Agreaves Posted August 19, 2012 Author Share Posted August 19, 2012 the code I posted before displayed it in an html table but not the way I wanted it. It was duplicating a picture 4 times in each row, but i want it to display 4 different pictures at a time in each row. Quote Link to comment https://forums.phpfreaks.com/topic/267244-how-to-display-database-data-to-a-table/page/2/#findComment-1370587 Share on other sites More sharing options...
jazzman1 Posted August 19, 2012 Share Posted August 19, 2012 I'll give you the correct table, but.... you have to know that I don't like this You have to learn every day, step by step. You don't need to use extract() in this particular case. /* Create a table to hold the data */ echo "<table width='1024' border='0' align='center' cellpadding='25' cellspacing='0' class='main_bdr'>"; echo "<tr><th>ID</th><th>Name</th><th>Cat</th><th>Description</th><th>Price</th><th>Pix</th></tr>"; /*Run a loop to retrieve all rows from the database */ while ($rows = mysqli_fetch_assoc($result)) { /* Extract all rows of data from the result */ //extract($rows); echo "<tr>"; echo "<td> {$rows['Id']}</td> <td> {$rows['Name']}</td> <td> {$rows['Cat']}</td> <td> {$rows['Description']}</td> <td> {$rows['Price']}</td> <td width='150' align='center'><img src='{$rows['Pix']}' width='150' height='150' /></td> "; echo "</tr>"; } echo "</table>"; Quote Link to comment https://forums.phpfreaks.com/topic/267244-how-to-display-database-data-to-a-table/page/2/#findComment-1370589 Share on other sites More sharing options...
Agreaves Posted August 19, 2012 Author Share Posted August 19, 2012 Ok thank you, ill try this and see how it works Quote Link to comment https://forums.phpfreaks.com/topic/267244-how-to-display-database-data-to-a-table/page/2/#findComment-1370590 Share on other sites More sharing options...
teng84 Posted August 19, 2012 Share Posted August 19, 2012 I suggest learn how to use jquery data table makes your life easier see: http://datatables.net/ Quote Link to comment https://forums.phpfreaks.com/topic/267244-how-to-display-database-data-to-a-table/page/2/#findComment-1370627 Share on other sites More sharing options...
Agreaves Posted August 23, 2012 Author Share Posted August 23, 2012 I created this simple shopping cart system but the after running the query the program wont show the result in the table, what amI doin wrong? <?php // Product Id from URL $product_id = $_GET['Id']; // Action from URL $action = $_GET['action']; //if there is an product_id and that product_id doesn't exist display an error message if($product_id && !productExists($product_id)) { die("Error. Product Doesn't Exist"); } switch($action) { //decide what to do case "add": $_SESSION['cart'][$product_id]++; //add one to the quantity of the product with id $product_id break; case "remove": $_SESSION['cart'][$product_id]--; //remove one from the quantity of the product with id $product_id if($_SESSION['cart'][$product_id] == 0) unset($_SESSION['cart'][$product_id]); //if the quantity is zero, remove it completely (using the 'unset' function) - otherwise it will show zero, then -1, -2 etc when the user keeps removing items. break; case "empty": unset($_SESSION['cart']); //unset the whole cart, i.e. empty the cart. break; } ?> <?php if($_SESSION['cart']) { //if the cart isn't empty //show the cart echo "<table border=\"1\" padding=\"3\" width=\"40%\">"; //format the cart using a HTML table //iterate through the cart, the $product_id is the key and $quantity is the value foreach($_SESSION['cart'] as $product_id => $quantity) { //get the name, description and price from the database - this will depend on your database implementation. //use sprintf to make sure that $product_id is inserted into the query as a number - to prevent SQL injection $sql = sprintf("SELECT Name, Description, Price FROM products WHERE Id = %d;", $product_id); $result = mysqli_query($con,$sql) or die("couldnt run query"); $row = mysqli_num_rows($result); $row2 = mysqli_fetch_assoc($result); $name = $row2['Name']; $description = $row2['Description']; $price = $row2['Price']; //Only display the row if there is a product (though there should always be as we have already checked) if($row > 0) { list($name, $description, $price) = mysqli_fetch_row($result); $line_cost = $price * $quantity; //work out the line cost $total = $total + $line_cost; //add to the total cost echo "<tr>"; //show this information in table cells echo "<td align=\"center\">$name</td>"; //along with a 'remove' link next to the quantity - which links to this page, but with an action of remove, and the id of the current product echo "<td align=\"center\">$quantity <a href=\"$_SERVER[php_SELF]?action=remove&id=$product_id\">X</a></td>"; echo "<td align=\"center\">$line_cost</td>"; echo "</tr>"; } } //show the total echo "<tr>"; echo "<td colspan=\"2\" align=\"right\">Total</td>"; echo "<td align=\"right\">$total</td>"; echo "</tr>"; //show the empty cart link - which links to this page, but with an action of empty. A simple bit of javascript in the onlick event of the link asks the user for confirmation echo "<tr>"; echo "<td colspan=\"3\" align=\"right\"><a href=\"$_SERVER[php_SELF]?action=empty\" onclick=\"return confirm('Are you sure?');\">Empty Cart</a></td>"; echo "</tr>"; echo "</table>"; }else{ //otherwise tell the user they have no items in their cart echo "You have no items in your shopping cart."; } //function to check if a product exists function productExists($product_id) { //use sprintf to make sure that $product_id is inserted into the query as a number - to prevent SQL injection $sql = sprintf("SELECT * FROM php_shop_products WHERE id = %d;", $product_id); return mysql_num_rows(mysql_query($sql)) > 0; } ?> Quote Link to comment https://forums.phpfreaks.com/topic/267244-how-to-display-database-data-to-a-table/page/2/#findComment-1371858 Share on other sites More sharing options...
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