I have a PDF uploader that is supposed to save a file to a file path that is based off of a username variable selected from a drop down menu.
Everything works but the uploader, as it displays the usernames and the directories are created upon registration. So there is no issue with that. My issue lies within the code below with the uploader:
<form action="" method="post" enctype="multipart/form-data">
<input type="file" name="pdf" /><br />
<select name="folder">
<?php
$con=mysqli_connect("host","user","pass","dbname");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT `first_name`, `last_name`, `username` FROM `cw_users` WHERE 1");
$user = 'username';
while($row = mysqli_fetch_array($result)) {
echo "<option value='". $row["username"] ."'>";
echo " $row[username] ";
echo "</option>";
}
mysqli_close($con);
?>
<?php
if (isset($_POST['submit'])) {
$pdfDirectory = "Users/".$_POST['folder']."/uploaded/";
//get the name of the file
$filename = basename( $_FILES['pdf']['name'], ".pdf");
//remove all characters from the file name other than letters, numbers, hyphens and underscores
$filename = preg_replace("/[^A-Za-z0-9_-]/", "", $filename).".pdf";
if (move_uploaded_file($_FILES['pdf']['tmp_name'], $pdfDirectory.$filename)) {
//the path to the PDF file
$pdfWithPath = $pdfDirectory.$filename;
}
}
?>
</select>
<input type="submit" value="Upload pdf" name="upload_pdf" />
</form>
P.S. if this could be adjusted to upload multiple files at the same time that would be great.
Your conditional statement if(isset($_POST['submit'])) is looking for a submit button named "submit", yet yours is named "upload_pdf".
<input type="submit" value="Upload pdf" name="upload_pdf" />
^^^^^^^^^^^^^^^^^
That should either read as if(isset($_POST['upload_pdf'])) or rename your submit button to:
<input type="submit" value="Upload pdf" name="submit" />
Related
Still a newb to coding . have almost no idea what im doing . i tried to make a php page which would let me upload and view an image . do not know what is wrong . i tried to do it as correctly as possible . could someone please help me out ?
<!doctype html>
<html>
<head>
</head>
<body>
<form method="post">
<input type="file" name="image"></input>
<input type="submit" name="submit" value="upload"></input>
<?php
if(isset($_POST['submit']))
echo "button has been clicked";
$con = mysqli_connect("127.0.0.1","root","","demo");
if(!$con)
echo "didnt connect to database ";
else echo "connected ";
$imagename= mysqli_real_escape_string($_FILES['image'] ['name']);
$imagefile =mysqli_real_escape_string(file_get_contents($_FILES['image']['tmp_name']));
$qry = "INSERT INTO image (name,file) VALUES ('$imagename','$imagefile')";
$result = mysqli_query($con,$qry);
if($result)
echo "image has been uploaded";
viewimage();
function viewimage()
{$recon = mysqli_connect("127.0.0.1","root","","demo");
$view = "SELECT * FROM image ";
$data =mysqli_query($recon,$view);
$res2 =mysqli_fetch_assoc($data);
$currimage =$res2['file'];
echo "$currimage <br/>";
}
?>
</body>
</html>
To be able to catch a post variable, you need to submit the form and handle the action. The first problem with your code is that your form is not complete - it's missing a closing tag. Second thing, to be able to send a file through the post, you'll need multipart form. You should add enctype="multipart/form-data" as an attribute of the form.
