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PHP - Retrieve Images From Database
I have this code which uoploads and resizes the images into a folder without a problem. My problem is I am unable to view the Thumb or Full image which has uploaded to the folder and the database with the other data. How do i correct this problem code below. Thanks
upload form and script Code: [Select] <?php //database properties $dbhost = 'localhost'; $dbuser = 'user'; $dbpass = 'pass'; $dbname = 'DB'; // make a connection to mysql here $conn = mysql_connect ($dbhost, $dbuser, $dbpass) or die ("I cannot connect to the database because: " . mysql_error()); mysql_select_db ($dbname) or die ("I cannot select the database '$dbname' because: " . mysql_error()); /******** start function ********/ function errors($error){ if (!empty($error)) { $i = 0; echo "<blockquote>\n"; while ($i < count($error)){ echo "<p><span class=\"warning\">".$error[$i]."</span></p>\n"; $i ++;} echo "</blockquote>\n"; }// close if empty errors } // close function // ----------------Create Smaller version of large image--------------------- function createthumbfull($src_filename, $dst_filename_full) { // Get information about the image list($src_width, $src_height, $type, $attr) = getimagesize( $src_filename ); // Load the image based on filetype switch( $type ) { case IMAGETYPE_JPEG: $starting_image = imagecreatefromjpeg( $src_filename ); break; case IMAGETYPE_PNG: $starting_image = imagecreatefrompng( $src_filename ); break; case IMAGETYPE_GIF: $starting_image = imagecreatefromgif( $src_filename ); break; default: return false; } // get the image to create thumbnail from $starting_image; // get image width $width = imagesx($starting_image); // get image height $height = imagesy($starting_image); // size to create thumnail width $thumb_width = 600; // divide iwidth by specified thumb size $constant = $width/$thumb_width; // round height by constant and add t thumb_height $thumb_height = round($height/$constant, 0); //create thumb with true colours $thumb_image = imagecreatetruecolor($thumb_width, $thumb_height); //create thumbnail resampled to make image smooth imagecopyresampled($thumb_image, $starting_image, 0, 0, 0, 0, $thumb_width, $thumb_height, $width, $height); $ran = rand () ; $thumb1 = $ran.".jpg"; global $thumb_Add_full; $thumb_Add_full = $dst_filename_full; $thumb_Add_full .= $thumb1; imagejpeg($thumb_image, "" .$dst_filename_full. "$thumb1"); } //-------------------- create thumbnail -------------------------------------------- //$src_filename = source of image //$dst_filename_thumb = location to store final image function createthumb($src_filename, $dst_filename_thumb) { $size = getimagesize($src_filename); // get image size $stype = $size['mime']; //get image type : gif,png,jpg $w = $size[0]; // width of image $h = $size[1]; // height of image //do a switch statment to create image from right image type switch($stype) { //if image is gif create from gif case 'image/gif': $simg = imagecreatefromgif($src_filename); break; //if image is jpeg create from jpeg case 'image/jpeg': $simg = imagecreatefromjpeg($src_filename); break; //if image is png create from png case 'image/png': $simg = imagecreatefrompng($src_filename); break; } $width = $w; // get image width $height = $h; // get image height // size to create thumnail width $thumb_width = 150; $thumb_height = 150; //use true colour for image $dimg = imagecreatetruecolor($thumb_width, $thumb_height); $wm = $width/$thumb_width; //width divided by new width $hm = $height/$thumb_height; //height divided by new height $h_height = $thumb_height/2; //ass new height and chop in half $w_height = $thumb_width/2; //ass