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PHP - Php, Js, Ajax Help With Uploading Photo..
I wonna to upload photo with js and ajax.. i can do inserting something in mysql, but i dont know how to do with files :/
Form: Code: [Select] <form enctype="multipart/form-data" method="get"> <center> <label>Title: <input type="text" name="s_name" value="" style="width:250px" /></label> <label>Image: <input type="file" name="s_image" style="width:250px" /></label> <input class="button" type="submit" name="s_add" value="Add" style="width:100px" OnClick="save();" /> </center> </form> This is js code: Code: [Select] function save() { col = document.getElementById("new"); if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp = new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp = new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange = function() { if (xmlhttp.readyState == 4 && xmlhttp.status == 200) { col.innerHTML = xmlhttp.responseText; } } xmlhttp.open("GET","sources/save.php?i dont know what to insert here,true); xmlhttp.send(); } this is save.php: Code: [Select] <?php $allowed_size = 204800; $name = $_GET['name']; $name = $_GET['image']; $date = time(); if (save_img($up_imgfile, $up_imgname, $allowed_size)) { $query = "INSERT INTO table (name, image, date) VALUES ('{$name}', '{$image}', '$date')"; $result = mysql_query($query, $connection); if ($result) { echo "<p class='warn-green'>$season_name has been successful added</p>"; } else { echo "<p class='warn-red'>Error: $season_name is not added, try again.</p>"; } } } ?>save_img() is php function to upload photo and make thumb.. How can i do this thing to do this function, but i need this: $up_imgname = $_FILES['s_image']['name']; $up_imgfile = $_FILES['s_image']['tmp_name']; Becouse of that i cant do this, how can i get temp name, image name of uploaded file ? Similar TutorialsI let my members to upload their photos, I m using the script I coded which checks the file extension, if the file extension is "jpg" ( if ($this->url['type'] == "image/jpg")) and less than 600kb it uploads. Otherwise it gives you a warning.. What I wonder is, do I face any unwanted results about this later ? Like someone uploads a virus or script then execute it and do something ? adding photo,Array ( [name] => profilepic.jpeg [type] => image/jpeg [tmp_name] => C:\xampp\tmp\phpB923.tmp [error] => 0 [size] => 152127 ) Hi, We have been using this code for ages, but suddenly for some reason we cannot upload an image using the code below. When we try to Print_r the $profilephoto variable, I get this error. The filename is correct, but what is that [error]?? And how do I resolve it? Oddly it does upload it locally, but LIVE, it won't This is the code.
if (isset($updatephoto))
{
echo "adding photo,";
print_r($profilephoto);
define ("MAX_SIZE","5000");
function getExtension($str) {
$i = strrpos($str,".");
if (!$i) { return ""; }
$l = strlen($str) - $i;
$ext = substr($str,$i+1,$l);
return $ext;
}
$errors=0;
if($_SERVER["REQUEST_METHOD"] == "POST")
{
$image =$_FILES["profilephoto"]["name"];
$uploadedfile = $_FILES['profilephoto']['tmp_name'];
if ($image)
{
$filename = stripslashes($_FILES['profilephoto']['name']);
$extension = getExtension($filename);
$extension = strtolower($extension);
if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif"))
{
echo "Unknown Extension..!";
}
else
{
$size=filesize($_FILES['profilephoto']['tmp_name']);
if ($size > MAX_SIZE*1024)
{
echo "File Size Excedeed..!!";
}
if($extension=="jpg" || $extension=="jpeg" )
{
$uploadedfile = $_FILES['profilephoto']['tmp_name'];
$src = imagecreatefromjpeg($uploadedfile);
}
else if($extension=="png")
{
$uploadedfile = $_FILES['profilephoto']['tmp_name'];
$src = imagecreatefrompng($uploadedfile);
}
else
{
$src = imagecreatefromgif($uploadedfile);
echo $scr;
}
list($width,$height)=getimagesize($uploadedfile);
$newwidth=600;
$newheight=($height/$width)*$newwidth;
$tmp=imagecreatetruecolor($newwidth,$newheight);
imagecopyresampled($tmp,$src,0,0,0,0,$newwidth,$newheight,$width,$height);
$pic=($_FILES['profilephoto']['name']);
$random = (rand()%99999999);
$newname="$random"."$pic";
$filename = "images/profiles/". $newname;
imagejpeg($tmp,$filename,100);
imagedestroy($src);
imagedestroy($tmp);
}}
}
$query = ("UPDATE users SET profilephoto =:newname WHERE id =:userid");
$result = $pdo->prepare($query);
$result->execute(array(':userid' => $userid, ':newname' => $newname));
echo "<script>
window.location.replace('/profile/')
</script>";}
what is the simple way to upload a photo to mysql or store in a dir, and update the database with the new url? Right now I redirect to index page after I delete a record. However I am looking to make it so that I can delete a record without redirecting the page. I know this can be accomplised using Ajax. I have spent countless hours before trying to make it work, but it did not work.
