I have following Form definition on PHP page:
<FORM action="formSave.php" method="post" enctype="multipart/form-data" data-ajax="false">
<!-- Some HTML controls here -->
<input type="file" name="foto" id="foto" />
</FORM>
formSave.php:
foreach ($_POST as $key => $value) {
if($key=="foto")
{
echo "Photo attached: ".$_FILES["foto"]["name"];
}
}
But it does not print file name.
How can I use file upload in addition to data-ajax="false".
Please note that I am using data-ajax=false for the first time in my life. I heard that it is used to enable/disable ajax call for mobile devices.
For file uploading, you should use $_FILES super-global variable:
<?php print_r($_FILES['foto']); ?>
<FORM action="formSave.php" method="post" enctype="multipart/form-data" data-ajax="false">
<!-- Some HTML controls here -->
<input type="file" name="foto" id="foto" />
</FORM>
formSave.php
$filename = $_FILES['foto']['tmp_name'];
Related
I'm trying to build a file upload form and I'm having trouble with the very basics. My form is this:
<html>
<body>
<form action="fileuploader.php" method="POST" enctype="multipart/form-data">
<input type="file" name="filename" />
<input type="submit"/>
</form>
</body>
</html>
My php code so far is one line and it doesn't do anything:
<?php
echo $_POST['filename'];
?>
The idea (at this point) is just to display the name of the file entered in the form. What am I doing wrong?
Based on your code I modified it. Have a try it.
HTML Part
<html>
<body>
<form action="fileuploader.php" method="POST" enctype="multipart/form-data">
<input type="file" name="filename" />
<input type="submit" name="submit" />
</form>
</body>
</html>
PHP
if (isset($_POST['submit'])) {
// Check if files array is not empty
if (!empty($_FILES)) {
$imageName = $_FILES['filename']['name'];
echo $imageName;
// Insert your code related to upload
}
}
You can print the filename using the following code:
<?php
echo $_FILES["filename"]["name"];
?>
All of my other form data is visible, but the name of the file is not showing up in the browser.
Here is a little portion of my code:
<form method="POST" action=<?php echo $_SERVER["PHP_SELF"];?> entype="multipart/form-data">
<input type="file" name="file">
<input type="submit" name="file" value="yoyo">
</form>
<?php
echo $name = $_FILES["file"]["name"];
echo "problem";
?>
and this is the output:
Notice: Undefined index: file in D:\xamp\htdocs\colgWeb\index.php on line 228
problem
Use a validator: You misspelled enctype (it has a c in it).
Consequently, the form is being submitted with the default (url based) encoding which doesn't support file uploads.
You need to think that as a two step page. First, you send your form, then you use the input.
<form method="POST" action="<?php echo $_SERVER["PHP_SELF"];?>" enctype="multipart/form-data">
<input type="file" name="file">
<input type="submit" name="file" value="yoyo">
</form>
<?php
if (isset($_FILES["file"]))
{
$name = $_FILES["file"]["name"];
echo "File: $name";
}
?>
Please try below code. You are using same name (ie "file") for both file and submit button."
<form method="POST" action=<?php echo $_SERVER["PHP_SELF"];?> entype="multipart/form-data">
<input type="file" name="file">
<input type="submit" name="submit" value="yoyo">
</form>
<?php
echo $name = $_FILES["file"]["name"];
echo "problem";
?>
I am trying to submit a form using PHP. I am trying to grab the value from two file inputs in my form, yet when I try to index them with my PHP code, I keep getting an error.
The error I get:
Undefined index: profile-pic in C:\xampp\htdocs\shareitme\form-test.php on line 5
Undefined index: cover-pic in C:\xampp\htdocs\shareitme\form-test.php on line 6
My Code:
<?php
if (isset($_POST['submit'])) {
$profile_pic= time() . $_FILES['profile-pic']['name'];
$cover_pic= time() . $_FILES['cover-pic']['name'];
}
?>
<form id="editprofile" method="post" action="<?php echo $_SERVER['PHP_SELF']
?>">
<input type="hidden" name="MAX_FILE_SIZE" value="5000000"/>
<input type="file" name="profile-pic"/>
<input type="file" name="cover-pic"/>
<input type="submit" name="submit" value="submit"/>
</form>
I know I have the names right, what am I doing wrong?
For uploading a file include enctype="multipart/form-data" in your form, like below:
<form id="editprofile" enctype="multipart/form-data" method="post" action="<?php echo $_SERVER['PHP_SELF']
?>">
The enctype attribute specifies how the form-data should be encoded when submitting it to the server.
