Learn VBA & Macros in 1 Week!

PHP - Image Not Displaying

Full Excel VBA Course - Beginner to Expert

Image Not Displaying	View Content

i have a db which store jpg image thru the upload prg code in php.But the image is not displayed properly in the <img> and also when i echo the blob data.

how to correct this code.i am pasting the code below

$result = mysql_query("select cover from Movies where movie_id=4");
$row = mysql_fetch_row($result);

$data = base64_decode($row[0]);

$im = imagecreatefromstring($row[0]);
imagejpeg($im);
header('Content-type: ' . $mime); // 'image/jpeg' for JPEG images
echo $data;
<img src=<?php echo $data;?>>

Full Excel VBA Course - Beginner to Expert

Image Not Displaying

Similar Tutorials

View Content

I am using this code

Code: [Select]
<?php echo IMAGES_HEADER . "header_02.jpg"; ?>
Instead of displaying the actual image on my site I am getting the path of the image. It is displaying "images/header/header_02.jpg" instead.

Thank You in advance

Displaying Image From Table?

Similar Tutorials

View Content

I just can't get anything right today! Now I am trying to have a list of images, and when the user clicks on the image, the next page will display the image along with other fields for that record. I am sending the id through the hyperlink to the next page, and I have echoed it to ensure it's comign through, but I cannot get anythign to display ont he next page. What am I doin wrong? Here is the link to the page. If you click on one of the images, you 'll see that the next page is empty:

http://webdesignsbyliz.com/wdbl_wordpress/test-submit/

Here is my code:

Code: [Select]

this is the gallery

<?php
$dbhost = 'localhost';
$dbuser = 'user';
$dbpass = 'pass';

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');

$dbname = 'jewelry';
mysql_select_db($dbname);

$all_records = "SELECT * FROM gallery";
$all_records_res = mysql_query($all_records);
$image = mysql_result($all_records_res, 0, 'image');
$id = mysql_result($all_records_res, 0, 'id');

while($nt=mysql_fetch_array($all_records_res)){
echo "<a href=http://webdesignsbyliz.com/wdbl_wordpress/test-display/?id=" .$nt['id']." ><img src=http://www.webdesignsbyliz.com/wdbl_wordpress/wp-content/themes/twentyten_2/upload/".$nt['image']." width=133 height=86></a>";
}

?>

display page:
Code: [Select]

<?php

$id = $_GET['id'];

$dbhost = 'localhost';
$dbuser = 'user';
$dbpass = 'pass';

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');

$dbname = 'jewelry';
mysql_select_db($dbname);

$all_records = "SELECT * FROM gallery WHERE id = $_GET[id]";
$all_records_res = mysql_query($all_records);
$image = mysql_result($all_records_res, 0, 'image');
$id = mysql_result($all_records_res, 0, 'id');

while($nt=mysql_fetch_array($all_records_res)){
echo "<img src=http://www.webdesignsbyliz.com/wdbl_wordpress/wp-content/themes/twentyten_2/upload/".$nt['image']." width=133 height=86></a>";
}

?>

What an I doing wrong this time?? :-(

Please Help Displaying Image After Upload

Similar Tutorials

View Content

Hi,
I am uploading an image to a folder and I need to display the uploaded image in a new page.
This is my code for the upload:
Code: [Select]
<?php
//define a maxim size for the uploaded images in Kb
define ("MAX_SIZE","2000000");

//This function reads the extension of the file. It is used to determine if the
// file is an image by checking the extension.
function getExtension($str) {
$i = strrpos($str,".");
if (!$i) { return ""; }
$l = strlen($str) - $i;
$ext = substr($str,$i+1,$l);
return $ext;
}

//This variable is used as a flag. The value is initialized with 0 (meaning no
// error found)
//and it will be changed to 1 if an errro occures.
//If the error occures the file will not be uploaded.
$errors=0;
//checks if the form has been submitted
if(isset($_POST['upload']))
{
//reads the name of the file the user submitted for uploading
$image=$_FILES['image']['name'];
//if it is not empty
if ($image)
{
//get the original name of the file from the clients machine
$filename = stripslashes($_FILES['image']['name']);
//get the extension of the file in a lower case format
$extension = getExtension($filename);
$extension = strtolower($extension);
//if it is not a known extension, we will suppose it is an error and
// will not upload the file,
//otherwise we will do more tests
if (($extension != "jpg") && ($extension != "jpeg") && ($extension !=
"png") && ($extension != "gif"))
{
//print error message
echo '<h1>Tipo de imagen no permitido.</h1>';
$errors=1;
}
else
{
//get the size of the image in bytes
//$_FILES['image']['tmp_name'] is the temporary filename of the file
//in which the uploaded file was stored on the server
$size=filesize($_FILES['image']['tmp_name']);