So here is a basic setup I created. Can you please update it with ajax code so that I can see how it's done properly?
<!DOCTYPE HTML> <html lang="en"> <head> <meta charset="UTF-8"> <title>Home Page</title> </head> <body> <div class="record" > <a href="record.php?id=<?php echo $record_id ?>"><?php echo $record_name; ?></a> <div class="delete-record"> <a href="delete.php">Delete Record</a> </div> </div> </body> </html> Edited by man5, 18 August 2014 - 08:55 PM. Now I'm having this strange issue with my website I'm currently working on a tester system and I've encountered a problem that I'm unable to find the issue, tho I'm thinking my ajax php part of the script to be the thing causing it even tho it seems strange that it would cause it. The first part which is connected to where the problem occurs is the echo"<form>"; and from there, It should take you to index.php?page=tester&select=answer, now that is where it in the browser goes there tho it still shows the page stuff from the last page which is index.php?page=tester&select=applications, so it's like showing both &select=answer and &select=applications on the same page. <?php $q=$_GET["q"]; include'../config/connection.php'; $result = mysql_query("SELECT * FROM applications WHERE id = '$q'"); echo "<center><table border='1'> <tr> <th>Account Name</th> <th>Character Name</th> <th>Gender</th> <th>Skin Color</th> </tr>"; $row = mysql_fetch_array($result); echo "<tr>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['charactername'] . "</td>"; echo "<td>" . $row['gender'] . "</td>"; echo "<td>" . $row['race'] . "</td>"; echo "</tr></table></center>"; echo"<br/>"; echo"<table><tr> <th>Description</th> <th>Metagaming</th> <th>Powergaming</th></tr>"; echo"<tr>"; echo "<td><textarea readonly='readonly' style='width:22em; height:20em;'>".$row['description']."</textarea></td>"; echo "<td><textarea readonly='readonly' style='width:22em; height:20em;'>".$row['mg']."</textarea></td>"; echo "<td><textarea readonly='readonly' style='width:22em; height:20em;'>".$row['pg']."</textarea></td>"; echo"</tr></table><table><br/><center><h1>Answer</h1><br/><form action='index.php?page=tester&select=answer' method='post'>"; echo"<textarea name='why' style='height:10em; width:60em;'></textarea><br/>"; echo"<input type='submit' name='answer' value='Accept' /><a/>"; echo"<input type='submit' name='answer' value='Decline' /></center>"; echo"<input type='hidden' name='id' value='$q'/>"; echo"</form></table>"; ?> Now on &select=answer it included a page which the script of that include consist of the stuff below, it outputs that the query was successfully, and all that. <? if(!empty($_POST['why'])) { $why = mysql_real_escape_string($_POST['why']); $answer = trim($_POST['answer']); $id = $_POST['id']; if($answer == "Accept") { $query1 = mysql_query("UPDATE characters SET accepted = '1' WHERE id = '".$id."'"); echo"Successfully accepted"; $answer = 1; } elseif($answer == "Decline") { echo"Successfully declined"; $answer = 0; } $query = mysql_query("UPDATE applications SET answer = '$why' AND tester = '".$_COOKIE['Username']."' AND accepted = '$answer' AND answered = '1' WHERE cid = '".$id."'") or die('Could not connect: ' . mysql_error()); if($query) { echo"<br/>Query went through without problems"; header("Refresh: 5;url=index.php?page=tester"); } } ?> This is the ajax part javascript of it which gets the information for index.php?page=tester&select=applications Code: [Select] <script type="text/javascript"> function showApplication(str) { if (str==""||str==0) { document.getElementById("txtHint").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("txtHint").