I am getting the img_description but my img_name remain empty.
Basic php code
$msg = "";
if(isset($_POST['upload'])) {
$target = "images/".basename($_FILES['image']['name']); // get error here
$image = $_FILES['image']['name']; // get error here
$text = $_POST['text'];
$sql = "INSERT into images (img_name, img_description) values ('$image', '$text')";
$smt = $heidisql->prepare($sql);
$smt->execute();
if(move_uploaded_file($_FILES['image']['tmp_name'], $target)) { //error here
$msg = "Image upload sucessfully";
}else {
$msg = "Image failed to upload properly";
}
My basic Form
<form method="POST" action="checkout.php" enctype="multipart/form-date">
<input type="hidden" name="size" value="1000000">
<div>
<input type="file" name="image">
</div>
<div>
<textarea name="text" cols="40" rows="4" placeholder="img text decription" ></textarea>
</div>
<div>
<input type="submit" name="upload" value="upload img">
</div>
</form>
Table: images, attribute -> img_id, img_name, img_description
Error: Undefined index: image
You have a typo in the enctype in your form tag.
Change
multipart/form-date
to
multipart/form-data
You have a write mistake enctype="multipart/form-date" use enctype="multipart/form-data" instead.
Related
I'm a php/mysql newbie and need your help!
I have an upload formula to upload multiple files to a database, selected one by one.
The files upload works fine, but I want to add a title field for each selected file. I have no idea how to implement this in my code.
Here is my code:
$db = mysqli_connect("localhost", "root", "", "xy");
if (isset($_POST['upload_docs'])) {
for($mi = 0; $mi < count($_FILES['sel_doc']['tmp_name']); $mi++) {
$doc = $_FILES['sel_doc']['name'][$mi];
$tmp_doc = $_FILES['sel_doc']['tmp_name'][$mi];
$target = "doc_uploads/". basename($doc);
if (move_uploaded_file($tmp_doc, $target)) {
echo "Upload successfully!";
}
else {
echo "Upload failed!";
}
$sql_f = "INSERT INTO docs (doc_file) VALUES ('$doc')";
mysqli_query($db, $sql_f);
}
}
<form method="POST" action="upload.php" enctype="multipart/form-data">
<div>
<input type="file" name="sel_doc[]">
<input type="file" name="sel_doc[]">
<input type="file" name="sel_doc[]">
<input type="file" name="sel_doc[]">
<input type="file" name="sel_doc[]">
</div>
<div id="upload_button">
<button type="submit" name="upload_docs">upload</button>
</div>
</form>
This is how 1 upload would looks like:
| id_doc | doc_file | doc_title|upload_id|
++++++++++++++++++++++++++++++++++++++++++++++++
|.....1......|...a.pdf...|....vvv....|....1....|
|.....2......|...b.pdf...|....www....|....1....|
|.....3......|...c.pdf...|....xxx....|....1....|
|.....4......|...d.pdf...|....yyy....|....1....|
|.....5......|...e.pdf...|....zzz....|....1....|
In my orginal code I insert in 2 tables for 1:n realationship (because of the "upload_id").
HTML
<div>
<?php for ($i=0; $i<5; $i++){ ?>
<b><?php echo $i ?></b><br>
File: <input type="file" name="sel_doc[<?php echo $i ?>]"><br>
Title: <input name="title[<?php echo $i ?>]"><br>
<?php } ?>
</div>
Will print this:
<div>
<b>0</b><br>
File: <input type="file" name="sel_doc[0]"><br>
Title: <input name="title[0]"><br>
<b>1</b><br>
File: <input type="file" name="sel_doc[1]"><br>
Title: <input name="title[1]"><br>
<b>2</b><br>
File: <input type="file" name="sel_doc[2]"><br>
Title: <input name="title[2]"><br>
<b>3</b><br>
File: <input type="file" name="sel_doc[3]"><br>
Title: <input name="title[3]"><br>
<b>4</b><br>
File: <input type="file" name="sel_doc[4]"><br>
Title: <input name="title[4]"><br>
</div>
PHP code
$db = mysqli_connect("localhost", "root", "", "xy");
if (isset($_POST['upload_docs'])) {
for($mi = 0; $mi < count($_FILES['sel_doc']['tmp_name']); $mi++) {
$doc = $_FILES['sel_doc']['name'][$mi];
$tmp_doc = $_FILES['sel_doc']['tmp_name'][$mi];
$target = "doc_uploads/". basename($doc);
$title = (isset($_POST['title'][$mi])?$_POST['title'][$mi]:"");
if (move_uploaded_file($tmp_doc, $target)) {
echo "Upload successfully!";
}
else {
echo "Upload failed!";
}
$sql_f = "INSERT INTO docs (doc_file, title) VALUES ('$doc', '$title')";
mysqli_query($db, $sql_f);
}
}
Btw; This example code is not safe for sql injection.
