Hello I'm just new to PHP I want to declare variables with the result of a while of an Array
<?php
include("conexion.php");
$query4= mysqli_query($conexion,"SELECT refprofesor FROM asigna_materia
WHERE refgrupo='3-A' AND refcarrera='09TICSI'");
while($row = mysqli_fetch_array($query4, MYSQLI_NUM))
{
echo $row[0].'<br>';
echo $row[1].'<br>';
}
$ejemplo1 = $row[0];
$ejemplo2 = $row[1];
?>
This is the Result Result
and this is the image of the database
BD
The problem is that the variable $ row [1] does not exist. This is due to the .iteration within the while, which each time takes one of the results from the base, which is the one that prints with $ row [0]. The correct way to create an array with all the values would be:
while ($ row = mysqli_fetch_array ($ query4, MYSQLI_NUM)) { $ results [] = $ row [0]; }
This creates the keys of the array with an autoincremental value and assigns it the value of each row. Greetings!
The error is simple, the query currently only returns one column, but you try to access a second column
echo $row[0].'<br>'; // correcto
echo $row[1].'<br>'; //incorrecto, indice no válido
Also, outside of while it would not make sense to access $row , the last two lines you must delete them.
Final Code
include("conexion.php");
$query4= mysqli_query($conexion,"SELECT refprofesor FROM asigna_materia
WHERE refgrupo='3-A' AND refcarrera='09TICSI'");
while($row = mysqli_fetch_array($query4, MYSQLI_NUM))
{
echo $row[0].'<br>';
}
It would not be convenient to have variables within while since with a query, we will not know exactly how many records will return, it will work as long as a single record is obtained (it would not need a while) .
For this, a array would be appropriate if you want to obtain the result and not print it directly.
$resultado = [];
while($row = mysqli_fetch_array($query4, MYSQLI_NUM))
{
$resultado[] = $row;
}
// Al terminar el while
//Aquí su variable $resultado, tendrá todos los registros devueltos
Try to name it
$row['refprofesor'];
would be
while($row = mysqli_fetch_array($query4, MYSQLI_NUM)) {
echo $ejemplo = $row['refprofesor']."<br>";
}