new height and chop in half //if original width is more then original height then modify by width if($w> $h) { $adjusted_width = $w / $hm; // add original width and divide it by modified height and add to var $half_width = $adjusted_width / 2; // chop width in half $int_width = $half_width - $w_height; // take away modified height from new width //make a copy of the image imagecopyresampled($dimg,$simg,-$int_width,0,0,0,$adjusted_width,$thumb_height,$w,$h); //else if original width is less or equal to original height then modify by height } elseif(($w <$h) || ($w == $h)) { $adjusted_height = $h / $wm; // diving original height by modified width $half_height = $adjusted_height / 2; // chop height in half $int_height = $half_height - $h_height; // take away modified height from new width //make a copy of the image imagecopyresampled($dimg,$simg,0,-$int_height,0,0,$thumb_width,$adjusted_height,$w,$h); } else { // don't modify image and make a copy imagecopyresampled($dimg,$simg,0,0,0,0,$thumb_width,$thumb_height,$w,$h); } $ran = "thumb_".rand (); // generate random number and add to a string $thumb2 = $ran.".jpg"; //add random string to a var and append .jpg on the end global $thumb_Add_thumb; //make this var available outside the function as it will be needed later. $thumb_Add_thumb = $dst_filename_thumb; // add destination $thumb_Add_thumb .= $thumb2; // append file name on the end //actually create the final image imagejpeg($dimg, "" .$dst_filename_thumb. "$thumb2"); } /********end functions ***********/ // if form submitted then process form if (isset($_POST['submit'])){ // check feilds are not empty $imageTitle = trim($_POST['imageTitle']); if (strlen($imageTitle) < 3 || strlen($imageTitle) > 255) { $error[] = 'Title must be at between 3 and 255 charactors.'; } // check feilds are not empty or check image file size if (!isset($_FILES["uploaded"])) { $error[] = 'You forgot to select an image.'; } // check feilds are not empty $imageDesc = trim($_POST['imageDesc']); if (strlen($imageDesc) < 3) { $error[] = 'Please enter a description for your image.'; } // location where inital upload will be moved to $target = "productimages/" . $_FILES['uploaded']['name'] ; // find the type of image switch ($_FILES["uploaded"]["type"]) { case $_FILES["uploaded"]["type"] == "image/gif": move_uploaded_file($_FILES["uploaded"]["tmp_name"],$target); break; case $_FILES["uploaded"]["type"] == "image/jpeg": move_uploaded_file($_FILES["uploaded"]["tmp_name"],$target); break; case $_FILES["uploaded"]["type"] == "image/pjpeg": move_uploaded_file($_FILES["uploaded"]["tmp_name"],$target); break; case $_FILES["uploaded"]["type"] == "image/png": move_uploaded_file($_FILES["uploaded"]["tmp_name"],$target); break; case $_FILES["uploaded"]["type"] == "image/x-png": move_uploaded_file($_FILES["uploaded"]["tmp_name"],$target); break; default: $error[] = 'Wrong image type selected. Only JPG, PNG or GIF accepted!.'; } // if valadation is okay then carry on if (!$error) { // post form data $imageTitle = $_POST['imageTitle']; $imageDesc = $_POST['imageDesc']; //strip any tags from input $imageTitle = strip_tags($imageTitle); $imageDesc = strip_tags($imageDesc); // add slashes if needed if(!get_magic_quotes_gpc()) { $imageTitle = addslashes($imageTitle); $imageDesc = addslashes($imageDesc); } // remove any harhful code and stop sql injection $imageTitle = mysql_real_escape_string($imageTitle); $imageDesc = mysql_real_escape_string($imageDesc); //add target location to varible $src_filename $src_filename = $target; // define file locations for full sized and thumbnail images $dst_filename_full = 'productimages/'; $dst_filename_thumb = 'productthumb/'; // create