Insert image MYSQL based on radio button choice with PHP - php

Question

I have a website. In this website I can upload images that are then placed into my SQL database. From here I select the images from the database and show them as thumbnails in the photo gallery, when clicked on the image it shows a large version where you can vote/like and comments etc.
Now what I am trying to do is make 3 category pages, basically 3x the photo gallery that shows the thumbnails.
I have 3 different tables in my database where I insert the images in to.
So I copied the photo gallery 3x and the original upload table in the database 3x.
How ever I do not want to create 3 upload.php files for each photo gallery.php file.
What I'm trying to do is have 3 radio button choices on my upload page and with the choice made there, the image gets uploaded into the matching database table (photo1, 2 or 3).
I have been trying to do this with Functions etc. but I just can't get it to work, I am probably doing something really simple, really stupid.
This is the code i have for the radio button and getting the image:
$titel = "Image";
$query = "SELECT * FROM `i268296_studie`.`fotos` ORDER BY foto_ID DESC";
$result = mysqli_query($conn, $query) or die("query error " . mysqli_error($conn) );
$fotos = array();
//create array from images in database
while($data = mysqli_fetch_assoc($result))
{
$fotos[] = array('src' => $data['src'], 'id' => $data['foto_ID']);
}
?>
<section id="upload">
<form method="post" action="upload.php" enctype="multipart/form-data">
<label for="bestand">Upload image:</label><br><br>
<input type="file" name="bestand" id="file"><br><br>
<label for="categorie"> Categorie: </label>
<input type="radio" name="cat" value="cat1">Portrait
<input type="radio" name="cat" value="cat2">Landscape
<input type="radio" name="cat" value="cat3">Other
<input type="submit" name="submit" value="Upload">
</form>
</section>
<?php
}
?>
<section class="images">
<?php
//show image thumbnails in photogallery
foreach($fotos as $foto)
{
?>
<img class="image" src="<?php echo 'upload/thumb/t_'.$foto['src'];?>">
<?php
}
?>
</section>
The above code I have 3 times (surrounded by HTML etc. as the photo gallery pages).
This is my Upload file (i'll leave most of the thumbnail making code out of it since it's only about the upload part).
$titel = "Image";
$dir='upload/';
$allowedExts = array("jpg", "jpeg", "gif", "png");
$answer= $_POST['cat'];
//Properties of the to be uploaded file
$fileName = $_FILES["bestand"]["name"]; //file name
$fileType = $_FILES["bestand"]["type"]; //file format
$fileSize = $_FILES["bestand"]["size"]; //file size
$tmpName = $_FILES["bestand"]["tmp_name"]; //temporary save location for file
$error = $_FILES["bestand"]["error"]; //error check for file
$fp = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);
if(!get_magic_quotes_gpc())
{
$fileName = addslashes($fileName);
}
//select image from database and check if it already exists
$sql = "SELECT * FROM `i268296_studie`.`fotos` WHERE src = '$fileName'";
$result = mysqli_query($conn, $sql) or die("query error " . mysqli_error($conn) );
$data = mysqli_fetch_assoc($result);
$num_rows=mysqli_num_rows($result);
if($num_rows > 0)
{
echo 'File already exists <br>';
echo '<a href="fotogallerij.php">Return to homepage/a>';
}
else
{
// if file doesn't exist move to database, create thumbnail path
if (move_uploaded_file( $tmpName,$dir.$fileName))
{
function category($cat, $titel, $filename)
{
global $conn;
$query = "INSERT INTO `i268296_studie`.`$cat` (`titel`, `src`) VALUES ('$titel', '$fileName')"; //INSERT file into database
$result = mysqli_query($conn, $query) or die("query error " . mysqli_error($conn) );
}
$tname = 't_'.$fileName;
$tpath = $dir.'thumb/';
$tnamestate = $tpath.$tname;
$tptype = substr($fileType,6);
$ttype = "imagecreatefrom$tptype";
$name = $fileName;
$path = $dir;
$namestate = $path.$name;
$width = 100;
$height = 100;
list($width_orig, $height_orig) = getimagesize("$namestate");
$ratio_orig = $width_orig/$height_orig;
if ($width/$height > $ratio_orig)
{
$width = $height*$ratio_orig;
}
else
{
$height = $width/$ratio_orig;
}
How ever I am staring myself blind on how to fix it or where to place it so it works.
I hope my explanation and question is clear to you guys trying to help me, if not please let me know what I can change or do to help :)
edit:
The errors I am getting are:
Undefined variable: titel
Undefined variable: fileName
Undefined variable: conn
mysqli_query() expects parameter 1 to be mysqli, null given
When I do not pick a radio button but just upload it directly i just get 1 error:
Undefined index: cat
And it uploads it into the 3rd category
Edit 2:
Changed the function with the global $conn in it.

Answer 1

1 function category($cat, $titel, $fileName) {
2 global $conn;
3 $query = "INSERT INTO `i268296_studie`.`$cat` (`titel`, `src`) VALUES ('$titel', '$fileName')";
4 $result = mysqli_query($conn, $query) or die("query error " . mysqli_error($conn) );
5 }
1: you need these variables from the context calling the function
2: you need the global $conn variable in the function context to be able to run the query
3: use a function parameter for the table to upate call
4: it would be better to return results instead of breaking inside the function
The calling code would be like the following:
if ($answer == "cat1") {
category("fotos", $titel, $fileName); // or "foto2", $titel, $fileName .....
}
Please mind the comments about injection vulnerabilities.
Also read this: http://php.net/manual/it/language.variables.scope.php