//compare the size with the maxim size we defined and print error if bigger
if ($size > MAX_SIZE*1024)
{
echo '<h1>La imagen es demasiado grande.</h1>';
$errors=1;
}

//we will give an unique name, for example the time in unix time format
$image_name=time().'.'.$extension;
//the new name will be containing the full path where will be stored (images
//folder)
$newname="banners/".$image_name;
//we verify if the image has been uploaded, and print error instead
$copied = copy($_FILES['image']['tmp_name'], $newname);
if (!$copied)
{
echo '<h1>Error al subir la imagen.</h1>';
$errors=1;
}}}

//If no errors registred, print the success message
if(isset($_POST['Submit']) && !$errors)
{
echo "<h1>La imagen se ha cargado correctamente</h1>";
}
$query = "INSERT INTO banner_rotator.t_banners (banner_path) ".
"VALUES ('$newname')";
mysql_query($query) or die('Error, query failed : ' . mysql_error());

//echo "<br>Files uploaded<br>";

header("Location: PC_cropbanner.php");
}Sowhat I need to do is to display the uploaded image in PC_cropbanner.php.
How can I do this?

Trouble With Displaying An Image From A Url

Similar Tutorials

View Content

I am having trouble with displaying an image from a url. In the following code if I echo out $test I get the url of the image. However, in the current code I am trying to display the image, but the only thing that gets displayed is a small broken image in the upper left.

$html = file_get_contents($website.$criteria); 

$dom = new DOMDocument;

@$dom->loadHTML($html);

$links = $dom->getElementsByTagName('img');

header('Content-Type: image/png');
foreach ($links as $link){
    $test = $link->getAttribute('src');
	echo file_get_contents($test);
}

image that gets displayed: http://imgur.com/epfF3wJ

Displaying Image Regardless Of Extension

Similar Tutorials

View Content

Hi everyone,

Sorry not sure if this is a php or html problem.

Im using php and a html form to upload images to my site, i have made it so that jpg, jpeg, gif and png images can be uploaded. The problem im having is displaying the image.

Code: [Select]
<img src="images/<?php echo $name ?>.jpg" />
That works fine if a jpg was uploaded, but what if a png or a gif was uploaded? The $name is going to be unique, there will not be more than one image with the same name, so what do i have to do to display the image regardless of what the extension is?

Thanks

Displaying The Same Blob Image

Similar Tutorials

View Content

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

Displaying Image From Table

Similar Tutorials

View Content

Now I am trying to display the images from my table and I have almost got it working, except for one small thing --- it seems to be displaying all the records, but instead of displaying the right image for each record, it's displaying the same image across all the records. Can anyone tell me what I have done wrong?

Code: [Select]

this is the gallery

<?php
$dbhost = 'localhost';
$dbuser = 'webdes17_lizkula';
$dbpass = 'minimoon';

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');

$dbname = 'webdes17_jewelry';
mysql_select_db($dbname);

$all_records = "SELECT * FROM gallery";
$all_records_res = mysql_query($all_records);
$image = mysql_result($all_records_res, 0, 'image');

while($nt=mysql_fetch_array($all_records_res)){
echo "<img src=http://www.webdesignsbyliz.com/wdbl_wordpress/wp-content/themes/twentyten_2/upload/$image>";
}

?>

I'm guessing I have the $image variable in the wrong place, but when I tried placing it within the while statement, the page never loaded, and instead acted like it was loading forever. What am I doing wrong? Here is the link to the page so you can see what is happening:

http://webdesignsbyliz.com/wdbl_wordpress/test-submit/

Displaying An Image From Mysqldb

Similar Tutorials

View Content

I must be doing something incredibly daft, because I'm incredibly new at this.

I have an image stored in a DB under a table called 'images' and I want to display it on my website but instead of that image I get the error: Warning: Cannot modify header information - headers already sent by (output started at /home/...

This is how I'm trying to achieve it.
Any ideas where I'm doing wrong?

Thanks.

