Im attempting to allows users to choose from a simple set of 4 avatars that they can use as their profile picture.
I've tried using:
<input type="image" class="profile-image-icon" src="img/avatar/avatar1.png" name="image" id="image">
But I finally found that using input type="image" is only for using an image as a submit button.
The avatars in question are already in my img/avatar folder, which I then only want to store the image path in the database.
I have echoed any error messages on changeProfile.php, which is returning "Nothing getting through" so I know it is an issue with my form, but I am unsure how to proceed.
Is there anyway to allow the user to simply click on the image and it 'POSTS' to changeProfile.php where I can then retrieve the file path to store in my database?
Thanks for any help provided!
edit_profile_image.php
<form class="profile-image-form" method="POST" action="profileChanged.php">
<fieldset>
<div class="image-control">
<input type="image" class="profile-image-icon"src="img/avatar/avatar1.png" name="image" id="image">
</div>
<?php echo "<input type='text' class='form-control' id='studentNumber' name='studentNumber' value='$studentID'>"; ?>
</fieldset>
</form>
changeProfile.php
if (isset($_POST['image'])) {
$image = mysqli_real_escape_string($conn, $_POST['image']);
$studentNumber = mysqli_real_escape_string($conn, $_POST['studentNumber']);
} else {
echo "Nothing coming thorugh";
}
Related
I want to view data stored in an uploading session but all I get is 'Null', am I going about this the wrong way?
session_start();
if(isset($_POST['submit'])){
$target = "test/";
$target = $target . basename('test') ;
$file = ($_FILES["uploaded"]["name"]);
$key = ini_get("session.upload_progress.prefix") . $_POST[ini_get("session.upload_progress.name")];
var_dump($_SESSION[$key]);
if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)){echo "done";}else echo "error";
}
and the html:
<form action="upload.php" method="POST" enctype="multipart/form-data">
<input type="hidden" name="<?php echo ini_get("session.upload_progress.name"); ?>" value="test" />
<input type="file" name="uploaded" />
<input type="submit" name='submit' />
</form>
You're trying to get upload progress status when upload is already done.
If you want to make it working, then you can for example send your form to iframe and during the upload ask server, using ajax, what is the status.
I would suggest to use it rather as a fallback for older browsers cause currently browsers are supporting ajax upload and you can display upload progress without making additional requests to server and creating some strange hidden iframes ;)
I am trying to get image file name by tag, But when I check it through isset($_FILES['imgFile']), it returns always false.
Here is my HTML tag for getting image file:
<input type="file" name="imgFile" accept="image/*" id="imgFile" src=""/>
Here is my php code to retrieve it:
if(isset($_FILES['imgFile']))
{
$img = $_FILES['imgFile']['name'];
echo $img";
}
else
{
echo "Image not set";
}
It always generate "Image not set" as an output though I have selected an image.
Are you using the correct enctype on the form?
<form enctype="multipart/form-data">
This is required when using a file upload element.
just use:
enctype="multipart/form-data"
Say this is your form:
<form action="same_page.php" method="post" enctype="multipart/form-data">
<input type="file" name="imgFile" accept="image/*" id="imgFile" src=""/>
<input type="submit" name="upload" value="Upload" />
</form>
and your php code to retrive name of the image is in the same page where the form is, then your code should be like this:
<?php
if (isset($_POST["upload"])) {
if (isset($_FILES['imgFile'])) {
$img = $_FILES['imgFile']['name'];
echo $img;
} else {
echo "Image not set";
}
}
?>
But if your php code is in another page, then you only need to use enctype="multipart/form-data" in the form as mentioned in the other answers.
I'm trying to post form data through ajax
form1.php
I use request to get all URL parameter data
$_REQUEST["Ename"];
$_REQUEST["eImg"];
To upload the image,i use this code http://www.9lessons.info/2011/08/ajax-image-upload-without-refreshing.html
In the above link,you can see the source code,in the place of $_FILES['photoimg']['name'];,i use $_FILES['image']['name']; but it is not uploading the file and giving success message.
include('db.php');
session_start();
$session_id='1'; // User session id
$path = "uploads/";
I removed script that is marked with **
$valid_formats = array("jpg", "png", "gif", "bmp","jpeg");
**if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST")
{**
$name = $_FILES['image']['name'];
$size = $_FILES['image']['size'];
if(strlen($name)) {
list($txt, $ext) = explode(".", $name);
if(in_array($ext,$valid_formats)) {
if($size<(1024*1024)) { // Image size max 1 Mb
$actual_image_name = time().$session_id.".".$ext;
$tmp = $_FILES['image']['tmp_name'];
if(move_uploaded_file($tmp, $path.$actual_image_name)) {
mysql_query("UPDATE users SET profile_image='$actual_image_name' WHERE uid='$session_id'");
echo "<img src='uploads/".$actual_image_name."' class='preview'>";
} else {
echo "failed";
}
} else {
echo "Image file size max 1 MB";
}
} else {
echo "Invalid file format..";
}
} **else {
echo "Please select image..!";
exit();
}**
you simply can't upload files via $.ajax().
you'll have to use some trycky iframe-stuff or something like that to make this work. luckily, there are ready-to-go plugins for jquery to handle this for you (like $.ajaxForm() for example wich seems to be the one that's used in the tutorial you're reading).
