>
<body bgcolor=white><h2>Here's the latest picture</h2>
<font color=red><?= $errmsg ?></font>
<center><img src=?gim=1><br>
<?= $title ?></center>
<hr>
<h2>Please upload a new picture and title</h2>
<form enctype=multipart/form-data method=post>
<input type=hidden name=MAX_FILE_SIZE value=150000>
<input type=hidden name=completed value=1>
Please choose an image to upload: <input type=file name=imagefile><br>
Please enter the title of that pictu  <input name=whatsit><br>
then: <input type=submit></form><br>
<hr>

</body>
</html>

Why wont it run properly?

Thanks in advance...
S

hello every body,

I am trying to connect the oracle 10g with PHP code but my all code get error and does not get connection from the server database so i need the help from your site.
can you help me

My Code is:-

<?php

$db_host="xxx.xxx.xx.xxx"; // Host name
$db_username="xxx"; // Mysql username
$db_password="xxx"; // Mysql password
$db_name="XE"; // Database name
$tbl_name="USER_DETAIL"; // Table name

// Connect to server and select database.
try
{
   $conn = "(DESCRIPTION=(ADDRESS_LIST = (ADDRESS = (PROTOCOL = TCP)(HOST = xxx.xxx.xx.xxx)(PORT = 1521)))(CONNECT_DATA=(SERVER=DEDICATED)(SERVER_NAME=MYDB)))" ;
   $c= oci_connect($db_username, $db_password, $conn);
   $d=new PDO("oci:db_name=$conn",$db_username, $db_password);   
}
catch(PDOException $e)
{   
      trigger_error("Could not connect to database",E_USER-ERROR);   
}

I have this code that will get the absent dates for all employees, how can i make it for one specific employee using oracle id?

code:

$res = $conn->query("SELECT s.oracleid
                          , s.staffname
                          , date_format(date, '%W %d/%m/%Y') as absent
                     	  FROM staff s
                          CROSS JOIN
                          date d
                          LEFT JOIN
                          attendance_records a ON s.oracleid = a.oracleid
                                              AND d.date = DATE(a.clockingindate)
                     	  WHERE a.oracleid IS NULL
                          ORDER BY s.oracleid, d.date    
                         ");

the records i am concerned about is for the attendance records table, I have oracle id in that table.

Hello,

I need to interrupt the execution of an oracle query if it is taking more than 10 seconds, and give user a message informing him about execution timeout. I googled a lot but i didn't find anything useful. Is there any way to set a time limit to oci_execute.

I cannot modify DB settings...

Any Idea?

Thnks

Hey guys,

I`m having problems finding information on how to insert a image on a blob column trough odbc on a oracle database. Does anyone knows where can i find it, or can anyone help me with sample code? I really need to use odbc functions (cant use oci8 )

Tanks for your time reading this and for the possible input you may add on this.

Sincerely

Arestas

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!