Code: [Select]
<?php
$user="###";
$password="###";
$database="###";
$con = mysql_connect(localhost,$user,$password);
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db($database, $con);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>MySite</title>
</head>
<body>
<div id="container">
<?php include("../navbar.php"); ?>
<div id="left-content"></div>
<div id="right-content">
<?php
   $item = $_GET['item'];
   $query = "SELECT * FROM main WHERE Ref='$item'";
   $result = mysql_query($query)
   or   die("Oops" .mysql_error());
   $row = mysql_fetch_array($result,MYSQL_BOTH)
   or   die("Oops" .mysql_error());
   extract($row);
   $query2 = "SELECT image FROM main WHERE Ref='$item'";

// the result of the query
$result2 = mysql_query($query2) or die("Invalid query: " . mysql_error());
header("Content-type: image/jpg");
echo mysql_result($result2, 0,'image');
echo "<p><strong>Name: </strong>".$FirstName." - ".$SecondName."</p>";
mysql_close($con);
?>
</div>
<br />
<?php include("../footer.php"); ?>
</div>
</div>
</body>
</html>

MOD EDIT: [code] . . . [/code] BBCode tags added.

Displaying Image From Folder

Similar Tutorials

View Content

Good day:

Im trying to display an image from a folder and the image name is retrieved by a query. The query is getting the image name all right but the image is not displaying. This is the code im using for the image display

Code: [Select]
<?php
$aid = 1;
  $connection = mysql_connect("localhost",
        "username",
        "password");

    mysql_select_db("articles", $connection);
$query="SELECT imagename, description FROM articles_description WHERE id='$aid'";
$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();
?>
<table width ="1000" border="1" cellspacing="2" cellpadding="2">

</tr>

<?php
$j=0;
while ($j < $num) {

$f8=mysql_result($result,$i,"description");
$f9=mysql_result($result,$i,"imagename")
?>

<tr>
<td valign="top"> <img src="images/'.$f9.'; ?>" alt="" name="picture" width="100" height="100" border="1" /></td>
<td><font face="Arial, Helvetica, sans-serif"><?php echo $f8; ?></font></td>

</tr>
<tr>
</tr>

<?php
$j++;
}
?>

Any help will be appreciated

Displaying Text On Image

Similar Tutorials

View Content

I've taken this eg from php manual site for displaying txt on image wonder why it is not working

Code: [Select]
<?php
// Create a 300x150 image
$im = imagecreatetruecolor(300, 150);
$black = imagecolorallocate($im, 0, 0, 0);
$white = imagecolorallocate($im, 255, 255, 255);

// Set the background to be white
imagefilledrectangle($im, 0, 0, 299, 299, $white);

// Path to our font file
$font = './arial.ttf';

// First we create our bounding box for the first text
$bbox = imagettfbbox(10, 45, $font, 'Powered by PHP ' . phpversion());

// This is our cordinates for X and Y
$x = $bbox[0] + (imagesx($im) / 2) - ($bbox[4] / 2) - 25;
$y = $bbox[1] + (imagesy($im) / 2) - ($bbox[5] / 2) - 5;

// Write it
imagettftext($im, 10, 45, $x, $y, $black, $font, 'Powered by PHP ' . phpversion());

// Create the next bounding box for the second text
$bbox = imagettfbbox(10, 45, $font, 'and Zend Engine ' . zend_version());

// Set the cordinates so its next to the first text
$x = $bbox[0] + (imagesx($im) / 2) - ($bbox[4] / 2) + 10;
$y = $bbox[1] + (imagesy($im) / 2) - ($bbox[5] / 2) - 5;

// Write it
imagettftext($im, 10, 45, $x, $y, $black, $font, 'and Zend Engine ' . zend_version());

// Output to browser
header('Content-Type: image/png');

imagepng($im);
imagedestroy($im);
?>

Displaying A Default Image

Similar Tutorials

View Content

Hi all,

I have this script below where I am trying to display a default image if an image can not be found.

For some reason though it is not working.

<?php
foreach (glob('./aircraft/' . $rowX['reg'] . '[0-8].jpg') as $file)
{
if (file_exists($file)) {
echo "<img src=\"" . $file . "\" /><br />";
}
else { echo "><img src=\"aircraft/wrightflyer.jpg\" /><br />";}
}
?>

Displaying An Image According To Session

Similar Tutorials

View Content

Hi guys, I have followed a tutorial and made a members only area using sessions. The user can upload an image and which gets renamed as their username. I was hoping to display all the users images that are logged in. I know how to do it with a single image by just setting the img src as the session username but I don't know how I would display multiple images if more than one person were logged in. Is it even possible?