the images createthumbfull($src_filename, $dst_filename_full); //call function to create full sized image createthumb($src_filename, $dst_filename_thumb); //call function to create thumbnail image // delete original image as its not needed any more. unlink ($src_filename); // insert data into images table $query = "INSERT INTO table (imageTitle, imageThumb, imageFull, imageDesc) VALUES ('$imageTitle', '$thumb_Add_thumb', '$thumb_Add_full', '$imageDesc')"; $result = mysql_query($query) or die ('Cannot add image because: '. mysql_error()); // show a message to confirm results echo "<h3 align='center'> $imageTitle uploaded</h3>"; } } //dispaly any errors errors($error); ?> <form enctype="multipart/form-data" action="" method="post"> <fieldset> <legend>Upload Picture</legend> <p>Only JPEG, GIF, or PNG images accepted</p> <p><label>Title<br /></label><input type="text" name="imageTitle" <?php if (isset($error)){ echo "value=\"$imageTitle\""; }?> /></p> <p><label>Image<br /></label><input type="file" name="uploaded" /></p> <p><label>Description<br /></label><textarea name="imageDesc" cols="50" rows="10"><?php if (isset($error)){ echo "$imageDesc"; }?></textarea></p> <p><label> </label><input type="submit" name="submit" value="Add Image" /></p> </fieldset> </form> view data Code: [Select] <?php // Connects to your Database mysql_connect("localhost", "user", "pass") or die(mysql_error()) ; mysql_select_db("DB") or die(mysql_error()) ; //Retrieves data from MySQL $data = mysql_query("SELECT * FROM table") or die(mysql_error()); //Puts it into an array while($info = mysql_fetch_array( $data )) { ?> <?php echo "<img src=http://www.web.com/productimages/" .$info['imagefull'] ." > <br>"; ?> <?php echo "<img src=http://www.web.com/productthumb/" .$info['imageThumb'] ." > <br>"; ?> <?php echo "<b>imageTitle:</b> ".$info['imageTitle'] . "<br> "; ?> <?php echo "<b>imageDesc:</b> ".$info['imageDesc'] . " <hr>"; ?> <?php }?> Similar Tutorials
Query About How To Retrieve A Password From The Database And Compare To The One The User Has Entered
hi, i have a form that requires image to be uploaded. i know how to store the form values in a database but i don't know how to store the image into the database. i want the user to upload picture and preview the data provided review (including the image) before submitting to the database. thanks CAN ANYONE TELL ME WHAT i AM DOING WRONG. I WANT TO RETRIEVE THE VERY FIRST RECORD IN MY DATABASE WHEN THE CODE EXECUTE IT ONLY SHOWS THE LAST RECORD IN THE DATABASE <?PHP $thisMonth = date('M'); $thisDay = date('j'); $eventMonth = array(); $eventDay = array(); $eventTime = array(); $eventName = array(); $eventLocation = array(); $dbMonth=""; $dbDay=""; $i=0; $conn = odbc_connect('eventsDB','',''); $sql= "SELECT month,day, time, event,location FROM Events"; $rs="$conn,$sql"; if (!$conn) { exit("Connection Failed: " . $conn); } $rs=odbc_exec($conn,$sql); if(!$rs) { exit("Error in SQL"); } echo "DATABASE OPEN"; while($i<3) { $dbMonth= odbc_result($rs,"month"); echo $eventMonth[$i]=odbc_result($rs,"month")."\n"; if($dbMonth<>$thisMonth) { odbc_fetch_row($rs); } echo $eventMonth[$i]=odbc_result($rs,"month")."\n"; echo $eventDay[$i]=odbc_result($rs,"day")."\n"; echo $eventTime[$i]=odbc_result($rs,"time")."\n"; echo $eventDay[$i]=odbc_result($rs,"event")."\n"; echo $eventLocation[$i]=odbc_result($rs,"location")."\n"; $i++; odbc_fetch_row($rs); echo $i; } //ends while loop odbc_close($conn); ?> I AM WORKING ON A PROJECT , I NEED YOUR HELP. I WANT TO USE EXISTING SRC OF IFRAME + RETRIVED VALUE FROM DATABASE INTO IFRAME This is my data.php to retrive values from database into iframe without refresh web page.