So, instead of
<form method="post">
<input type="file" name="image"></input>
<input type="submit" name="submit" value="upload"></input>
You'll need
<form method="post" enctype="multipart/form-data">
<input type="file" name="image"></input>
<input type="submit" name="submit" value="upload"></input>
</form>
you must move the uploaded image to server
try this code -- create directory /uploads/
<!doctype html>
<html>
<head>
</head>
<body>
<form method="post" enctype="multipart/form-data">
<input type="file" name="image"></input>
<input type="submit" name="submit" value="upload"></input>
<?php
if(isset($_POST['submit']))
echo "button has been clicked";
$con = mysqli_connect("127.0.0.1","root","","demo");
if(!$con)
echo "didnt connect to database ";
else echo "connected";
$uploads_dir = '/uploads';
$tmp_name = $_FILES["image"]["tmp_name"];
$name = $_FILES["image"]["name"];
move_uploaded_file($tmp_name, "$uploads_dir/$name");
$qry = "INSERT INTO image (name,file) VALUES ('$name','$tmp_name')";
$result = mysqli_query($con,$qry);
if($result)
echo "image has been uploaded";
viewimage();
function viewimage()
{$recon = mysqli_connect("127.0.0.1","root","","demo");
$view = "SELECT * FROM image ";
$data =mysqli_query($recon,$view);
$res2 =mysqli_fetch_assoc($data);
$currimage =$res2['file'];
echo '<img src="'.$currimage.'" /> <br/>';
}
?>
</body>
</html>
I have checked out other questions of same topic on this site and tried to find the solution but unsuccessful. Images are stored in database and loaded in folder successfully but are not displayed
Here is my code:
<html>
<body>
<form action="image.php" method="post" enctype="multipart/form-data">
<input type="text" name="image_description" placeholder="Enter name" required>
<input type="file" name="myfile">
<input type="submit" name="upload" value="upload">
</form>
</body>
</html>
<?php
include("db.php");
if(isset($_POST['upload'])) {
$image_description = $_POST['image_description'];
$name = $_FILES["myfile"]["name"];
$type = $_FILES["myfile"]["type"];
$size = $_FILES["myfile"]["size"];
$temp = $_FILES["myfile"]["tmp_name"];
$error = $_FILES["myfile"]["error"];
$upload=move_uploaded_file($temp, "uploaded/" . $name);
$query= "INSERT INTO image(image_description,image_name,image_type,image_size) VALUES ('$image_description','$name','$type','$size')";
if(mysqli_query($conn,$query) && $upload) {
echo "successfully uploaded";
}
else
die(mysqli_error($conn));
}
$query = mysqli_query($conn,"SELECT * FROM image");
while($row = mysqli_fetch_array($query))
{?>
<img style="width: 200px;height: 200px;" src="<?php echo 'uploaded/' .$row['image_name'] ?>">
<?php
echo $row['image_description'] . "<br>";
}?>
Images are displayed as in picture
This is database table
The URL of your page is index.php/; notice the trailing slash.
A relative URL (e.g. src="uploaded/..") will resolve to index.php/uploaded/...
That folder obviously does not exist on your disk.
Use root-relative URLs: src="/uploaded/.."
or use relative URLs but go to the right folder: src="../uploaded/.."
or fix your weird URL and make it index.php, from which even relative URLs will resolve correctly.
I have created a PHP signup form for visitors to fill and submit that asks for their basic information.
I am trying to accomplish the following two tasks;
Add Image/File Upload Field
Redirect them to a confirmation page
I have been unable to make it work. Below is what I have.
My Code
HTML Form
<form name="form1" method="post" action="signup.php">
Username: <input type="text" name="user">
<br>Email: <input type="text" name="mail">
<br>Experience: <select name="exp"> <option value="beginner">Beginner</option>
<option value="intermediate">Intermediate</option> <option value="advanced">Advanced</option>
</select><br> <input type="submit" name="Submit" value="Sign Up">
</form>
Signup.php
<?php
$username = $_POST['user'];
$email = $_POST['mail'];
$experience = $_POST['exp'];
//the data
$data = "$username | $email | $experience\n";
//open the file and choose the mode
$fh = fopen("users.txt", "a");
fwrite($fh, $data);
//close the file
fclose($fh);
print "User Submitted";
?>
It seems like you lack an input field in your HTML to begin with.
here's an example of a form for uploading files.
<form action="upload.php" method="post" enctype="multipart/form-data">
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form>
Once you've done that you're not quite there yet because your file is stored in a temporary folder, you will need to move the file to your uploads folder like so:
$target_file = "uploads/" . basename ($_FILES["fileToUpload"]["name"]);
move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)
I hope this helps!
Ta add an upload you need to add enctype="multipart/form-data" to your form tag,then add the upload field. For the Email field change the type to HTML5 type="email", this will do a little validaation check that it is in the correct format. At the bottom of the php file add a location header if all went well. You could put it all in one file with an if statement at the top.You should also sanitize your inputs
this is a upload file script which will loop through all the data of a file and insert
if(isset($_POST['submit'])) {
if (is_uploaded_file($_FILES['filename']['tmp_name'])) {
echo "<h1>" . "File ". $_FILES['filename']['name']. "uploaded successfully." . "</h1>";
}
$csv_file=$_FILES['filename']['tmp_name'];
$type=$_FILES['filename']['type'];
$handle = fopen($csv_file, "r");
$i=0;
while (($data = fgetcsv($handle)) !== FALSE) {
if($i>0) {
$import="insert into `table_name`(col1,col2,col3,col4,col5,col6,col7)values('".addslashes($data[0])."','".addslashes($data[1])."','".addslashes($data[2])."','".addslashes($data[3])."','".addslashes($data[4])."','$data[5]','$data[6]')";
mysql_query($import) or die(mysql_error());
}
$i=1;
}
echo "Success";
echo "<br>";
echo $_FILES['filename']['type'];
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" enctype="multipart/form-data" name="" id="">
Choose File:<br>
<input name="filename" type="file" />
<input type="submit" name="submit" value="submit" />
</form>
I'm having problem with my code. I can't update my image.