I try upload image to mysql database and display it along with the description of image using php. After i upload the image and display it , a broken image was displayed but the description of the image was displayed without any error. How can I solve this problem ? Appreciate your help
<?php
$msg = "";
//if upload button is pressed
if(isset($_POST['upload']))
{
// the path to store the uploaded image
$target = "images/".basename($_FILES['image']['name']);
// connect to database
$db = mysqli_connect("localhost","root","","product");
// Get all the submitted data from the form
$image = $_FILES['image']['name'];
$text = $_POST['text'];
$sql = "INSERT INTO product_list (image, text) VALUES ('$image','$text')";
mysqli_query($db,$sql); // stores the submitted data into the database table : product_list
// move uploaded image to the folder : image
if (move_uploaded_file($_FILES['image']['tmp_name'],$target))
{
$msg = "Image and text uploaded successfully";
}else
{
$msg = "There was a problem uploading image";
}
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Image Upload With Description</title>
<link rel="stylesheet" type="text/css" href="formstyle.css">
</head>
<body>
<div id="content">
<?php
$db = mysqli_connect("localhost","root","","product");
$sql = "SELECT * FROM product_list";
$result = mysqli_query($db, $sql);
while ($row = mysqli_fetch_array($result))
{
echo "<div id='img_div'>";
echo "<img src='".$row['image']."'>";
echo "<p>".$row['text']."</p>";
echo "</div>";
}
?>
<form method="post" action="try.php" enctype="multipart/form-data">
<input type="hidden" name="size" value="1000000">
<div>
<input type="file" name="image">
</div>
<div>
<textarea name="text" cols="40" rows="4" placeholder="Details of product"></textarea>
</div>
<div>
<input type="submit" name="upload" value="Upload Image">
</div>
</form>
</div>
</body>
</html>
This is my result :
You are storing it in the DB without the images directory. You either need to store it with that, or always remember to call it that way in your image calls.
echo "<img src='images/".$row['image']."'>";
or make your the record you are writing the same as the filesystem location.
$image = 'images/' . $_FILES['image']['name'];
Note you are open to SQL injections and file inclusion injections with this code.
try this
<?php
$msg = "";
//if upload button is pressed
if(isset($_POST['upload']))
{
// the path to store the uploaded image
$destination_path = getcwd().DIRECTORY_SEPARATOR;
$target_path = $destination_path . basename( $_FILES["image"]["name"]);
// connect to database
$db = mysqli_connect("localhost","root","","product");
// Get all the submitted data from the form
$image = $_FILES['image']['name'];
$text = $_POST['text'];
$sql = "INSERT INTO product_list (image, text) VALUES ('$image','$text')";
mysqli_query($db,$sql); // stores the submitted data into the database table : product_list
//#move_uploaded_file($_FILES['image']['tmp_name'], $target_path)
// move uploaded image to the folder : image
if (move_uploaded_file($_FILES['image']['tmp_name'],$target_path))
{
$msg = "Image and text uploaded successfully";
}else
{
$msg = "There was a problem uploading image";
}
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Image Upload With Description</title>
<link rel="stylesheet" type="text/css" href="formstyle.css">
</head>
<body>
<div id="content">
<?php
$db = mysqli_connect("localhost","root","","product");
$sql = "SELECT * FROM product_list";
$result = mysqli_query($db, $sql);
while ($row = mysqli_fetch_array($result))
{
echo "<div id='img_div'>";
echo "<img src='".$row['image']."'>";
echo "<p>".$row['text']."</p>";
echo "</div>";
}
?>
<form method="post" action="index.php" enctype="multipart/form-data">
<input type="hidden" name="size" value="1000000">
<div>
<input type="file" name="image">
</div>
<div>
<textarea name="text" cols="40" rows="4" placeholder="Details of product"></textarea>
</div>
<div>
<input type="submit" name="upload" value="Upload Image">
</div>
</form>
</div>
</body>
</html>
I am uploading the description, the number and the image path through the PHP database.
When trying to call the image from the controller, an error is displayed which states: Undefined property: stdClass::$imgPath .... on line 9. On the other hand, the text elements are uploaded successfully.