<?php
$conn = new mysqli('localhost', 'root', '', 'x');
if ($conn->connect_error) {
die("Connection error: " . $conn->connect_error);
}
$result = $conn->query("SELECT number1 FROM users");
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
echo $row['number1'] . '<br>';
}
}
?>
<?php include('data.php') ?>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
</head>
<body>
<div id="show"></div>
<script type="text/javascript" src="jquery-3.1.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
setInterval(function () {
$('#show').load('data.php')
}, 1000);
});
</script>
<iframe id="w3" src="DATABASE RETRIVED VALUE" height="80px" width="300px" frameborder="0" style="border: 0; width:300px; height:80px; background-color: #FFF;"></iframe>
</body>
</html>
INDEX FILE CODES CAN SUCCESSFULLY RETRIVE VALUES FROM DATABASE BUT, I WANT TO STORE IT IN IFRAME SRC . VAR= URLS = 'HTTP://WWW.GOOGLE.COM' IFRAME SRC WILL BE VAR URLS + RETRIEVED VALUE FROM DATABASE
Hi, Im trying to retrieve HTML from a mysql database but nothing i've tried seems to work. The HTML im trying to retrieve is an iframe with a link and styles (code from amazon associates). Im trying to display links to specific products on amazon from the product page on my site. All data about the product is retrieved from the database so i have code to select the amazon link row in my database table but i cant get it to display. It says the html isnt a string so i cant echo it, fair enough. I have tried using the following code: Code: [Select] $get_buylink_sql = "SELECT mobo_buylink FROM mobo WHERE mobo_id = $mobo_id"; $get_buylink_res = mysqli_query($mysqli, $get_buylink_sql) or die(mysqli_error($mysqli)); while ($buylink = mysqli_fetch_array($get_buylink_res)) { echo"<iframe src=\"".$buylink."\" style=\"width:120px;height:240px;\" scrolling=\"no\" marginwidth=\"0\" marginheight=\"0\" frameborder=\"0\"></iframe>"; } mysqli_free_result($get_buylink_res); mysqli_close($mysqli);I have also tried putting the whole iframe code in the database which didnt work either. The mobo_id variable works fine for retrieving the rest of the data and i need to get the amazon link from the same record. I hope i've put this in a way you can understand, but if not i'll try and explain better and give you a link to my site if needed. Thanks, Alex I have code for displaying result from database:
foreach ($_SESSION["products"] as $cart_itm)
{
$product_code = $cart_itm["code"];
$results = $mysqli->query("SELECT * FROM products WHERE product_code='$product_code' LIMIT 1");
$obj = $results->fetch_object();
Now, I display data with: $obj->product_name
When I get the list, how to put inline numbers (1 2 3 4 ...)?
1 products1
2 products1
3 products1
4 products1..
Hello people, I have created an application in Code Charge where my site users can login and play games online.The game code is run via flash and html. The Game has 5 stages and from a page in my application called transfertogame.php , i transfer them to the game page. Now i want to be able to track each players progress via database so that when they log back in and they get to the transfertogame.php page, it can check the page they were on before and immediately take them to the page without starting from page 1. Any help will be appreciated as i am very poor with php scripting This is the session ID i guess is being used from my application page //Initialize Method @4-537EA73F function Initialize() { if(!$this->Visible) return; $this->ds->Parameters["sesUserID"] = CCGetSession("UserID"); } //End Initialize Method //Validate Method @4-7E1FC38C function Validate() { $Validation = true; $Where = ""; $this->CCSEventResult = CCGetEvent($this->CCSEvents, "OnValidate"); return (($this->Errors->Count() == 0) && $Validation); } //End Validate Method Code: [Select] <?php echo $row_Recordset1['name']; ?> i want to convert the name in to like <a href=main.php?name= $row_Recordset1['name'] >> in databse i has 3 name Steve Mark Lisa i want that names into link like main.php?name=mark and so on how to do it help me h Am a newbie in php. Since I can't insert values to the database with respect to a user Id or with any other token using WHERE clause. I.e "INSERT INTO receipts(date) VALUES(example) where id="**....." If I need to fetch several values of column for a particular user, how do I go about it? Thank you!!! Hi, after following lots of advice and changing to MySqli I am running into a few probs. This is me just probably missing something stupid, I know what I want, but can't figure out what query I should use and where I should place it. All the queries I have tried have failed.