Assistance With Displaying Blob Image

Similar Tutorials

View Content

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> </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.

Convert Image Into Thumbnail Before Displaying Them.

Similar Tutorials

View Content

I have a form where you can enter a image url and it's title. It calls display.php which displays the image as well as the title.
But I want the image to be displayed as a thumbnail of size 150X100.
Do i have to store it in directory before I convert it and display? If so, can somebody tell me how to convert the image into a thumbnail?

Here is my code:

Code: [Select]
<table align="center" bgcolor="silver">
<form action="display.php" method ="post">
<tr>
<td align="center"><b><font color="#214754" size="3">Reg</font></b></td>
</tr>
<tr>
<td> Title: </td>
<td><input type="text" size ="40" name="registry_title1" /></td>
</tr>
<tr>
<td> Image URL: </td>
<td><input type="text" size ="40" name="registry_img1" /></td>
</tr>

<tr>
<td>
<input type="submit" name="submit" value="submit" />
</td>
</tr>
</form>
</table>
</div>
here is the code for display.php
Code: [Select]
<?
/* Get Registry details */

$registry_img1=$_POST['registry_img1'];
$registry_title1=stripslashes($_POST['registry_title1']);
?>

<html>
<body>
<table width='600' cellspacing='0' cellpadding='0' border='0' bgcolor='#ffffff' align='center'>
<tr>
<td style='padding-bottom:8px';><font style='font-size:18px;color:rgb(0, 0, 0);text-decoration:none'><a style='font-size:18px;color:rgb(0, 0, 0);text-decoration:none' href='#'><? echo ucfirst($url); ?> Registry</font></a></td>
</tr>

<tr>
<td style='padding-bottom:10px;'>
<a style='color:rgb(206, 0, 0);font-size:14px;font-weight:bold;line-height:18px;text-decoration:none' href='<? echo $registry_url1; ?>' target='_blank'><img style='padding-right:10px' border='0' align='left' alt='' src='<? echo $registry_img1; ?>'><? echo $registry_title1;?>
</a>
</td>

</table>

</body>
</html>

Am Having Problem Displaying Image From My Database

Similar Tutorials

View Content

the script succesfuly insert image to the database bt, i cant be able to display it on my pages, any help i will appreciate Attached Files saveimage.php 1.15KB 2 downloads images_tbl.php 192bytes 3 downloads

Trying To Get Image From Ftp Server With Php And Displaying It In Webpage.

Similar Tutorials

View Content

Hi guys,
I have a page that lets you upload an image to a an ftp server. Now I am trying to create a page that will get those images and display them in a web page. I am trying to do this with php.

Does anyone know if it can be done?

Thanks guys,

Wordpress- Displaying Date Via. Image

Similar Tutorials

View Content

Hi, everyone. I am trying to put together a WordPress for a client and although I hate asking for help - I am at a total standstill! She wants date tabs on the left of each post (on the main homepage and then each post; NOT pages- obviously, since those aren't 'dated', per say...).

Here is the function.php code I am using:
Code: [Select]

// custom date tab - creates two divs
// first div pulls in a background png image named by date
// and adds year dynamically. Placed via css to left of post as
// a hanging tab.

remove_action('genesis_before_post_content', 'genesis_post_info');
add_action('genesis_before_post_content', 'family_tree_post_info');
function family_tree_post_info() {
if(is_page()) return; // don't do post-info on pages
echo '<div class="post-date-wrap">';
echo '<div class="post-date post-date-background" style="background:url(';
echo CHILD_URL;;
echo '/images/date';
echo the_time('F');
echo '/';
echo the_time('M');
echo '.png) no-repeat 4px -3px">';
echo the_time('Y');
echo '</div>';
echo '</div>';
echo '<div class="post-info">';
genesis_post_comments_link(__('Leave a Comment', 'child'), __('1 Comment', 'child'), __('% Comments', 'child'));
edit_post_link(__('(Edit)', 'child'), '', ''); // if logged in
echo '</div>';
}

This shouldn't matter [much] - but here is the style.css for the post-dat:

Code: [Select]

