How to display id as form counter

Question

I’m creating a form counter and I’d like to know if I’m going the right way ?

This is my select query

<?php
error_reporting(-1);
ini_set('display_errors', 'On');

//Criar a conexao
$link = new mysqli ("localhost", "root", "", "peixaria");
if($link->connect_errno){
     echo"Nossas falhas local experiência ..";
     exit();
}



         $sql= "SELECT MAX(m.id_pedido)
                FROM mpedido as m WHERE id_pedido
                ORDER BY m.id_pedido DESC
                LIMIT 1";
        $resultado_id= mysqli_query($link,$sql);

?>

and this is the php code that displays the id as counter . but type is giving an error

Notice: Array to string Conversion in C: xampp htdocs Requested projects.php on line 61

  <span>Comanda:</span>
                    <?php


                while($row = mysqli_fetch_array($resultado_id)){
                echo '<span>'.["id_pedido"].'</span>';

                }

                ?>

How could I make a counter from the id

Answer 1

The mistake is in:

echo '<span>'.["id_pedido"].'</span>';

There is no such thing as.

When you use the code:

while($row = mysqli_fetch_array($resultado_id)){

}

Are you saying that as long as there are lines, you will get the array $row, or should be used:

while($row = mysqli_fetch_array($resultado_id)){

 echo '<span>'.$row["id_pedido"].'</span>';

}

But... that won’t work either. The reason is that there is no index id_pedido and the reason it doesn’t exist is because there’s no such thing defined in SELECT.

---

You don’t need to use the while() if there is the LIMIT 1, if the limit is 1 it will run only once.

---

Possible corrections:

Method 0. Use numerical index:

Part of While:

$row = mysqli_fetch_array($resultado_id);
echo '<span>'.$row[0].'</span>';

Method 1. Name the oxen:

Part of the Query:

$sql= "SELECT MAX(m.id_pedido) as ultimo FROM mpedido as m WHERE id_pedido ORDER BY m.id_pedido DESC LIMIT 1";

$resultado_id= mysqli_query($link,$sql);

Soon, now the SELECT has a result with the name ultimo for the result of MAX(m.id_pedido), which is easier.

Part of While:

$row = mysqli_fetch_array($resultado_id);
echo '<span>'.$row['ultimo'].'</span>';

Answer 2

I’m sure I understand, but I believe you want something like:

//Criar a conexao
$link = new mysqli ("localhost", "root", "", "peixaria");
if($link->connect_errno){
 echo"Nossas falhas local experiência ..";
 exit();
}   

$sql= "SELECT MAX(m.id_pedido) as ultimo_id FROM mpedido as m";
$resultado_id= mysqli_query($link,$sql);
?>

As you used MAX(), it will return the highest value of the requested id_field, so if this is a sequential field it will already return the desired value.

Already in php you should do so

<span>Comanda:</span>
<?php
  while($row = mysqli_fetch_array($resultado_id)){
   echo '<span>'.$row["ultimo_id"].'</span>';
  }

?>

When you do this while, you receive a line with the returned information, but it comes as an array, so you need to access the index you created in SQL, in the case "ultimo_id".

I hope that’s what you’re looking for.