<a href="report_product.php?id=<?php echo $data['category']; ?>">
this line shows the name of category for example stationary..but this should be the id of selected category.is there any need to declare the id?$category_id = $_POST['category_id'];
like this?
i tried with this also but it gives an error of undefined index
. through this link i want to show the list of products on another page associated with selected category
i am on the learning phase and stuck on this issue kindly help me to sort out that
if(isset($_POST['search_category'])){ // $category_id = $_POST['category_id']; $records = mysqli_query($db,"SELECT DISTINCT category , (SELECT COUNT(products.name) FROM products where category_id=categories.id) AS total_products, ( SELECT SUM(quantity) FROM productstock a LEFT JOIN products b ON a.product_id=b.id LEFT JOIN categories c ON b.category_id=c.id where c.deleted=0 AND b.category_id = categories.id GROUP BY category_id ) AS available_stock, SUM(product_qty*orignalCost) AS SaleWise_cost, SUM(product_qty*saleprice) AS SaleWise_price, SUM(product_qty*saleprice) AS total_sale , SUM((product_qty*saleprice)-(product_qty*orignalCost)) AS profit FROM categories INNER JOIN products ON categories.id = products.category_id INNER JOIN sales ON sales.product_id = products.id INNER JOIN productstock ON productstock.product_id = products.id WHERE categories.deleted=0 GROUP BY category_id;"); // fetch data from database while($data = mysqli_fetch_array($records)) { ?> <td><a href="report_product.php?id=<?php echo $data['category']; ?>"><?php echo $data['category']; ?></a></td> <td><?php echo $data['total_products']; ?></td> <td><?php echo $data['available_stock']; ?></td> <td><?php echo $data['SaleWise_cost']; ?></td> <td><?php echo $data['SaleWise_price']; ?></td> <td><?php echo $data['total_sale']; ?></td> <td><?php echo $data['profit']; ?></td> </tr> <?php } } ?> </table>
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Answer
You’re not selecting the category ID in you query add the ID column to the SELECT list.
For example, instead of
SELECT DISTINCT category, ...
Try doing
SELECT DISTINCT category, id ...
and then the ID should be returned in the data for use like…
<a href="report_product.php?id=<?php echo $data['id']; ?>"><?php echo $data['category']; ?></a>