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","tester/applications.php?q="+str,true); xmlhttp.send(); } </script>If you need any more information feel free to ask for it. Thanks in advance. Hello, I've been trying this for hours now, looking at different examples and trying to change them to work for me, but with no luck... This is what I am trying to do: I have a simple form with: - 1 input field, where I can enter a number - 1 Submit Button When I enter a number into the field and click submit, I want that number to be send to the php file that is in the ajax call, then the script will take that number and run a bunch of queries and then return a new number. I want that new number to be used to call the php script via ajax again, until no number is returned, or something else is returned like the word "done" or something like that, at which point is simply makes an alert or populated a div with a message... The point is, that depending on the number entered it could take up to an hour to complete ALL the queries, so I want the script that is called to only run a fixed amount of queries at a time and then return the number it is currently at (+1), so that it can continue with the next number when it is called again. I would like to use jquery, but could also be any other way, as long as I get this to work. I already have the php script completed that needs to be called by the ajax, it returns a single number when being called. Thank you, vb I'm having a problem understanding how to import more than one photo in a list of up to 20 photos. I can only import 1 photo from a comma delimited csv with a pipe between each photo url. <? function getImg($value,$userid, $stocknum){ $photoStr = ''; if($value !=''){ $photoURLs = explode('|', $value); for($i=0; $i<count($photoURLs);$i++){ $url = $photoURLs[$i]; $j = $i+1; if (@copy($url, "../upload/".$userid."/".$userid."_".$stocknum."_".$j.".jpg")) { $photoStr .= $userid."_".$stocknum."_".$j.".jpg|"; } } } return $photoStr; } ?> And after csv and sql <? if($img){ $img = rtrim($img, '|'); $imgArray = explode('|', $img); $imgIndex = 1; foreach ($imgArray AS $imgItem){ $db->query("UPDATE `stockitem` SET img".$imgIndex."= '$imgItem' where stocknum='$stocknum'","" , __FILE__, __LINE__); $imgIndex++; } } ?> Any help would be appreciated. Hi all, i was wondering if someone could point me in the right direction on how to create the following. After someone uploaded a photo than Cut out only his head in a fixed square say 90x120 pix (of a larger picture). I have seen some on the internet but most of them use flash and i rather use javascript or no javascript at all. Can this be done with GD library for instance where you just give the coordinates and the size of a square. (if this can be done automated i love to hear it) I have seen one where someone should pinpoint the eyes. IF someone has experience with this or a general idea on what techniques to use let me know I bet there are loads of way to do this but the most light weight technique would be great. Something like the attached image i am in the process of creating an auction site. However the images do not show , just the placeholder. Can anyone indicate where I have gone wrong ?
i am using this script to enter the items into the Database
html>
<link rel="stylesheet" type="text/css" href="style.css">
if(isset($_GET["item"]))
if($_GET["item"]=="duplicate")
else if($_GET["item"]=="successful")
}
echo "<label for='item_name' class='label'>Item Name:</label>";
echo "</form>";
</body>
This script displays the items :
<html>
<link rel="stylesheet" type="text/css" href="style.css">
<body> session_start();
if(!isset($_SESSION["username"]))
else
$DBHOST = "localhost";
while($row = $result->fetch_assoc())
$iid = $row["item_id"];
$conn->close();
Hey guys, I am a newb but trying to figure out how I can upload multiple associated photo files. My code is working for one photo, however I would like to have multiple photos that are associated uploaded at the same time. I would like the photos save in a format such as: photo_filename1 = 100, photo_filename2 = 100_a I am using the following code with no luck, not sure how I can work this. Any help is much appreciated!!! Thanks in advance. <code> <?php include("config.inc.php"); // initialization $result_final = ""; $counter = 0; // List of our known photo types $known_photo_types = array( 'image/pjpeg' => 'jpg', 'image/jpeg' => 'jpg', 'image/gif' => 'gif', 'image/bmp' => 'bmp', 'image/x-png' => 'png' ); // GD Function List $gd_function_suffix = array( 'image/pjpeg' => 'JPEG', 'image/jpeg' => 'JPEG', 'image/gif' => 'GIF', 'image/bmp' => 'WBMP', 'image/x-png' => 'PNG' ); // Fetch the photo array sent by preupload.php $photos_uploaded1 = $_FILES['photo_filename1']; $photos_uploaded2 = $_FILES['photo_filename2']; // Fetch the photo caption array $photo_caption = $_POST['photo_caption']; while( $counter <= count($photos_uploaded1) ) { if($photos_uploaded1['size'][$counter] > 0) { if(!array_key_exists($photos_uploaded1['type'][$counter], $known_photo_types)) { $result_final .