.post-date-wrap {
width: 65px;
height: 73px;
}

.content-sidebar .post-date-wrap {
float: left;
margin-top: -42px;
margin-left: -107px;
background: url(images/dates/bgd-left.png) no-repeat top left;
}

.content-sidebar-sidebar .post-date-wrap {
float: left;
margin-top: -42px;
margin-left: -84px;
background: url(images/dates/bgd-left.png) no-repeat top left;
}

.full-width-content .post-date-wrap {
float: left;
margin-top: -42px;
margin-left: -89px;
background: url(images/dates/bgd-left.png) no-repeat top left;
}

.sidebar-content .post-date-wrap {
width: 65px;
height: 73px;
float: right;
margin-top: -42px;
margin-right: -89px;
background: url(images/dates/bgd-right.png) no-repeat top left;
}

.sidebar-sidebar-content .post-date-wrap {
width: 65px;
height: 73px;
float: right;
margin-top: -42px;
margin-right: -86px;
background: url(images/dates/bgd-right.png) no-repeat top left;
}

.sidebar-content-sidebar .post-date-wrap, .sidebar-content-sidebar .post-date {
visibility: collapse;
height: 0;
width: 0;
}

.post-date {
width: 62px;
height: 17px;
display: inline;
text-align: center;
color: #4e260d;
padding: 50px 0 0 5px;
font-size: 9px;
}

.content-sidebar .post-date, .content-sidebar-sidebar .post-date, .full-width-content .post-date {
float: left;
}

.sidebar-content .post-date, .sidebar-sidebar-content .post-date {
float: right;
}

Here are two pictures:
* I have one of these for every day; for each month - found in images/dates - thereafter organized by months (in written out for October, November; then 1.png; 2.png, etc. for each DAY of the month)

This is how it is coming up right now; only the year (that really small lettering that says 2011):

I need the images with the month & date to appear inside of that box/tab. I have been working on this for too long and cannot get it to go. My 'thing' is design, so I really hope you can all help me! I just can't get PHP down.. I am sure it is something so ridiculous that I cannot see.

Thanks for your help!!

- Nicole

Storing Then Displaying Image From Database

Similar Tutorials

View Content

Hi, wondering if somebody can tell me where I'm going wrong (I'm new to all of this). I have the following php code which uploads an image file into my database:

Code: [Select]
//Connect to database
include 'Resources/Include/db.inc.php';

$tmp=$_FILES['image']['tmp_name'];
//get users IP
$ip=$_SERVER['REMOTE_ADDR'];

//Don't do anything if file wasn't selected
if (!empty($tmp)) {

//Copy file to temporary folder
copy($tmp, "./temporary/".$ip."");

//open the copied image, ready to encode into text to go into the database
$filename1 = "./temporary/".$ip;
$fp1 = fopen($filename1, "rb");

//record the image contents into a variable
$contents1 = fread($fp1, filesize($filename1));

$contents1 = addslashes($contents1);

//close the file
fclose($fp1);

$ftype = $_FILES['image']['type'];

//insert information into the database
if(!mysql_query("INSERT INTO LetterImages (Data,Type,LetterID,Page)"." VALUES ( '$contents1', '$ftype',1,1)")){
echo mysql_error();
}

//delete the temporary file we made
unlink($filename1);

}

This seems to work ok, as when I go to the LetterImages table there is now an additional row with a file in the blob field. I then have the following code which is supposed to display the image:

Code: [Select]
$result=mysql_query("SELECT * FROM LetterImages WHERE LetterID=1 AND Page=1");

//fetch data from database
$sqldata=mysql_fetch_array($result);

$encoded=stripslashes($sqldata['Data']);
$ftype=$sqldata['Type'];

//tell the browser what type of image to display
header("Content-type: $ftype");

//decode and echo the image data
echo $encoded;

Instead of displaying an image, however, this just displays pages and pages of incomprehensible data. Can anybody tell me where I'm going horribly wrong?

Displaying Information When Image Is Clicked

Similar Tutorials

View Content

Hi

I'm trying to create a page where there is a grid of images and if you click on one of those images it will expand. Once the image has expanded and you click on the i icon that is below it, the i icon will display the information about that image.

