PHP MYSQL $_GET['ID'] from table on dynamically generated page - php

Question

I have a few products stored in a table with auto-incremented ID entitled "product_id".
I have managed to create a page that displays a list of the products names as selected from the db and dynamically created links for each one. Let's say I have product apple. When I click on apple it takes me to view_product_details.php?id=9
But when I click apple, the view_product_details.php page tells me
"Notice: Undefined index: product_id in C:\xampp\htdocs\working\product-website-exercise\view_product_details.php on line 16"
<?php
//$result = mysql_query("SELECT * FROM temaproduct.products WHERE ID = '".mysql_real_escape_string($_GET['product_id'])."'");
$id = $_GET['product_id']; //This is line 16
$result = mysqli_query($conn,"SELECT * FROM products WHERE ID = $id");
echo $result['product_description'];
echo "<br>";
var_dump($result);
?>
I have tried with different queries but can't figure it out, please help me establish the connection properly so I can read the other fields from the table on the view_product_details page, based on product_id.
EDIT: Thank you guys, with your help, here is the code that works now, if everybody needs this snippet:
<?php
$id = intval($_GET['id']);
$sql = mysqli_query($conn,"SELECT * FROM products WHERE product_id = ".$id);
if(mysqli_num_rows($sql)){
$product_data = mysqli_fetch_array($sql);
echo "<h2><center>".$product_data['title']."</h2></center>";
}
?>

Answer 1

You are using id as a query string in this URL as:
view_product_details.php?id=9
So, you need to get id as:
$id = $_GET['id']; //This is line 16
Second issue in your code is that, you can not get result from database without using mysqli_fetch_* function.
echo $result['product_description']; // this will return nothing
Your Modified Code:
<?
$id = intval($_GET['id']);
$sql = mysqli_query($conn,"SELECT * FROM products WHERE ID = ".$id);
if(mysqli_num_rows($sql)){
$result = mysqli_fetch_array($sql);
echo $result['product_description'];
}
else{
echo "No record found";
}
?>
Suggestion:
You need to do one more thing, please use intval() function if any one pass string or anything else in the query string than your query will not return an error only return 0 record like:
Example:
view_product_details.php?id=abcdJUNK
Than convert it into 0 as:
$id = intval($_GET['id']);
For Future Visitors:
After debugging, found this error "Unknown Column ID"
so correct query was this as OP mentioned (column name was product_id):
SELECT * FROM hangouts WHERE product_id = 9

Answer 2

You are actually passing id in view_product_details.php?id=9 query params not product_id in the url.
To use product_id you can change the url like view_product_details.php?product_id=9 or you can use $_GET['id']
& replace this line
$id = $_GET['product_id'];
with this
$id = $_GET['id'];
You can use get_defined_vars(http://php.net/manual/en/function.get-defined-vars.php) to check which variables are
available.
Also I suggest you suppress your errors for production & show all errors in developement.
To hide errors in production
ini_set("display_errors", 0);
ini_set("log_errors", 1);
To show errors in developement
error_reporting(E_ALL);
ini_set('display_errors', 1);

Answer 3

According to your code and URL that you provide you are passing value in variable id not in product_id. This is your URL view_product_details.php?id=9 here value is in variable id i.e id=9.
$id = $_GET['id']; //This is line 16

Answer 4

two problems here first your name attribute in html form is different then what you are using in php i changed the php code for that issue and second problem is you are missing to convert the result in to assosiative array and directly calling the product description just read the comments in modified version of code down below.
this is your current code
$id = $_GET['product_id']; //This is line 16
$result = mysqli_query($conn,"SELECT * FROM products WHERE ID = $id");
echo $result['product_description'];
this is how it should be
// this is the solution for first issue
$id = $_GET['id']; //This is line 16
$result = mysqli_query($conn,"SELECT * FROM products WHERE ID = $id");
// this is the solution for second issue
// covert it to associative array
$resultData = mysqli_fetch_assoc($result);
echo $resultData['product_description'];