= "File ".($counter+1)." is not a photo<br />"; } else { mysql_query( "INSERT INTO gallery_photos(`photo_filename1`, `photo_caption`, `photo_category`) VALUES('0', '".addslashes($photo_caption[$counter])."', '".addslashes($_POST['category'])."')" ); $new_id = mysql_insert_id(); $filetype = $photos_uploaded1['type'][$counter]; $extention = $known_photo_types[$filetype]; $filename = $new_id.".".$extention; mysql_query( "UPDATE gallery_photos SET photo_filename1='".addslashes($filename)."' WHERE photo_id='".addslashes($new_id)."'" ); // Store the orignal file copy($photos_uploaded1['tmp_name'][$counter], $images_dir."/".$filename); // Let's get the Thumbnail size $size = GetImageSize( $images_dir."/".$filename ); if($size[0] > $size[1]) { $thumbnail_width = 100; $thumbnail_height = (int)(100 * $size[1] / $size[0]); } else { $thumbnail_width = (int)(100 * $size[0] / $size[1]); $thumbnail_height = 100; } // Build Thumbnail with GD 1.x.x, you can use the other described methods too $function_suffix = $gd_function_suffix[$filetype]; $function_to_read = "ImageCreateFrom".$function_suffix; $function_to_write = "Image".$function_suffix; // Read the source file $source_handle = $function_to_read ( $images_dir."/".$filename ); if($source_handle) { // Let's create an blank image for the thumbnail $destination_handle = ImageCreate ( $thumbnail_width, $thumbnail_height ); // Now we resize it ImageCopyResized( $destination_handle, $source_handle, 0, 0, 0, 0, $thumbnail_width, $thumbnail_height, $size[0], $size[1] ); } // Let's save the thumbnail $function_to_write( $destination_handle, $images_dir."/tb_".$filename ); ImageDestroy($destination_handle ); // $result_final .= "<img src='".$images_dir. "/tb_".$filename."' /> File ".($counter+1)." Added<br />"; } } $counter++; } //file2 while( $counter <= count($photos_uploaded2) ) { if($photos_uploaded2['size'][$counter] > 0) { if(!array_key_exists($photos_uploaded2['type'][$counter], $known_photo_types)) { $result_final2 .= "File ".($counter+1)." is not a photo<br />"; } else { mysql_query( "INSERT INTO gallery_photos(`photo_filename2`) VALUES('0')" ); $new_id = mysql_insert_id(); $filetype = $photos_uploaded2['type'][$counter]; $extention = $known_photo_types[$filetype]; $filename = $new_id."_1.".$extention; mysql_query( "UPDATE gallery_photos SET photo_filename2='".addslashes($filename)."' WHERE photo_id='".addslashes($new_id)."'" ); // Store the orignal file copy($photos_uploaded2['tmp_name'][$counter], $images_dir."/".$filename); $result_final2 .= "<img src='".$images_dir. "/tb_".$filename."' /> File ".($counter+1)." Added<br />"; } } $counter++; } // Print Result echo <<<__HTML_END <html> <head> <title>Photos uploaded</title> </head> <body> $result_final $result_final2 </body> </html> __HTML_END; ?> </code> hello all, i have a photo uploader that clients use. i want this to be able to put clients photos in there own folder within a specified path. $(function() { $('#custom_file_upload').uploadify({ 'uploader' : '/uploadify.swf', 'script' : 'uploadify.php?user=<?php echo("$logged[ip]"); ?>', 'cancelImg' : '/cancel.png', 'folder' : '/wuploads/{$logged('email')}', 'multi' : true, 'auto' : true, 'fileExt' : '*.ZIP;*.zip;*.rar', 'queueID' : 'custom-queue', 'queueSizeLimit' : 1000, 'simUploadLimit' : 1000, 'removeCompleted': true, 'onSelectOnce' : function(event,data) { $('#status-message').text(data.filesSelected + ' files have been added to the queue.'); }, 'onAllComplete' : function(event,data) { $('#status-message').text(data.filesUploaded + ' files uploaded, ' + data.errors + ' errors.'); } }); }); i have tried the code above with 'folder' : '/wuploads/{$logged('email')}', but this doesnt seem to work, im new to php and tried a few different combinations but it wont budge! thanks, gavin Hi, I am making login page with database and php. I need to be able when the person clicks theogin button to take photo of the person and add it to that same database based on the info entered.