Please find the PHP code below:
<?php
$prov = json_decode(file_get_contents("php://input"));
require_once("connection.php");
$connection = connectToMySQL();
// Check if image file is a actual image or fake image
$proverbDescription = $prov->proverbDescription;
$proverbNumber = $prov->proverbNumber;
$imgPath = $prov->imgPath;
$query = "INSERT INTO tbl_proverb (proverbDescription, proverbNumber, imgPath) VALUES ('$proverbDescription', '$proverbNumber', '$imgPath')";
echo($proverbDescription);
echo($proverbNumber);
echo($imgPath);
if (move_uploaded_file($_FILES['imgPath']['tmp_name'], $target_file)) //for uploading file
{
echo("File uploaded to: " . $target_dir);
}
$result = mysqli_query($connection, $query)
or die("Error in query: ". mysqli_error($connection));
if(mysqli_affected_rows($connection) > 0){
$success = true;
}else{
$success = false;
}
?>
The AngularJS Controller:
app.controller('proverbCtrl', function($scope, $http, $location) {
$scope.addProverb = function(prov) {
$http.post('php/addProverbs.php', prov).success(function(prov) {
console.log(prov);
// $location.path("/home");
});
};
});
And my HTML form:
<form id="demo" class="collapse" ng-submit="addProverb(prov)" enctype="multipart/form-data" method="POST">
<label>Image:</label>
<input type="file" ng-model="prov.imgPath" name="imgPath" id="imgPath" accept="image/*">
<label>Proverb Description:</label>
<input type="text" ng-model="prov.proverbDescription" ><br><br>
<label>Proverb Number:</label>
<input type="text" ng-model="prov.proverbNumber"><br><br>
<input type="submit" name="submit"><br>
See your post
</form>
If all you need is to get the file name to PHP try something like this:
HTML:
<input type="file" onchange="angular.element(this).scope().fileName(this)"
ng-model="prov.imgPath" name="imgPath" id="imgPath" accept="image/*">
Controller:
$scope.fileName= function(element) {
$scope.$apply(function(scope) {
scope.prov.imgPath = element.name;
});
};
So I'm trying to upload music files with php but the files aren't going to the folder, here's my code what are my mistakes and what am i missing
<form action="" method="post" enctype="multipart/form-data">
<input type="file" name="file"/>
<input type="text" placeholder="File Name" name="file_name">
<input type="text" placeholder="Artist Name" name="artist_name">
<textarea rows="5" placeholder="Description" name="description"></textarea>
<input type="submit" class="btn btn-warning bgcolor" name="submit" value="Upload"/>
</form>
<?php
if(isset($_POST['submit'])){
function GetMusicExtension($musictype)
{
if(empty($musictype)) return false;
switch($musictype){
case 'audio/mp3': return '.mp3';
default: return false;
}
}
if (!empty($_FILES["file"]["name"])) {
$file_name=$_FILES["file"]["name"];
$temp_name=$_FILES["file"]["tmp_name"];
$music_type=$_FILES["file"]["type"];
$size = $_FILES['file']['size'];
$ext= GetMusicExtension($music_type);
$musicname = $_POST['file_name'].$ext;
$target_path = "music/".$musicname;
if(move_uploaded_file($temp_name, $target_path)) {
$query = "insert into music(name,artist,description)
values('".$musicname."','".$_POST['artist_name']."','".$_POST['description']."')";
mysqli_query($link,$query) or die("error in $query == ----> ".mysqli_error($link));
}
}else{
exit('<script>alert("Error Saving Data... try again");</script>');
}
}
?>
PS: If i change the mime to an image mime and try uploading it uploads
UPDATE:
If i change post_max_size and upload_max_size it works. Thanks all.
I am uploading an image using the following php code but the file is not getting upload.
if(isset($_POST['submit'])){
$title = $_POST['title'];
$target_folder = "../newsimageuploads/";
$bannerimagelink = "http://example.com/newsimageuploads";
$bannerimage = addslashes(file_get_contents($_FILES['bannerimage']['tmp_name']));
$bannerimage_name = addslashes($_FILES['bannerimage']['name']);
$bannerimage_size = getimagesize($_FILES['bannerimage']['tmp_name']);
if ($bannerimage!=""){
$rand = rand(111111, 9999999);
$fname = $_FILES["bannerimage"]["name"];
$newname = "Image ".$rand.".png";
move_uploaded_file($_FILES["bannerimage"]["tmp_name"], $target_folder.$newname);
$bannerimage_location = $bannerimagelink."/".$newname;
}
$query =mysql_query("INSERT INTO mytable (title,image) VALUES ('$title','$bannerimage_location')")or die(mysql_error());
if (($query) === TRUE) {
echo "<p style='color:green;'>Added Successfully</p>";
} else {
echo "Some Error Occured :(";
}
}
And my HTML part is
<form action="#" method="post">
<input type="text" name="title">
<input type="file" name="bannerimage" accept="image/jpeg,image/png,image/gif">
<button type="submit" name="submit">Add</button>
</form>
My title is getting insert in the MySQL table but image does not.
You are missing enctype='multipart/form-data' in your form
<form action="#" method="post" enctype="multipart/form-data">
Look here for more details
Add
enctype="multipart/form-data"
to the form tag. Without this attribute you will only get the name of the file. But the file itself won't be uploaded.
<form action="#" method="post" enctype="multipart/form-data">
<input type="text" name="title">
<input type="file" name="bannerimage" accept="image/jpeg,image/png,image/gif">
<button type="submit" name="submit">Add</button>