I just need a query that gets the $current_stored_password from the password field on the database, to confirm the last check
elseif
($current_password !== $current_stored_password) {
include 'includes/overall/header.php';
echo $current_password . ' AND ' . $_POST['current_password'] . ' Password and password again do not match';
include 'includes/overall/header.php';
}
Here is the whole script.
<?php
session_start();
error_reporting(0);
//ini_set('display_errors', '1');
require( 'database.php' );
$username = $_SESSION['loggedinuser'];
$current_stored_password = $_SESSION['password'];
$current_password = $_POST['current_password'];
$password = mysqli_real_escape_string($con, md5( $_POST['password']));
$password_again = mysqli_real_escape_string($con, md5( $_POST['password_again']));
// Run checks
if (isset($_POST['current_password'], $_POST['password'], $_POST['password_again'])) {
if( strlen( $_POST['current_password'] ) < 8 )
{
include('includes/overall/header.php');
echo "Password Must Be 8 or More Characters.";
include('includes/overall/footer.php');
}
elseif( strlen( $_POST['password'] ) < 8 )
{
include('includes/overall/header.php');
echo "Password Must Be 8 or More Characters.";
include('includes/overall/footer.php');
}
elseif
( strlen( $_POST['password_again'] ) < 8 )
{
include('includes/overall/header.php');
echo "Password Must Be 8 or More Characters.";
include('includes/overall/footer.php');
}
elseif
($password !== $password_again) {
include 'includes/overall/header.php';
echo ' Password and password again do not match';
include 'includes/overall/header.php';
}
elseif
($current_password !== $current_stored_password) {
include 'includes/overall/header.php';
echo $current_password . ' AND ' . $_POST['current_password'] . ' Password and password again do not match';
include 'includes/overall/header.php';
} else {
// Define a query to run
$query = "UPDATE `user` SET `password` = '$password' WHERE `username` = '$username'";
// Query the database
$result = mysqli_query($con,$query);
// Check if the query failed
if( !$result )
{
die('There was a problem executing the query ('.$query.'):<br>('.mysqli_errno($con).') '.mysqli_error($con));
}
else {
include 'includes/overall/header.php';
echo 'Password has been changed';
include 'includes/overall/footer.php';
}
}
}
// Close the connection
mysqli_close($con);
?>
At the moment the message displayed when the form is submitted is
echo $current_password . ' AND ' . $_POST['current_password'] . ' Password and password again do not match';How do I retrieve the password from the database to compare against the current password entered by the user? Any help is much appreciated. PS. Yes I know I have repeated code and that md5 is not secure, but I am just building onto a template I got and will be making changes to shorten the code and secure the password soon As a complete newbie to php and webdesigning i have a following problem.I would like to retrieve the data from database and display it in a drop down menu.Then i should allow the user to select the values from drop down list along with other details,in other words i have to embed the drop down output as the form input for the user and store the form data in another table.I am running a xampp server and i am using php 5.4 version.Please help.My code is as follows.In this case project_name is displayed as the drop down output.but how do i use the same drop down output as a input in the form. <html> <head></head> <body> <?php error_reporting(E_ALL ^ E_DEPRECATED); include 'connect.php' ; $tbl_name="projects"; $sql="SELECT project_name FROM $tbl_name "; $result=mysql_query($sql); if($result === FALSE) { die(mysql_error()); } ?> <form name="resources" action="hourssubmit.php" method="post" > <?php echo "<select name='project_name'>"; while ($row = mysql_fetch_array($result)) { echo "<option value='" . $row['project_name'] ."'>" . $row['project_name'] ."</option>"; } echo "</select>"; ?> </form> </body> </html> i have 8 division (div), i want to display 4 rows in 4 division and the remain 4 rows in the next 4 division here is my code structure for carousel
<div class="nyie-outer"> second row third row
fourth row fifth row sixth row seven throw eighth row
</div><!--/.second four rows here-->
sql code
CREATE TABLE product( php code
<?php how can i echo that result in those rows
Hello.