I am doing this to know if that person that logged in is the person in reality.
Is it possible Hai i got a site where i take the details of the visitors that enter my office premises. Now what i wanted is to add the photo of the visitor into that. i have a webcam with me through which i can take pictures. I dont have an idea how to do this on the website. I need some one to help me on this.. I dont know really where to start.. Any help would be greately appreciatd.. Hi all, i've set this script for my gallery, it works, it display me the name of the albums, but when i click to see the pictures inside, the page is blank, and also don't appear the "Back to the albums" link, i can't find the error in the code... Code: [Select] <html> <head> <script type="text/javascript" src="../lightbox2.05/js/lightbox.js"> </script> </head> <body> <?php $page = $_SERVER['PHP_SELF']; //settings $column = 5; //directories $base = "carrelli_noleggio"; $thumbs = "thumbs"; // get album $get_album = $_GET['album']; //if no album selected if (!$get_album) { echo "<b>Select an album:</b><p />"; //find each album and display as links $handle = opendir($base); while (($file = readdir($handle))!==FALSE) { if (is_dir($base."/".$file) && $file != "." && $file != ".." && $file != $thumbs) { echo "<a href='$page?album=$file'>$file</a><br />"; } } closedir($handle); } else { //check if album exist, and additional security checks if (!is_dir($base."/".$get_album) || strstr($get_album,".") !=NULL || strstr($get_album,"/") !=NULL || strstr($get_album,"\") !=NULL) { echo "Album doesn't exist."; } else { $x = 0; echo "<b>$get_album</b><p />"; $handle = opendir($base."/".$get_album); while (($file = readdir($handle)) !== FALSE) { if ($file != "." && $file != "..") { echo "<table style='display:inline;'><tr><td><a href='$base/$get_album/$files' ref='lightbox'<img src='$base/$thumbs/$file' height='100' width'100'></a></td></tr></table>"; $x++; if ($x==$column) { echo"<br />"; $x = 0; } } } closedir($handle); echo "<p /><a href='$page>Back to albums</a>"; } } ?> </body> </html> Hello, can you help me on my news article? I would like to be able to add photo on the article. Below is my codes Quote ======= html ======= <?php include_once ('post_news.php'); ?> <form Action="add_news.php" Method="post"> <input name="postdate" type="text" id="postdate"> <input name="title" type="text" id="title"> <textarea name="newstext" cols="60" rows="15" id="newstext"></textarea> <input type="submit" name="submit" value="Submit"> <input type="reset" name="reset" value="Reset"> </form> ============== php - post_news.php ============== <?php ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); if (isset ($_POST['submit'])) { require_once ('inc/dbConfig.php'); // Define the query. $query = "INSERT INTO news (id, postdate, title, newstext) VALUES (0,'{$_POST['postdate']}', '{$_POST['title']}','{$_POST['newstext']}')"; // Execute the query. if (@mysql_query ($query)) { print "<p>Data has been added.</p>"; } else { print "<p>Could add the entry because: <b>" . mysql_error() . "</b>. The query was $query.</p>"; } mysql_close(); } ?> ===== db structure ====== id int(10) unsigned NOT NULL auto_increment, postdate timestamp(14), title varchar(50) NOT NULL, newstext text NOT NULL, PRIMARY KEY (id), KEY postdate (postdate) Hope you guys can help me. Thanks! I have used the same method for uploading photos for ages but feel there may be a better way. For both methods, please assume all uploads are large digital camera uploads of a few megabytes and thousands of pixels wide (as you would expect from large digital camera images). CURRENT METHOD (1) receive uploaded file and use ImageMagick to convert to 4 file sizes, then store these file sizes (also set so ImageMagick reduces the quality to something like a setting of 80 out of 100 which will further reduce file size. - file sizes (widths): 800 240 120 60 (mobile phones etc.) (2) Based on the type of size and view required I then use <img src="/photos/240/file.jpg to display (240 being the required image size of course) POSSIBLE BETTER METHOD (1) Use ImageMagick to reduce to 800 wide as will never display bigger than this anyway, reduce quality