For more info on forms : http://www.w3schools.com/tags/att_form_enctype.asp
For renaming a uploaded file, it better to upload it to server with move_uploaded_file and than rename it.
try:
<?php
if (isset($_REQUEST['go']) && $_REQUEST['go'] == "1") {
$profile_pic= time() . $_FILES['profile-pic']['name'];
$cover_pic= time() . $_FILES['cover-pic']['name'];
}
?>
<form id="editprofile" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input type="hidden" name="MAX_FILE_SIZE" value="5000000"/>
<input type="file" name="profile-pic"/>
<input type="file" name="cover-pic"/>
<input type=hidden name="go" value="1">
<input type="submit" name="submit" value="submit"/>
</form>
Whenever you are want to upload file. you need to add enctype in from.
`enctype="multipart/form-data"'
This should work
<form id="editprofile" enctype="multipart/form-data" method="post" action="<?php echo $_SERVER['PHP_SELF']
?>">
What you're trying to do here, if I'm not mistaken, is upload images on your server, right? You can simply use a third-party upload tool for that. I'd recommend AjaxUpload as it is based on AJAX and pretty easy to implement.
For simplicity, I'm applying the upload functionality on one image. You can simply put it inside a function and reuse it.
<p id="showMsg"></p>
<input type="file" name="profile-pic" id="profilePic" />
JQuery:
$("#profilePic").ajaxUpload({
url : $_SERVER['PHP_SELF'],
name: "file",
onSubmit: function() {
$('#showMsg').html('Uploading ... ');
},
onComplete: function(result) {
$('#showMsg').html('File uploaded with result' + result);
}
});
PHP:
<?php
if (isset($_FILES))
{
$file = $_FILES['file'];
move_uploaded_file($_FILES["file"]["tmp_name"], 'upload_dir/' . $_FILES["file"]["name"]); // This function will upload the image to 'upload_dir' folder. You can modify it as per your requirements.
}
?>
1) if you are using file type then your form attribute should be enctype="multipart/form-data".
<form id="editprofile" enctype="multipart/form-data" method="post" action="<?php echo $_SERVER['PHP_SELF'];?>">
2) Before using your post value it is best practice to check whether its setted or not.
$pic=isset($_FILES['profile-pic']['name']) ? $_FILES['profile-pic']['name'] : '';
Use enctype="multipart/form-data" in form tag.
Whenever you want to post binary datas you must have to set this enctype attribute.
Hi Hi trying to upload a file via a form move it to another folder and then print its name. But doesnt work don't know why.
<form method="post" action='exercice.php' id="form1">
<input type="file" name="files" id="files" onChange="submitForm();">
</form>
<?php
if (isset($_FILES['files']))
{
move_uploaded_file($_FILES['files']['tmp_name'], "uploaded/");
echo $_FILES['files']['name'];
}
?>
form missing:
enctype="multipart/form-data"
ref:
spec
<form method="post" action='exercice.php' id="form1">
<input type="file" name="files" id="files" onChange="submitForm();">
</form>
<?php
if (isset($_FILES['files']))
{
move_uploaded_file($_FILES['files']['tmp_name'], "uploaded/".$_FILES['files']['name']);
echo $_FILES['files']['name'];
}
?>
You need to include the file name in the upload path, like shown above.
Also, where is your upload (ajax) code?
Did isset($_FILES['files']) return true?
i'm following this tutorial: http://net.tutsplus.com/tutorials/javascript-ajax/uploading-files-with-ajax/comment-page-1/#comments to learn how to upload multiple files via ajax.
This is my html:
<form class="form-horizontal" id="settingsChangeAvatar" method="post" enctype="multipart/form-data" action="<?php echo $AJAX."/ajaxUpload.php"?>">
<input class="input-xlarge input-file" id="settingsUploadAvatar" name="settingsUploadAvatar" type="file" multiple />
<button class="btn" id="uploadAvatarButton" type="submit">Upload</button>
</form>
And this is my ajaxUpload.php:
foreach($_FILES["settingsUploadAvatar"]["error"] as $key => $error){
if($error == UPLOAD_ERR_OK) {
$name = $_FILES["settingsUploadAvatar"]["name"][$key];
move_uploaded_file($_FILES["settingsUploadAvatar"]["tmp_name"][$key], $_SERVER["DOCUMENT_ROOT"]."/webname/".$_FILES["settingsUploadAvatar"]["name"][$key]);
}
}
echo("File uploaded");
My code should be the same as the one in the tutorial.
Thanks for helping.
Change the 'name' attribute of your input from settingsUploadAvatar to settingsUploadAvatar[].