I have a bit of a problem. When I fetch the link field from the database.i don't see an actual link on the page.
One more thing, what type of field should I use to store the link in the database? Probably there is where I went wrong.
All help is
hi guys, im new to this forum I'm new also to php, I need help from you guys: I want to display personal information from a certain person (the data is on the mysql database) using his name as a link: example: (index.php) names 1. Bill Gates 2. Mr. nice Guy i want to click Bill Gates (output.php) Name: Bill Gates Country:xxxx Age: xx etc. How can i make this or how to learn this? Hi all, while I was coding a script for my website something struck me which ive not really got an idea how to code it, but anyway, in my Database ive got a table which holds image URL's but how could I echo them out in PHP which will show the image? Not just the writing. Thanks for your help. I want users to be allowed to upload images to a table in a database but I cannot seem to find a suitable way to implementing using the code I already have, here is the code below; Code: [Select] <?php error_reporting(E_ALL ^ E_NOTICE); ini_set("display_errors", 1); require_once ('./includes/config.inc.php'); require_once (MYSQL); $add_cat_errors = array(); if ($_SERVER['REQUEST_METHOD'] == 'POST') { if (!empty($_POST['product'])) { $prod = mysqli_real_escape_string($dbc, strip_tags($_POST['product'])); } else { $add_cat_errors['product'] = 'Please enter a Product Name!'; } if (filter_var($_POST['prod_descr'])) { $prod_descr = mysqli_real_escape_string($dbc, strip_tags($_POST['prod_descr'])); } else { $add_cat_errors['prod_descr'] = 'Please enter a Product Description!'; } // Check for a category: if (filter_var($_POST['cat'], FILTER_VALIDATE_INT, array('min_range' => 1))) { $catID = $_POST['cat']; } else { // No category selected. $add_page_errors['cat'] = 'Please select a category!'; } if (filter_var($_POST['price'])) { $price = mysqli_real_escape_string($dbc, strip_tags($_POST['price'])); } else { $add_cat_errors['price'] = 'Please enter a Product Description!'; } if (filter_var($_POST['stock'])) { $stock = mysqli_real_escape_string($dbc, strip_tags($_POST['stock'])); } else { $add_cat_errors['stock'] = 'Please enter a Product Description!'; } if (empty($add_cat_errors)) { $query = "INSERT INTO product (product, prod_descr, catID, price, image, stock) VALUES ('$prod', '$prod_descr', '$catID', '$price', '$image', '$stock')"; $r = mysqli_query ($dbc, $query); if (mysqli_affected_rows($dbc) == 1) { echo '<p>Record Successfully Added!!!!!</p>'; $_POST = array(); } else { trigger_error('OH NOOOOO!!!!'); } } } require_once ('./includes/form_functions.inc.php'); ?> <form action="add_product.php" method="post"> Product Name: <?php create_form_input('product', 'text', $add_cat_errors);?> Product Description: <?php create_form_input ('prod_descr', 'text', $add_cat_errors);?> Category: <select name="cat"<?php if (array_key_exists('cat', $add_cat_errors))?>> <option>Select a Category</option> <?php $q = "SELECT catID, cat FROM category ORDER BY cat ASC"; $r = mysqli_query($dbc, $q); while ($row = mysqli_fetch_array($r, MYSQLI_NUM)) { echo "<option value=\"$row[0]\""; if (isset($_POST['cat']) && ($_POST['cat'] == $row[0])) echo 'selected="selected"'; echo ">$row[1]</option>\n"; } ?> Price: <?php create_form_input('price', 'text', $add_cat_errors);?> Upload an Image: <?php if(array_key_exists('image', $add_cat_errors)) { echo '<input type="file" name="image" />'; } ?> Stock: <?php create_form_input('stock', 'text', $add_cat_errors);?> <input type="submit" name="submit_button" value="ADD RECORD" /> </form> Everything else writes perfectly but I have been trying ways to create a file upload and write successfully to the database. The 'image' field has a data type of 'varchar'. I apologise immensely for the long winded explanation but if anyone could help me please that would be much appreciated. Hello, I am new to PHP coding, and I have some issues trying to upload multiple images into database. The code works only for 1 image, but when I try to add foreach, I receive this message: Warning: Invalid argument supplied for foreach() . Can anyone help me ?