This isn't like a popup or anything like that, because when you click on the image and it expands, it expands in the same spot and it doesn't popup. Here is the html

$str.='<div class="portfolio-banner-wrapper'.$langClass.'" style="position: relative;">';
$str.=      '<a class="portfolio-arrow-left'.$langClass.' portfolio-navigate'.$langClass.'" href="javascript:void(0);"><img src="images/left-arrow.png" onmouseover="this.src=\'images/blue-left-arrow.png\'" onmouseout="this.src=\'images/left-arrow.png\'" /></a>';
$str.=      '<a class="portfolio-arrow-right'.$langClass.' portfolio-navigate'.$langClass.'" href="javascript:void(0);"><img src="images/right-arrow.png" onmouseover="this.src=\'images/blue-right-arrow.png\'" onmouseout="this.src=\'images/right-arrow.png\'"/></a>';
$str.=      '<div class="portfolio-banner-inner">';
$str.=          $this->getPortfolioTest();
$str.=      '</div>';
$str.='</div>';

$str.='<div class="overlay-bg'.$langClass.'">';
$str.=      '<div class="additional-navigation-wrapper'.$langClass.'">';
$str.=          '<div class="info'.$langClass.'" style="display: none;">';
$str.=              '<a class="border-bottom-white padding-level-one inactive additional-nav-info1" href="javascript:void(0);">';
$str.=                  '<img class="icon" src="images/i_icon.png" />';
$str.=                  '<img class="nav-arrow no-action floatright" src="images/nav-arrow-white.png" />';
$str.=                  '<span class="clearboth"></span>';
$str.=              '</a>';
$str.=              '<div class="additional-nav-info-wrapper1">';
$str.=                  $this->i_icon();                   
$str.=              '</div>';
$str.=          '</div>';
$str.=      '</div>';

Here is the php

function getPortfolioTest(){
        global $_PRODUCTS_TABLE, $_HTTP_ADDRESS, $_PRODUCTS_IMAGES_DIR;
        $i=0;
        $cpt = 1;
        $str = '';
        $whichCount = 1;
        $jobSearch='';
        $query = mysql_query("SELECT * FROM $_PRODUCTS_TABLE WHERE `active`='1' AND `image` LIKE '%.%'".$jobSearch." ORDER BY `client` ASC");
        $combineArr = mysql_num_rows($query);
        while( $result = mysql_fetch_object($query) ){
        	$product = new Product($result->id);
        	$product->setFromDatabase();

        	$linkOut = getSEOLink($product->id);
        	$target = "";
        	if(trim($product->linkout) != ""){
        		$linkOut = $product->linkout;
        		$target = ' target="_blank"';
        	}

        	if($whichCount == 1){
				$portfolioClass="portfolio-active";
				$style = "position: absolute; left:0%; top:0; width:100%;";	
			}else{
				$portfolioClass="portfolio-inactive";	
				$style = "position: absolute; left:-100%; top:0; width:100%;";	
			}

        	if($whichCount == 1){
				$str.='<div id="portfolio-slide'.$i.'" class="portfolio-slide '.$portfolioClass.'" style="'.$style.'">';
	            $str.=		'<div class="portfolio-slide-inner">';
	            $str.=			'<div class="portfolio-banner-content portfolio-banner-left">';
	            $str.=				'<div class="portfolio-banner-header">';
	            $str.=				'</div>';
	            $str.=				'<div class="portfolio-banner-copy">';
	            $str.=					'<ul id="gallery">';	
			}

			$str.=						'<li>';
			$str.=							'<a'.$target.' href="javascript:void(0);">';
            $str.=								'<img src="'.$_HTTP_ADDRESS.'products_images/'.$result->image.'">';
            								echo "this is getPortfolioTest";
            $str.=							'</a>';
            $str.=							'<span>';
            $str.=								'<h3>'.$result->name.'</h3>';
            $str.=								$result->description;
            $str.=							'</span>';	
            $str.=						'</li>';

			if($whichCount % 9 == 0 && $whichCount < $combineArr){

				$i++;
				$str.=						'<div class="clearboth"></div>';
				$str.=					'</ul>';
	            $str.=				'</div>';
	     
	            $str.=			'</div>';
	            $str.=		'</div>';
	            $str.='</div>';		

