How to echo the array of a sql consult into a html

Question

I am having trouble with an array, I don't know why it isn't showing on the HTML index.

$db is the stablished connection

The table structure is:

pedidos:
id, cliente_id, creacion, id, modificacion, precio_total, and status

productos:
id, nombre, descripcion, precio, descripcion

Here's the PHP code:

<?PHP
$sql=mysqli_query($db,"
    SELECT 'p.id', 'p.nombre', 'p.precio', 'a.cantidad', 'p.precio' * 'a.cantidad' AS total, 'a.id'
        FROM articulos_pedidos a INNER JOIN productos p 
        ON 'a.producto_id' = 'p.id'
            INNER JOIN pedidos pe
        on 'a.pedido_id' = 'pe.id' 
        WHERE pedido_id = 23")OR die("query is incorrect-........................");
            $a = $sql->fetch_assoc();
            $data_pedido = array(
                    'id' => $a['pedido_id'],
                'nombre' => $a['p.nombre'],
                'precio' => $a['p.precio'],
                'cantidad' => $a['a.cantidad'],
                'total' => $a['ptotal']
                );
?>

<?php echo $data_pedido['id']; ?>
<?php echo $data_pedido['nombre']; ?>
<?php echo $data_pedido['precio']; ?>
<?php echo $data_pedido['cantidad']; ?>
<?php echo $data_pedido['total']; ?>

When you say *html index*, do you mean you have a file called `index.html`? — Qirel, Dec 02 '16 at 05:37
yes the html file, actually is name index.php, down de code i write i start the html code — Alan Daniel Estrada Martinez, Dec 02 '16 at 05:40
Alright - if you inspect the source code, do you see anything there? Does `error_reporting(E_ALL);` `ini_set('display_errors', 1);` output anything? Or [`mysqli_error`](http://php.net/manual/en/mysqli.error.php)? — Qirel, Dec 02 '16 at 05:44
if you have a mysql gui like phpmyadmin, try running your query there and see what it outputs. — Kamrul Khan, Dec 02 '16 at 05:47
when i inspect the src, does not show anything. And when i run the query on sql console it actully shows up what im lookin for. — Alan Daniel Estrada Martinez, Dec 02 '16 at 05:50
it really does not show nothing, not on the browse nor the sql.log — Alan Daniel Estrada Martinez, Dec 02 '16 at 06:07
i tried it outputs, "Notice: Array to string conversion" and also the word "Array", with a capital letter — Alan Daniel Estrada Martinez, Dec 02 '16 at 06:12
use print_r($data_pedido); You cant use echo for printing array. — Jomol MJ, Dec 02 '16 at 06:13
Array ( [id] => [nombre] => [precio] => [cantidad] => [total] => [test] => ) — Alan Daniel Estrada Martinez, Dec 02 '16 at 06:14
I think you cannot access `p.precio` as `$a['p.precio']`. You have to use [aliases](http://www.w3schools.com/sql/sql_alias.asp). Try to access `$a['total']` not `$a['ptotal']` in `$data_pedido` definition. — rationalboss, Dec 02 '16 at 06:18
@Alan Whoops, I don't know why I haven't seen this before... You're using singlequotes around your column-names! Remove all of your `'` in the query - you aren't using any strings anyway. This query really shouldn't work in phpMyAdmin or other GUIs though because of this. — Qirel, Dec 02 '16 at 06:21
Possible duplicate of [When to use single quotes, double quotes, and backticks?](http://stackoverflow.com/questions/11321491/when-to-use-single-quotes-double-quotes-and-backticks) — Qirel, Dec 02 '16 at 06:21
After removing the single qoutes it show up Use of undefined constant a - assumed 'a' in — Alan Daniel Estrada Martinez, Dec 02 '16 at 06:34
Progress! Sounds like you're missing a variable identifier `$` in front of a variable somewhere. The error gives you a line to look at. — Qirel, Dec 02 '16 at 06:48
it show that its where the array start, should i enter a $before? — Alan Daniel Estrada Martinez, Dec 02 '16 at 07:06
$query=mysqli_query($db," SELECT p.id, p.nombre, p.precio, a.cantidad, p.precio * a.cantidad AS total, a.id FROM articulos_pedidos a INNER JOIN productos p ON a producto_id = p.id INNER JOIN pedidos pe on a pedido_id = pe.id WHERE pedido_id = 23")OR die("incorrect"); $row = $query->fetch_assoc(); $data_pedido = array( 'id' => $row['pedido_id'], 'nombre' => $row['pnombre'], 'precio' => $row['p.precio'], 'cantidad' => $row['a.cantidad'], 'total' => $row['ptotal'], 'test' => $row[''] ); it show query is incorrect — Alan Daniel Estrada Martinez, Dec 02 '16 at 07:11

Vijaya Vignesh Kumar · Answer 1 · 2016-12-03T14:19:36.203

-2

$data_pedido=array();
while($a = $sql->fetch_assoc()) {       
    echo $data_pedido['id']=$a['pedido_id'];
    echo $data_pedido['nombre']=$a['p.nombre'];
    echo $data_pedido['precio']=$a['p.precio'];
    echo $data_pedido['cantidad']=$a['a.cantidad'];
    echo $data_pedido['total']=$a['ptotal'];
}

edited Dec 03 '16 at 14:19

answered Dec 02 '16 at 06:15

Vijaya Vignesh Kumar

444
3
14

at the end of while loop print the array and comment the output here – Vijaya Vignesh Kumar Dec 02 '16 at 06:43

How to echo the array of a sql consult into a html

1 Answers1