Here is my full code:
index.php
//this link will post my data to next page
update
updatepage.php
//this page will get all data that want to update...
<?php
include("config.php");
$name=$_GET['code'];
$sql1 = "select * from imagename where name='$name'";
$result = mysql_query($sql1) or die(mysql_error());
$row=mysql_fetch_array($result);?>
<form name="form1" method="post" action="processupdatepage.php" class="formposition" enctype="multipart/form-data" >
<table>
<tr><td>
<input type="hidden" name="id" value="<?php echo $row['id']; ?>" /></td></tr>
<tr><td width="98">Image</td><td width="288">
//This is where I have problem, I can't get my image value but other data value work very well.....
<input type="file" name="image" value="<?php echo $row['image'];?>" /></td></tr><br/>
<tr><td>nane</td><td><input type="text" name="name" value="<?php echo $row['name']; ?>"/></td></tr><br/>
<tr><td></td> <td colspan="2">
<input type="submit" name="submit" value="Save" /> </td></tr>
</table>
</form>
OK. Now this my process file:
processupdatepage.php
<?php
include("config.php");
$id=$_POST['id'];
$image=$_FILES['image']['name'];
$name=$_POST['name'];
$target = "images/";
$target = $target . basename( $_FILES['image']['name']);
$query = "UPDATE imagename SET name='$name', image='$image' WHERE id='$id'";
$bb = mysql_query($query) or die(mysql_error());
if($bb)
{
//Writes the photo to the server
if(move_uploaded_file($_FILES['image']['tmp_name'], $target))
{
$sql001="UPDATE imagename SET image='$image' WHERE image='$image'";
mysql_query($sql001);
} else { }
header("Location:index.php");
}
else
{
echo "Could not be updated";
}
?>
I can update BUT must select image. If I'm not selecting an image, my image field in database will be empty BUT other data updates are OK...
The value of an <input type="file"> cannot be programatically changed. Only by triggering its default behaviour that pops up the select file dialogue.
What I see from you example code is that you want to associate the value to the file form control element straight as it was retrieved from the database. So why saving the same back. Still, if you want to do this, you can use a hidden element with the value, that will be saved. But if you don't want the user to change it, just don't render it into the form and leave it out from your update query so it will remain unchanged.
Also, you could add some logic when you process the post and leave the field 'image' out of your update query if no file was selected.
I think the problem is here:
<input type="file" name="image" value="<?php echo $row['image'];?>" />
You can upload the image by clicking the browse button. so what do you trying to do like that?
If you want to show the image already uploaded you can use :
$img_path = 'get image path from database entry';
<img src= "<?php echo $img_path; ?>"/>
Here is full code to upload an Image, this is tested and works but you need to have a directory called 'upload' (this code does not create the folder).
It is advisable to only store image path on database.
<?php
if(isset($_POST['submit'])){
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/png")
|| ($_FILES["file"]["type"] == "image/pjpeg"))
&& ($_FILES["file"]["size"] < 200000))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
if (file_exists("upload/" . $_FILES["file"]["name"])){
echo $_FILES["file"]["name"] . " already exists. ";
}
else{
move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $_FILES["file"]["name"]);
echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
// HERE YOU CAN WRITE YOU INSERT INTO MYSQL CODE
//SOMETHING LIKE THIS:
$path = 'assets/article_images/'.$article_id.'/'.$_FILES["image_file"]["name"];
$statement1 = "INSERT INTO image_article (article_id, image_path) values ('$article_id', '$path') ";
$query_result = mysql_query($statement1) or die($statement1."mysql error<br/><br/>".mysql_error());
if($query_result==1){
echo 'Image uploaded';
}
}
}
}
else
{
echo "Invalid file";
}
}
?>
<html>
<body>
<form action="index.php" method="post"
enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file"><br>
<input type="submit" name="submit" value="Submit">
</form>
</body>
</html>
I am making a gallery that uses a MySQL database (yeah I know it's a bad practice but it's the requirement for the moment.) I can upload multiple images but I'm having trouble displaying all images stored inside the database. The FORM allows five images to be uploaded. Then the user must proceed to another page where all the images in database (including the ones uploaded recently) will be displayed together with the description of the images. I have code already but the one that will work on the display is not working or I think is wrong.