	            $str.='<div id="portfolio-slide'.$i.'" class="portfolio-slide '.$portfolioClass.'" style="'.$style.'">';
	            $str.=		'<div class="portfolio-slide-inner">';
	            $str.=			'<div class="portfolio-banner-content portfolio-banner-left">';
	            $str.=				'<div class="portfolio-banner-header">';
	            $str.=				'</div>';
	            $str.=				'<div class="portfolio-banner-copy">';
	            $str.=					'<ul id="gallery">';					
			}
			
			if($whichCount == $combineArr){
				$str.=						'<div class="clearboth"></div>';
				$str.=					'</ul>';
	            $str.=				'</div>';
	          
	            $str.=			'</div>';
	            $str.=		'</div>';
	            $str.='</div>';				
	        }
			$whichCount++;

            $cpt++;
          
        }
        return $str;
    }

    function i_icon(){
    	global $_PRODUCTS_TABLE, $_HTTP_ADDRESS, $_PRODUCTS_IMAGES_DIR;

    	$str = '';
    	$i = 0;

    	$query = mysql_query("SELECT * FROM $_PRODUCTS_TABLE WHERE `active`='1'");
                    while($result = mysql_fetch_object($query)){
		$str.=                  '<div class="additional-nav-info-inner'.$i.' overlay-bg" style="display:none;">';
		$str.=                      '<ul>';
		$str.=                          '<h3>'.$result->name.'</h3>';
		$str.=                          '<p>';
		$str.=                             $result->overview;
		$str.=                          '</p>';
		$str.=                      '</ul>';
		$str.=                  '</div>';
		$i++;  
		                    }

		return $str;                     
		    }

Here is the Jquery

function galleryInit(){
	$('#gallery li').hover(
	    function(){$('span',this).slideToggle('fast');},
	    function(){$('span',this).slideToggle('fast');
    });

    	$(".portfolio-banner-inner li").click(function(e){

		if($(".activeExpand").length > 0){

			$(".portfolio-active").css({"left":"0%"});
			$(".portfolio-active").prevAll().css({"left":"-100%"});
			$(".portfolio-active").nextAll().css({"left":"100%"});
			$(".portfolio-banner-inner li").removeClass("inactiveExpand").removeClass("activeExpand").removeAttr("style").find("img").removeAttr("style");
			$(".portfolio-arrow-left-scroll").addClass("portfolio-arrow-left").removeClass("portfolio-arrow-left-scroll");
	    	$(".portfolio-arrow-right-scroll").addClass("portfolio-arrow-right").removeClass("portfolio-arrow-right-scroll");


			return false;	
		} 

		$this = $(this);
		$(".portfolio-banner-inner li").addClass("inactiveExpand")
		$this.removeClass("inactiveExpand").addClass("activeExpand");
		
		
		$(".portfolio-slide").css({"left":"0"})

		$this.parents(".portfolio-slide:first").prevAll().find(".portfolio-slide-inner li").css({"position":"absolute","width":"100%","height":"100%","top":"0","left":"-100%"}).find("img").css({"width":"952px","height":"502px"});
		$this.parents(".portfolio-slide:first").nextAll().find(".portfolio-slide-inner li").css({"position":"absolute","width":"100%","height":"100%","top":"0","left":"100%"}).find("img").css({"width":"952px","height":"502px"});
		

		$this.siblings().prevAll().css({"position":"absolute","width":"100%","height":"100%","top":"0","left":"-100%"}).find("img").css({"width":"952px","height":"502px"});
		$this.siblings().nextAll().css({"position":"absolute","width":"100%","height":"100%","top":"0","left":"100%"}).find("img").css({"width":"952px","height":"502px"});


		$this.css({"position":"absolute","width":"100%","height":"100%","top":"0","left":"0"});
		$this.find("img").animate({
	        width: '952px',//what the width of the image to be expanded is
	        height: '502px'//what the width of the image to be expanded is
	    }, 200);


	    $(".portfolio-arrow-left").addClass("portfolio-arrow-left-scroll").removeClass("portfolio-arrow-left");
	    $(".portfolio-arrow-right").addClass("portfolio-arrow-right-scroll").removeClass("portfolio-arrow-right");

	});

}

function imageSlider(direction){

	$activeExpandWhere = $(".activeExpand");
	if($(".activeExpand").css("left") != "0px") return false;

	if(direction == "right"){
		$(".inactiveExpand").css({"left":"100%"});

		$(".activeExpand").animate({"left":-100+"%"},500,"easeInCubic",function(){
		$(this).removeClass("activeExpand").addClass("inactiveExpand");