EDIT:
the plugin also allows to add extra data thats not present in the form itself. to quote the documentation:
data
An object containing extra data that should be submitted along with the form.
data: { key1: 'value1', key2: 'value2' }
For upload image by ajax you should use an iframe and set its id to form target.
Please have a look at
http://www.coursesweb.net/ajax/upload-images
It is very simple code to upload image
That won't work!
Images are handled differently from the text data in Ajax so you would have to do more than just post it using the $.ajax({}) method.
You can however use the jquery.form.js plugin it works perfect http://jquery.malsup.com/form/#download there is a tutorial on how to use it
here
Any ways I have used it my self so let me elaborate for you.
The JavaScript code is here
$('.uploadForm').live('click', function(evt){
$('#feedback').html(' ');
$('#feedback').html('<img src="images/loader_image.gif" alt="Uploading...."/>');
$("#formID").ajaxForm({
target: '#feedback'
}).submit();
evt.preventDefault();
});
If your PHP code is fine this should work .....
Just post the rest of the form fields in the normal way way
This should work for you. If the PHP code is fine
For example if you had other form fields like firstname and lastname in form like this one
<div class="form">
<fieldset class="ui-corner-all">
<h3 class="ui-widget-header ui-corner-top" align="center">Client information</h3>
<form action="save_new_client.php" enctype="multipart/form-data" id="clientForm" method="POST">
<label>First Name</label>
<input name="firstname" type="text" id="firstname" class="required" minlength="3"/>
<label>Lastname</label>
<input name="date_added" type="text" id="date_added" class="dateEst" />
<label>Image</label>
<input name="photo" type="file" id="photo" />
<input type="submit" name="button" id="button" value="Save" class="uploadForm"/>
<input type="reset" name="reset" id="button" value="Cancel" /></td>
</form>
</fieldset>
<div id="feedback"></div>
</div>
Below it you'll just need to add a div or paragraph to hold your feedback message ....
then the rest will be just fine (like I said if your PHP code is okay)I have not looked through it alot
I am trying to create an avatar uploader, but I want to be able to click only one button that opens the browsing window and then automatically calls the upload method from the AvatarUpload class.
The problem I have identified is that the form doesn't seem to be posting anything.
Any help would be great!
var_dump(isset($_POST['uploaded']));
if( isset($_POST['uploaded']) )
{
$img = new AvatarUpload();
$img->startUpload();
}
else
{
?>
<form method="post" enctype="multipart/form-data" name="uploadAvatar">
<p>
<input type="file" name="uploaded" id="file" onchange="this.form.submit()" />
<p>
</form>
<?php
}
?>
when you upload files they are listed under the $_FILES, not $_POST.
I'm using this form to add the title,link of the image and the text of the article to the database.
I'm using type="text" for the image link,now,it's getting borring to upload an image on a external image upload service and copy the link.
I want to upload the image with this for and store THE LINK of the image in the database.
The form:
<?php if (!$_POST["go"]){ ?>
<form method="post" action="">
<input name="article_title" type="text">
<input name="article_image_url" type="text"> <!-- i want here type="file" -->
<textarea name="article_text"></textarea>
<input type="submit" name="go" value="Submit">
</form>
<?php
} else {
$date=date("Y.m.d");
$title = $_POST["article_title"];
$image_url = $_POST["article_image_url"];
$text = $_POST["text"];
$sql="INSERT INTO articles (title,image_url,text,date) VALUES ('$title', '$image_url', '$text', '$date')";
if (mysql_query($sql)){
echo "done";}
else {echo "error<br>" . mysql_error();}}
?>
Please help me with this :)
ps:sorry for my English :$
The first thing you should do, and it seems you are clueless about SQL escaping, is add following before you access the first $_POST var:
$_POST = array_map("mysql_real_escape_string", $_POST);
Then you seemingly want to use a file upload for the image. If so change the url field to:
<input type=file name=image>
This uploaded file will show up in $_FILES. Use it like this, preferrably after you've read the other fields from $_POST:
if ($img = $_FILES["image"]["tmp_name"]) {
$image_url = md5_file($img) . ".jpeg";
move_uploaded_file($img, "./upload/$image_url");
$image_url = "http://www.example.org/where/$image_url";
}
There are lot's of security concerns with that. But that's out of scope here, so I hardwired it to .jpeg. There's lots of information in the manual and its comments: http://de2.php.net/manual/en/features.file-upload.php