Here is the form code:
<html>
<head>
<title> Upload image</title>
</head>
<body>
<div align="center">
<form action="fUpload.php" method="POST" enctype="multipart/form-data">
All forms must be filled. <br />
File: <br />
<input type="file" name="image[]"/> <input type="text" name="imageDescription[]" size="30" /> <br />
<input type="file" name="image[]"/> <input type="text" name="imageDescription[]" size="30" /> <br />
<input type="file" name="image[]"/> <input type="text" name="imageDescription[]" size="30" /> <br />
<input type="file" name="image[]"/> <input type="text" name="imageDescription[]" size="30" /> <br />
<input type="file" name="image[]"/> <input type="text" name="imageDescription[]" size="30" /> <br />
<input type="submit" value="Upload image" />
</form>
</div>
</body>
</html>
Here is the script that would upload:
<?php
//connect to the database//
$con = mysql_connect("localhost","root", "");
if(!$con)
{
die('Could not connect to the database:' . mysql_error());
echo "ERROR IN CONNECTION";
}
$sel = mysql_select_db("imagedatabase");
if(!$sel)
{
die('Could not connect to the database:' . mysql_error());
echo "ERROR IN CONNECTION";
}
//file properties//
$file = $_FILES['image']['tmp_name'];
echo '<br />';
/*if(!isset($file))
echo "Please select your images";
else
{
*/for($count = 0; $count < count($_FILES['image']); $count++)
{
//$image = file_get_contents($_FILES['image']['tmp_name']);
$image_desc[$count] = addslashes($_POST['imageDescription'][$count]);
$image_name[$count] = addslashes($_FILES['image]']['name'][$count]); echo '<br \>';
$image_size[$count] = #getimagesize($_FILES['image']['tmp_name'][$count]);
$error[$count] = $_FILES['image']['error'][$count];
if($image_size[$count] === FALSE || ($image_size[$count]) == 0)
echo "That's not an image";
else
{
// Temporary file name stored on the server
$tmpName[$count] = $_FILES['image']['tmp_name'][$count];
// Read the file
$fp[$count] = fopen($tmpName[$count], 'r');
$data[$count] = fread($fp[$count], filesize($tmpName[$count]));
$data[$count] = addslashes($data[$count]);
fclose($fp[$count]);
// Create the query and insert
// into our database.
$results = mysql_query("INSERT INTO images( description, image) VALUES ('$image_desc[$count]','$data[$count]')", $con);
if(!$results)
echo "Problem uploding the image. Please check your database";
//else
//{
echo "";
//$last_id = mysql_insert_id();
//echo "Image Uploaded. <p /> <p /><img src=display.php? id=$last_id>";
//header('Lcation: display2.php?id=$last_id');
}
//}
}
mysql_close($con);
header('Location: fGallery.php');
?>
And finally the one that should display:
<html>
<body>
</body>
<?php
//connect to the database//
mysql_connect("localhost","root", "") or die(mysql_error());
mysql_select_db("imagedatabase") or die(mysql_error());
//requesting image id
$id = addslashes($_REQUEST['id']);
$image = mysql_query("SELECT * FROM images WHERE id = $id");
while($datum = mysql_fetch_array($image, MYSQL_ASSOC))
{
printf("Description %s $image = $image['image'];
header("Content-type: image/jpeg");
}
mysql_close();
?>
Your help is much appreciated. I need it badly to move on.
From what i understand from your post is that uploading and storing isn't a problem, but showing the images is. That's probably because you're using vars that are not set, so no results kan be found in the database. If i misunderstood let me know.
<?php
// No ID
$image = mysql_query("SELECT * FROM images ORDER BY id DESC");
?>
Also look at what Prof83 says. Ignore my post if your script works with just one image.
Last but not least, if you're using different filetypes, also echo the correct MIME format in the header.
Update
I combined both answers.
Edit your loop:
<?php
while($row = mysql_fetch_assoc($image))
{
echo '<img src="img.php?id='.$row["id"].'">';
}
?>
Create a page name img.php
<?php
$query = mysql_query("SELECT image FROM images WHERE id = ".$_GET['id']);
$row = mysql_fetch_assoc($query);
header("Content-type: image/jpeg");
echo $row['image'];
?>
Ok you can't display multiple images within a image/jpeg page...
You're telling the browser that the page is image/jpeg (in other words, the page is AN IMAGE) but you're echoing out multiple image data
You should rather use the gallery page to show all images like this:
<?php
// $images = result from database of all image rows
foreach ($images as $img) echo '<img src="img.php?id='.$img["id"].'">';
?>
and in img.php:
// Load the image data for id in $_GET['id'];
header("Content-type: image/jpeg");
echo $data;