I want to display 3 clickable images in a single row which repeats as long as there is data in the database, so far it is displaying a single clickable image from the database. below is all the code.. <table width="362" border="0"> <?php $sql=mysql_query("select * from `publication` GROUP BY `catsue`") or die(mysql_error()); $num=mysql_num_rows($sql); while($rowfor=mysql_fetch_array($sql)) { $cat=$rowfor['catsue']; $pic=mysql_query("select * from `category` where `catsue`='$cat'") or die(mysql_error()); $picP=mysql_fetch_array($pic); $base=basename($picP['title']); ?> <tr> <td width="352" height="88"><table width="408" border="0"> <tr> <td width="113" rowspan="5"><a href="archive_detail.php?id=<?php echo $rowfor['id'];?>&category=<?php echo $rowfor['catsue'];?>"><img src="ad/pic/<?php echo $base;?>" width="100" height="100" border="0"/></a></td> <td width="94">Title</td> <td width="179" height="1"><?php echo $rowfor['catsue'];?> </td> </tr> <tr> <td> </td> <td width="179" height="3"> </td> </tr> <tr> <td> </td> <td width="179" height="8"> </td> </tr> <tr> <td> </td> <td width="179" height="17"> </td> </tr> <tr> <td> </td> <td width="179" height="36"> </td> </tr> </table></td> </tr> <?php }?> </table> Hi guys its me again, I am having a problem that I cant figure out... Here is my code: <?php $sqlCommand = "SELECT image FROM background"; $query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error()); $sqlCommand2 = "SELECT backgroundimage FROM site"; $query2 = mysqli_query($myConnection, $sqlCommand2) or die (mysqli_error()); while ($row = mysqli_fetch_array($query)) { while ($row2 = mysqli_fetch_array($query2)) { if($row['image'] == $row2['backgroundimage']){ echo '<img src="site_background/'.$row['image'].'" width="75px" height="75px" style="border:2px solid red;" /><br /><br />'; } if($row['image'] != $row2['backgroundimage']){ echo '<img src="site_background/'.$row['image'].'" width="50px" height="50px" style="border:2px solid black;" />'; } } } mysqli_free_result($query); mysqli_free_result($query2); ?> What is will do is get the images from the "backgrounds" table and the image from the "site" table (the current image). I am then wanting to pick out the current image and give it a red border and then display the other left over images smaller with a black border. I can get the images to all display with the black border or the current image to display with a red border but the other images dont show... I have tried mixing things around but I have not been able to get all the images to display with the formatting I want. I dont know if it is a simple syntax error or I am doing things completely wrong... I have been looking at it for so long its just become one big mess of code to me lol Any help to get this working as I want would be great! Cheers Ben this is the line in my script that I have to show the image: Code: [Select] $output .= "<img>{$row['disp_pic']}</img></br>\n"; As you can see I added the image tag, but it wont show the actual image. IE shows it as a small square with another small square picture icon in the middle of it (i'm sure you guys know what i mean). Is it better to store uploaded images as 'content' in the database? (like he http://www.techsupportforum.com/forums/f49/tutorial-upload-files-to-database-176804.html).. Or just as a regular file which is referenced in the database..? |
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