also. But don't resize further and only need one file. (2) Use PHP commands to display an image, and set height/width/quality then. Eg: <img src="showImage.php?file=file.jpg&width=240&quality=60 So both would do the same thing but the second seems better as you don't need to create 4 images for every photo upload. Why make lots of thumbnails when you can make one and then get PHP to create at the time of display? Although I suppose if you regularly displayed these images in a search result then PHP would need to do that extra processing on every search. Creating all 4 at the time of upload only requires the one bit of processing. Have I just found the answer? G'day, on the home page of a website i'm building, i'm wanting to have a random photo that looks at all the subfolders of the gallery directory, but the best I can achieve is just one subfolder. Here is the code, if anyone can assist, that'll be appreciated. <?php function getRandomFromArray($ar) { mt_srand( (double)microtime() * 1000000 ); $num = array_rand($ar); return $ar[$num]; } function getImagesFromDir($path) { $images = array(); if ( $img_dir = @opendir($path) ) { while ( false !== ($img_file = readdir($img_dir)) ) { // checks for gif, jpg, png if ( preg_match("/(\.gif|\.jpg|\.png)$/", $img_file) ) { $images[] = $img_file; } } closedir($img_dir); } return $images; } $root = ''; // If images not in sub directory of current directory specify root //$root = $_SERVER['DOCUMENT_ROOT']; $path = 'gallery/topic1/'; // Obtain list of images from directory $imgList = getImagesFromDir($root . $path); $img = getRandomFromArray($imgList); ?> <center><a href="?GoTo=Photo Gallery"><img src="<?php echo $path . $img ?>" alt="" height="267" width="400"/></a></center> Hey Guys, I have a php script that update's franchise information using a mysql table. Note: Ter = Territory, ie. the territory that franchise covers. <?php include('config.php'); if (isset($_GET['Ter']) ) { $ter = (int) $_GET['Ter']; if (isset($_POST['submitted'])) { //Photo Upload //This is the directory where images will be saved $target = "images/"; $target = $target . basename( $_FILES['photo']['name']); //This gets all the other information from the form $photo =($_FILES['photo']['name']); //Pause Photo Upload foreach($_POST AS $key => $value) { $_POST[$key] = mysql_real_escape_string($value); } $sql= "UPDATE `ter` SET `Ter` = '{$_POST['Ter']}' , `BranchName` = '{$_POST['BranchName']}' , `BranchAddress` = '{$_POST['BranchAddress']}' , `BranchTel` = '{$_POST['BranchTel']}' , `BranchEmail` = '{$_POST['BranchEmail']}' , `BranchLink` = '{$_POST['BranchLink']}' , `Theme` = '{$_POST['Theme']}' , `LocalInfo` = '{$_POST['LocalInfo']}' , `BranchInfo` = '{$_POST['BranchInfo']}' , `photo` = '{$_POST['photo']}' WHERE `Ter` = '$ter' "; mysql_query($sql) or die(mysql_error()); //Unpause Photo Upload //Writes the photo to the server move_uploaded_file($_FILES['photo']['tmp_name'], $target); //End of Photo Upload echo (mysql_affected_rows()) ? "Edited Branch.<br />" : "Nothing changed. <br />"; } $row = mysql_fetch_array ( mysql_query("SELECT * FROM `ter` WHERE `Ter` = '$ter' ")); ?> In phpmyadmin I can see my table and it has the correct image name displayed in the photo column. So you would assume its worked. But when I look in the 'images/' location no image has been uploaded. So I think there is an error with the upload part but cant figure out whats wrong. Cheers, S As the title say, I can not for the life of me get the "$bank" content to display, no matter HOW much I try... Does anyone see any errors. I am sooooooo wiped out at this! main page <? $body = ' <script type="text/javascript" src="change-content.js"></script> <div id="addSold"> <form action="'.$_SERVER['REQUEST_URI'].'" method="post" name="form" autocomplete="off"> <fieldset id="Vehicle"> <legend>Vehicle</legend> <ul> <li><label for="Year">Year</label>'.$Year.'</li> <li><label for="Make">Make</label>'.$Make.'</li> <li><label for="Model">Model</label>'.$Model.'</li> <li><label for="Trim">Trim</label><input type="text" name="Trim" id="Trim" size="10" value="'.$trim.'" disabled="disabled"></li> </ul> <ul> <li><label for="Mileage">Mileage</label><input type="text" name="Mileage" id="Mileage" size="5" maxlength="6" value="'.