		});
		if($(".activeExpand").next("li").length == 0){
			if($activeExpandWhere.parents(".portfolio-slide:first").next().length == 0){
				$(".portfolio-slide").removeClass("portfolio-active").addClass("portfolio-inactive");
				$(".portfolio-slide:first").removeClass("portfolio-inactive").addClass("portfolio-active");
				$(".portfolio-slide:first").find(".portfolio-slide-inner li").eq(0).animate({"left":0+"%"},500,"easeInCubic",function(){
					$(this).removeClass("inactiveExpand").addClass("activeExpand");
				});
				return false;
			}
			$(".portfolio-slide").removeClass("portfolio-active").addClass("portfolio-inactive");
			$activeExpandWhere.parents(".portfolio-slide:first").next().removeClass("portfolio-inactive").addClass("portfolio-active");
			$activeExpandWhere.parents(".portfolio-slide:first").next().find(".portfolio-slide-inner li").eq(0).animate({"left":0+"%"},500,"easeInCubic",function(){
				$(this).removeClass("inactiveExpand").addClass("activeExpand");
			});
			return false;
		}
		$(".activeExpand").next("li").animate({"left":0+"%"},500,"easeInCubic",function(){
			$(this).removeClass("inactiveExpand").addClass("activeExpand");
		});
		return false;
	}

	$(".inactiveExpand").css({"left":"-100%"});

	$(".activeExpand").animate({"left":100+"%"},500,"easeInCubic",function(){
		$(this).removeClass("activeExpand").addClass("inactiveExpand");

	});
	if($(".activeExpand").prev("li").length == 0){
		if($activeExpandWhere.parents(".portfolio-slide:first").prev().length == 0){
			$(".portfolio-slide").removeClass("portfolio-active").addClass("portfolio-inactive");
			$(".portfolio-slide:last").removeClass("portfolio-inactive").addClass("portfolio-active");
			$(".portfolio-slide:last").find(".portfolio-slide-inner li").eq(($(".portfolio-slide:last").find(".portfolio-slide-inner li").length-1)).animate({"left":0+"%"},500,"easeInCubic",function(){
				$(this).removeClass("inactiveExpand").addClass("activeExpand");
			});
			return false;
		}

		$(".portfolio-slide").removeClass("portfolio-active").addClass("portfolio-inactive");
		$activeExpandWhere.parents(".portfolio-slide:first").prev().removeClass("portfolio-inactive").addClass("portfolio-active");

		$activeExpandWhere.parents(".portfolio-slide:first").prev().find(".portfolio-slide-inner li").eq(0).animate({"left":0+"%"},500,"easeInCubic",function(){
			$(this).removeClass("inactiveExpand").addClass("activeExpand");
		});
		return false;
	}
	$(".activeExpand").prev("li").animate({"left":0+"%"},500,"easeInCubic",function(){
		$(this).removeClass("inactiveExpand").addClass("activeExpand");
	});


	
	}

$(document).ready(function(e) {

	galleryInit();

	
	$(".additional-nav-info1").click(function(){
		if($(".additional-nav-info-inner1").css("display") == "block"){
			$(this).removeClass("active");
			$(this).addClass("inactive");
			$(".additional-nav-info-inner1").stop().slideToggle(250);
		} else {
			$(this).removeClass("inactive");
			$(this).addClass("active");
			$(".additional-nav-info-inner1").stop().slideToggle(250);
		}
	});

	$(".additional-nav-info-inner1 a").hover(function(){
		$(this).find("img").eq(0).fadeOut(250);
		$(this).find("img").eq(1).fadeIn(250);
	}, function(){
		$(this).find("img").eq(0).fadeIn(250);
		$(this).find("img").eq(1).fadeOut(250);
	});

}

Issues Displaying One Image From 2 Tables

Similar Tutorials

View Content

Hello guys,

I'm trying to figure out how to display and image that is located in one table (profile) and make it correspond to a variable on another table (latest_post) to where you echo out the variable from the table(latest_post) and the image from table(profile) appears next to the post.

any how here is my code:
Code: [Select]
<?php

  $query = mysql_fetch_array(mysql_query("SELECT person FROM latest_post AND avatar_img FROM profile WHERE id FROM latest_post = id FROM profile  "));
  echo $query['avatar_img'];

?>

Also when echoing out the query I only get the name of the image and not the image this is most likely because my image is stored in a different folder right?

Thanks in advance, your help is much appreciated since I'm barely a beginner at php and MySQL.