saving images in PHP with variable name - php

Question

I am currently developing a web application and need some help.
The functionality is the following: A website connected to a mysql database. You need to select two times brand and then products from that brand solved via interlinked dynamic dropdowns. Based on that selections happen two queries selecting color values from the database, showing them in textfields.
Solved via jquery, PHP and JSON.
Now I have created a section, which will show an image in the color from the selected RGB values from the DB. PHP needs to create the JPG image using the RGB values from the DB and save the image on the server.
When I implement imagecreate in php and hand over the path to jquery to allocate it to a '"#target", I get the issues.
I considered two ways:
a) generate image on the fly, name it 1.jpg and overwrite the image in every instance. Then I run into issues with the browser cache, loading the previous image although the new image is already stored in the folder.
b) generate new images and save them for better performance named with the RGB values (i.e 151|50|22.jpg) and load if already existing, create if not, but there I cannot save the path when a variable is in the imagejpeg() function in PHP?
I can save i)a combination of path elements and a string '1.jpg' and ii)a path made from a pure string but not a path made from a combination of text and a variable carrying the right image name.
Propably it is done with right escaping of the variable???
Errors are:
Warning: imagejpeg(images\rgb\225|151|169.jpg): failed to open stream: No such file or directory in C:\xampp\htdocs\COLOMETRIX\get_swatch_1a.php on line 39
{"decision":"images\rgb\225|151|169.jpg"} this is the Json output
Code is a follows:
<?php
$prod1 = $_GET['q'];
include('db_config.php');
// clear variables
$decision='';
$data='';
// select dataset
$result=$db->query("SELECT R, G, B, Get_Col, Selection
FROM farbdatenbank.farbdatenbank
WHERE Name ='".$prod1."' ")
or die($db->error);
$datensatz=$result->fetch_array(MYSQLI_ASSOC);
// define paths and filename
$file=$datensatz['Get_Col'];
$filename=$file.".jpg";
$path_rgb='images'.DIRECTORY_SEPARATOR.'rgb'.DIRECTORY_SEPARATOR.$filename;
if($datensatz['Selection'] == "Excel"){
// get color values
$R=$datensatz['R'];
$G=$datensatz['G'];
$B=$datensatz['B'];
//check if file already exists
if (file_exists($path_rgb) == true) {
$decision=$path_rgb;
}
else
{
// create new image
$im1=imagecreatetruecolor(150, 70);
$colim1=imagecolorallocate($im1, $R, $G, $B);
imagefill($im1,0,0,$colim1);
$old = 'images'.DIRECTORY_SEPARATOR.'rgb'.DIRECTORY_SEPARATOR.'zzz.jpg';
imagejpeg($im1,$path_rgb,100);
$decision=$path_rgb;
}
}
Else
{
$decision='images'.DIRECTORY_SEPARATOR.'pics'.DIRECTORY_SEPARATOR.$filename;
}
// build output array and encode in JSON format
$data=array("decision"=>$decision);
echo json_encode($data);
mysqli_close($db);
?>
Thanks for your time
Michael