$row['mileage'].'"></li> <li><label for="VIN">VIN</label><input type="text" name="VIN" id="VIN" size="23" maxlength="17" value="'.$row['vin'].'" disabled="disabled"></li> <li><label for="Color">Color</label>'.$Exterior.'</li> </ul> </fieldset> <fieldset id="Deal"> <legend>Deal</legend> <ul> <li> <label for="soldDte1">Date</label> <input type="text" name="soldDte1" id="soldDte1" size="1" maxlength="2" onkeyup="return autoTab(this, 2, event)" value="08"> / <input type="text" name="soldDte2" id="soldDte2" size="1" maxlength="2" onkeyup="return autoTab(this, 2, event)" value="30"> / <input type="text" name="soldDte3" id="soldDte3" size="1" maxlength="2" value="'.$year.'"> <a href="#"><img id="date_'.$row[stock].'" src="images/Icons/dateOff.png" onfocus="this.select();lcs(this)" onmouseover="MM_swapImage(\'date_'.$row[stock].'\',\'\',\'images/Icons/dateOn.png\',1)" onmouseout="MM_swapImgRestore()" alt="Choose Date"></a> </li> <li> <label for="salesman">Salesman</label> <select name="salesman" id="salesman"> <option></option> '.$salesmen.' </select> </li> </ul> <ul> <li> <label for="dealType">Deal Type</label> <select name="dealType" class="select-content" onchange="getFile(this.value)"> <option></option> <option value="AL">Auto Loan</option> <option value="Cash">Cash</option> <option value="CAC">Credit Acceptance</option> <option value="IH">In House</option> <option value="SAL">Sensible Auto</option> </select> </li> <li> <label for="tradeDrop">Trade</label> <select name="tradein" id="tradeDrop" onchange="show_hide_trade(this.value);"> <option value="No">No</option> <option value="Yes">Yes</option> </select> </li> </ul> </fieldset> <div id="Bank" class="view">'.$bank.'</div> </form> </div> '; ?> get_Bank.php <? if ($_GET['dealType'] == "AL") { $bank = ' <fieldset id="AL"> <legend>Auto Loan Figures</legend> <ul> <li><label for="price">Price</label><input type="text" name="price" id="price" class="price" size="7" onchange="currency(this)"></li> <li><label for="down">Down</label><input type="text" name="down" id="down" class="price" size="6" onchange="currency(this)"></li> <li><label for="tax">Tax</label><input type="text" name="tax" id="tax" class="price" size="6" onchange="currency(this)"></li> <li><label for="reg">Plates</label><input type="text" name="reg" id="reg" class="price" size="4" onchange="currency(this)"></li> <li><label for="gap">Gap</label><input type="text" name="gap" id="gap" class="price" size="4" onchange="currency(this)"></li> </ul> <ul> <li> <label for="pymtNum">--------------- Payment ---------------</label> <input type="text" name="pymtNum" id="pymtNum" size="3" maxlength="3" onkeyup="return autoTab(this, 3, event)"> @ <input type="text" name="pymtAmnt" id="pymtAmnt" class="price" size="5" onchange="currency(this)"> per <select name="pymtType"> <option value="Weekly" selected="selected">Week</option> <option value="Monthly">Month</option> </select> </li> <li><label for="APR">APR</label><input type="text" name="APR" id="APR" class="rate" size="6" value="19.00"></li> </ul> </fieldset> '; } elseif ($_GET['dealType'] == "CAC") { $bank = ' Credit Acceptance stuff goes here '; } else { $bank = 'You must choose a bank before continuing'; } ?> change-content.js Code: [Select] window.onload = init; // finds all <select> tags will class="select-content" and activates function function init() { var sel = document.getElementsByTagName("select"); for (var i=0; i<sel.length; i++){ if (sel[i].className == "select-content") { sel[i].onchange = getFile; } sel[i].selectedIndex = 0; } } function getFile (url) { var url = "AJAX/get_Bank.php?dealType="+ this.value; if (window.XMLHttpRequest) {xmlhttp=new XMLHttpRequest();} else {xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");} xmlhttp.open("GET",url,false); xmlhttp.send(); // a loop that looks through all <div>s on the page // and then replaces the id with the value and gets that file var divs = document.getElementsByTagName("div"); for (var i=0; i<divs.length; i++) { if(divs[i].id == "bank") { divs[i].id = this.value; divs[i].innerHTML=xmlhttp.responseText; } } } |
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