Get URL value to update spreadsheet

[MHatrak]

I have a .txt file, that I include in my php file. The line that has $sofas = "the url to my spreadsheet";  I would like to change the $csvFile = "$path"; dynamically using the url.

http://domain.com/furniture.php?category=sofas  and have the below code to update the line $csvFile = "$path"; and use the sofas url value instead. This is my first time using php so any help would be appreciated.

<?PHP
if (isset($_GET['category']))
{
$category = $_GET['category'];
}
$category = (isset($category)) ? $category : '';

$path = "$myarray[0] $category";
?>


<?PHP
function readCSV($csvFile){
	$file_handle = fopen($csvFile, 'r');
	while (!feof($file_handle) ) {
		$line_of_text[] = fgetcsv($file_handle, 1024);
	}
	fclose($file_handle);
	return $line_of_text;
}

	$csvFile = "$path";
                
$csv = readCSV($csvFile);
//print_r($csv);
?>

[Barand]

Not sure I fully understand what you are saying. Do you mean this...?

 

file1.php (to be included)
 

<?php

$sofas = 'url to my spreadsheet';

furniture.php

<?php

include 'file1.php';

...

$csv = readCSV($sofas);

 

[MHatrak]

3 hours ago, Barand said:

Not sure I fully understand what you are saying. Do you mean this...?

 

file1.php (to be included)
 

<?php

$sofas = 'url to my spreadsheet';

furniture.php

<?php

include 'file1.php';

...

$csv = readCSV($sofas);

 

Yes, the included file is a text file. That part works fine, I am trying to change $csv = readCSV ($sofas); value using url ie. http://domain.com/index.php?category=bedroom

and have that change value of $csv = readCSV ($bedroom);

[Barand]

Sorry, still don't know what you are working with.

• What's in the text file?
• What does $myarray contain?
• What should the value you are passing to readCSV($value) look like?

[MHatrak]

I don't think I am explaining it right, I just want to take the url http://domain.com/index.php?category=bedroom and replace $sofa with $bedroom and have the $bedroom pull the correct url.  I think a print or echo is correct not sure? If it helps here is what the included php file contains

<?php
$sofa = "http://pathtospreadsheet.com";
$bedroom = "http://pathtospreadsheet.com";
$dining = "http://pathtospreadsheet.com";
?>

ignore $myarray was attempting to use to make work as it gets the value from text file.  The value is a plain url.

how do you change this code

<?PHP
function readCSV($csvFile){
	$file_handle = fopen($csvFile, 'r');
	while (!feof($file_handle) ) {
		$line_of_text[] = fgetcsv($file_handle, 1024);
	}
	fclose($file_handle);
	return $line_of_text;
}

	$csvFile = "$sofas";
                
$csv = readCSV($csvFile);
//print_r($csv);
?>

to this just using value in the url

<?PHP
function readCSV($csvFile){
	$file_handle = fopen($csvFile, 'r');
	while (!feof($file_handle) ) {
		$line_of_text[] = fgetcsv($file_handle, 1024);
	}
	fclose($file_handle);
	return $line_of_text;
}

	$csvFile = "$bedroom";
                
$csv = readCSV($csvFile);
//print_r($csv);
?>

 

[Barand]

I would replace

<?php
$sofa = "http://pathtospreadsheet.com";
$bedroom = "http://pathtospreadsheet.com";
$dining = "http://pathtospreadsheet.com";
?>

with an array...

$categories = [ 
                "sofa"    => "http://pathtospreadsheet.com",
                "bedroom" => "http://pathtospreadsheet.com",
                "dining"  => "http://pathtospreadsheet.com"
             ];

then

// get the required category from the url
$cat = $_GET['category'] ?? '';         // blank if no category provided

if ($cat) {
    $path = $categories[$cat];          // get path from the array
    $csv = readCSV($path);
    
    echo '<pre>', print_r($csv, 1), '</pre>';
}